ml aggarwal class 9 solutions pdf download

ML Aggarwal Class 9 Solutions for ICSE Maths Chapter 13 Rectilinear Figures

Exercise 13.1

Question 1.
If two angles of a quadrilateral are 40° and 110° and the other two are in the ratio 3 : 4, find these angles.
Solution:

Question 2.
If the angles of a quadrilateral, taken in order, are in the ratio 1 : 2 : 3 : 4, prove that it is a trapezium.
Solution:

Question 3.
If an angle of a parallelogram is two-thirds of its adjacent angle, find the angles of the parallelogram.
Solution:

Question 4.
(a) In figure (1) given below, ABCD is a parallelogram in which ∠DAB = 70°, ∠DBC = 80°. Calculate angles CDB and ADB.
(b) In figure (2) given below, ABCD is a parallelogram. Find the angles of the AAOD.
(c) In figure (3) given below, ABCD is a rhombus. Find the value of x.

Solution:

Question 5.
(a) In figure (1) given below, ABCD is a parallelogram with perimeter 40. Find the values of x and y.
(b) In figure (2) given below. ABCD is a parallelogram. Find the values of x and y.
(c) In figure (3) given below. ABCD is a rhombus. Find x and y.
Solution:

Question 6.
The diagonals AC and BD of a rectangle > ABCD intersect each other at P. If ∠ABD = 50°, find ∠DPC.
Solution:

Question 7.
(a) In figure (1) given below, equilateral triangle EBC surmounts square ABCD. Find angle BED represented by x.
(b) In figure (2) given below, ABCD is a rectangle and diagonals intersect at O. AC is produced to E. If ∠ECD = 146°, find the angles of the ∆ AOB.
(c) In figure (3) given below, ABCD is rhombus and diagonals intersect at O. If ∠OAB : ∠OBA = 3:2, find the angles of the ∆ AOD.

Solution:

Question 8.
(a) In figure (1) given below, ABCD is a trapezium. Find the values of x and y.
(b) In figure (2) given below, ABCD is an isosceles trapezium. Find the values of x and.y.
(c) In figure (3) given below, ABCD is a kite and diagonals intersect at O. If ∠DAB = 112° and ∠DCB = 64°, find ∠ODC and ∠OBA.


Solution:

Question 9.
(i) Prove that each angle of a rectangle is 90°.
(ii) If the angle of a quadrilateral are equal, prove that it is a rectangle.
(iii) If the diagonals of a rhombus are equal, prove that it is a square.
(iv) Prove that every diagonal of a rhombus bisects the angles at the vertices.
Solution:

Question 10.
ABCD is a parallelogram. If the diagonal AC bisects ∠A, then prove that:
(i) AC bisects ∠C
(ii) ABCD is a rhombus
(iii) AC ⊥ BD.
Solution:

Question 11.
(i) Prove that bisectors of any two adjacent angles of a parallelogram are at right angles.
(ii) Prove that bisectors of any two opposite angles of a parallelogram are parallel.
(iii) If the diagonals of a quadrilateral are equal and bisect each other at right angles, then prove that it is a square.
Solution:

Question 12.
(i) If ABCD is a rectangle in which the diagonal BD bisect ∠B, then show that ABCD is a square.
(ii) Show that if the diagonals of a quadrilateral are equal and bisect each other at right angles, then it is a square.
Solution:

Question 13.
P and Q are points on opposite sides AD and BC of a parallelogram ABCD such that PQ passes through the point of intersection O of its diagonals AC and BD. Show that PQ is bisected at O.
Solution:

Question 14.
(a) In figure (1) given below, ABCD is a parallelogram and X is mid-point of BC. The line AX produced meets DC produced at Q. The parallelogram ABPQ is completed. Prove that:
(i) the triangles ABX and QCX are congruent;
(ii)DC = CQ = QP
(b) In figure (2) given below, points P and Q have been taken on opposite sides AB and CD respectively of a parallelogram ABCD such that AP = CQ. Show that AC and PQ bisect each other.

Solution:

Question 15.
ABCD is a square. A is joined to a point P on BC and D is joined to a point Q on AB. If AP=DQ, prove that AP and DQ are perpendicular to each other.
Solution:

Question 16.
If P and Q are points of trisection of the diagonal BD of a parallelogram ABCD, prove that CQ || AP.
Solution:

Question 17.
A transversal cuts two parallel lines at A and B. The two interior angles at A are bisected and so are the two interior angles at B ; the four bisectors form a quadrilateral ABCD. Prove that
(i) ABCD is a rectangle.
(ii) CD is parallel to the original parallel lines.

Solution:

Question 18.
In a parallelogram ABCD, the bisector of ∠A meets DC in E and AB = 2 AD. Prove that
(i) BE bisects ∠B
(ii) ∠AEB = a right angle.
Solution:

Question 19.
ABCD is a parallelogram, bisectors of angles A and B meet at E which lie on DC. Prove that AB
Solution:

Question 20.
ABCD is a square and the diagonals intersect at O. If P is a point on AB such that AO =AP, prove that 3 ∠POB = ∠AOP.
Solution:

Question 21.
ABCD is a square. E, F, G and H are points on the sides AB, BC, CD and DA respectively such that AE = BF = CG = DH. Prove that EFGH is a square.
Solution:

Question 22.
(a) In the Figure (1) given below, ABCD and ABEF are parallelograms. Prove that
(i) CDFE is a parallelogram
(ii) FD = EC
(iii) Δ AFD = ΔBEC.
(b) In the figure (2) given below, ABCD is a parallelogram, ADEF and AGHB are two squares. Prove that FG = AC
Solution:

Question 23.
ABCD is a rhombus in which ∠A = 60°. Find the ratio AC : BD.
Solution:

Exercise 13.2

Question 1.
Using ruler and compasses only, construct the quadrilateral ABCD in which ∠ BAD = 45°, AD = AB = 6cm, BC = 3.6cm, CD = 5cm. Measure ∠ BCD.
Solution:

Question 2.
Draw a quadrilateral ABCD with AB = 6cm, BC = 4cm, CD = 4 cm and ∠ ABC = ∠ BCD = 90°
Solution:

Question 3.
Using ruler and compasses only, construct the quadrilateral ABCD given that AB = 5 cm, BC = 2.5 cm, CD = 6 cm, ∠BAD = 90° and the diagonal AC = 5.5 cm.
Solution:

Question 4.
Construct a quadrilateral ABCD in which AB = 3.3 cm, BC = 4.9 cm, CD = 5.8 cm, DA = 4 cm and BD = 5.3 cm.
Solution:

Question 5.
Construct a trapezium ABCD in which AD || BC, AB = CD = 3 cm, BC = 5.2cm and AD = 4 cm
Solution:

Question 6.
Construct a trapezium ABCD in which AD || BC, ∠B= 60°, AB = 5 cm. BC = 6.2 cm and CD = 4.8 cm.
Solution:

Question 7.
Using ruler and compasses only, construct a parallelogram ABCD with AB = 5.1 cm, BC = 7 cm and ∠ABC = 75°.
Solution:

Question 8.
Using ruler and compasses only, construct a parallelogram ABCD in which AB = 4.6 cm, BC = 3.2 cm and AC = 6.1 cm.
Solution:

Question 9.
Using ruler and compasses, construct a parallelogram ABCD give that AB = 4 cm, AC = 10 cm, BD = 6 cm. Measure BC.
Solution:

Question 10.
Using ruler and compasses only, construct a parallelogram ABCD such that BC = 4 cm, diagonal AC = 8.6 cm and diagonal BD = 4.4 cm. Measure the side AB.
Solution:

Question 11.
Use ruler and compasses to construct a parallelogram with diagonals 6 cm and 8 cm in length having given the acute angle between them is 60°. Measure one of the longer sides.
Solution:

Question 12.
Using ruler and compasses only, draw a parallelogram whose diagonals are 4 cm and 6 cm long and contain an angle of 75°. Measure and write down the length of one of the shorter sides of the parallelogram.
Solution:

Question 13.
Using ruler and compasses only, construct a parallelogram ABCD with AB = 6 cm, altitude = 3.5 cm and side BC = 4 cm. Measure the acute angles of the parallelogram.
Solution:

Question 14.
The perpendicular distances between the pairs of opposite sides of a parallelogram ABCD are 3 cm and 4 cm and one of its angles measures 60°. Using ruler and compasses only, construct ABCD.
Solution:

Question 15.
Using ruler and compasses, construct a rectangle ABCD with AB = 5cm and AD = 3 cm.
Solution:

Question 16.
Using ruler and compasses only, construct a rectangle each of whose diagonals measures 6cm and the diagonals intersect at an angle of 45°.
Solution:

Question 17.
Using ruler and compasses only, construct a square having a diagonal of length 5cm. Measure its sides correct to the nearest millimeter.
Solution:

Question 18.
Using ruler and compasses only construct A rhombus ABCD given that AB 5cm, AC = 6cm measure ∠BAD.
Solution:

Question 19.
Using ruler and compasses only, construct rhombus ABCD with sides of length 4cm and diagonal AC of length 5 cm. Measure ∠ABC.
Solution:

Question 20.
Construct a rhombus PQRS whose diagonals PR and QS are 8cip and 6cm respectively.
Solution:

Question 21.
Construct a rhombus ABCD of side 4.6 cm and ∠BCD = 135°, by using ruler and compasses only.
Solution:

Question 22.
Construct a trapezium in which AB || CD, AB = 4.6 cm, ∠ ABC = 90°, ∠ DAB = 120° and the distance between parallel sides is 2.9 cm.
Solution:

Question 23.
Construct a trapezium ABCD when one of parallel sides AB = 4.8 cm, height = 2.6cm, BC = 3.1 cm and AD = 3.6 cm.
Solution:

Question 24.
Construct a regular hexagon of side 2.5 cm.
Solution:

Multiple Choice Questions

Choose the correct answer from the given four options (1 to 12):
Question 1.
Three angles of a quadrilateral are 75°, 90° and 75°. The fourth angle is
(a) 90°
(b) 95°
(c) 105°
(d) 120°
Solution:
Sum of 4 angles of a quadrilateral = 360° Sum of three angles = 75° + 90° + 75° = 240° Fourth angle = 360° – 240° = 120° (d)

Question 2.
A quadrilateral ABCD is a trapezium if
(a) AB = DC
(b) AD = BC
(c) ∠A + ∠C = 180°
(d) ∠B + ∠C = 180°
Solution:
A quadrilateral ABCD is a trapezium if ∠B + ∠C= 180°
(Sum of co-interior angles) (d)

Question 3.
If PQRS is a parallelogram, then ∠Q – ∠S is equal to
(a) 90°
(b) 120°
(c) 0°
(d) 180°
Solution:
PQRS is a parallelogram ∠Q – ∠S = 0
(∵ Opposite angles of a parallelogram, are equal) (c)

Question 4.
A diagonal of a rectangle is inclined to one side of the rectangle at 25°. The acute angle between the diagonals is
(a) 55°
(b) 50°
(c) 40°
(d) 25°
Solution:
In a rectangle a diagonal is inclined to one side of the rectangle is 25°

Question 5.
ABCD is a rhombus such that ∠ACB = 40°. Then ∠ADB is
(a) 40°
(b) 45°
(c) 50°
(d) 60°
Solution:

Question 6.
The diagonals AC and BD of a parallelogram ABCD intersect each other at the point O. If ∠D AC = 32° and ∠AOB = 70°, then ∠DBC is equal to
(a) 24°
(b) 86°
(c) 38°
(d) 32°
Solution:

Question 7.
If the diagonals of a square ABCD intersect each other at O, then ∆OAB is
(a) an equilateral triangle
(b) a right angled but not an isosceles triangle
(c) an isosceles but not right angled triangle
(d) an isosceles right angled triangle
Solution:

Question 8.
If the diagonals of a quadrilateral PQRS bisect each other, then the quadrilateral PQRS must be a
(a) parallelogram
(b) rhombus
(c) rectangle
(d) square
Solution:
Diagonals of a quadrilateral PQRS bisect each other, then quadrilateral must be a parallelogram.
(∵ A rhombus, rectangle and square are also parallelogram) (a)

Question 9.
If the diagonals of a quadrilateral PQRS bisect each other at right angles, then the quadrilateral PQRS must be a
(a) parallelogram
(b) rectangle
(c) rhombus
(d) square
Solution:
Diagonals of quadrilateral PQRS bisect each other at right angles, then quadrilateral PQRS [ must be a rhombus.
(∵ Square is also a rhombus with each angle equal to 90°) (c)

Question 10.
Which of the following statement is true for a parallelogram?
(a) Its diagonals are equal.
(b) Its diagonals are perpendicular to each other.
(c) The diagonals divide the parallelogram into four congruent triangles.
(d) The diagonals bisect each other.
Solution:
For a parallelogram an the statement ‘The diagoanls bisect each other’ is true. (d)

Question 11.
Which of the following is not true for a parallelogram?
(a) opposite sides are equal
(b) opposite angles are equal
(c) opposite angles are bisected by the diagonals
(d) diagonals bisect each other
Solution:
The statement that in a parallelogram, .the opposite angles are bisected by the diagonals, is not true in each case. (c)

Question 12.
A quadrilateral in which the diagonals are equal and bisect each other at right angles is a
(a) rectangle which is not a square
(b) rhombus which is not a square
(c) kite which is not a square
(d) square
Solution:
In a quadrilateral, if diagonals are equal and bisect each other at right angles, is a square. (d)

Chapter Test

Question P.Q.
The interior angles of a polygon add upto 4320°. How many sides does the polygon have ?
Solution:

Question P.Q.
If the ratio of an interior angle to the exterior angle of a regular polygon is 5:1, find the number of sides.
Solution:

Question P.Q.
In a pentagon ABCDE, BC || ED and ∠B: ∠A : ∠E =3:4:5. Find ∠A.
Solution:

Question 1.
In the given figure, ABCD is a parallelogram. CB is produced to E such that BE=BC. Prove that AEBD is a parallelogram.

Solution:

Question 2.
In the given figure, ABC is an isosceles triangle in which AB=AC. AD bisects exterior angle PAC and CD || BA. Show that
(i) ∠DAC=∠BCA
(ii) ABCD is a parallelogram.
Solution:

Question 3.
Prove that the quadrilateral obtained by joining the mid-points of an isosceles trapezium is a rhombus.
Solution:

Question 4.
Find the size of each lettered angle in the Following Figures.

Solution:

Question 5.
Find the size of each lettered angle in the following figures :

Solution:

Question 6.
In the adjoining figure, ABCD is a rhombus and DCFE is a square. If ∠ABC = 56°, find
(i) ∠DAG
(ii) ∠FEG
(iii) ∠GAC
(iv) ∠AGC.

Solution:

Question 7.
If one angle of a rhombus is 60° and the length of a side is 8 cm, find the lengths of its diagonals.
Solution:

Question 8.
Using ruler and compasses only, construct a parallelogram ABCD with AB = 5 cm, AD = 2.5 cm and ∠BAD = 45°. If the bisector of ∠BAD meets DC at E, prove that ∠AEB is a right angle.
Solution:

ML Aggarwal Class 9 Solutions for ICSE Maths Chapter 19 Coordinate Geometry

EXERCISE 19.1

Question 1.
Find the co-ordinates of points whose
(i) abscissa is 3 and ordinate -4.
(ii) abscissa is – \(\frac { 3 }{ 2 }\)and ordinate 5.
(iii) whose abscissa is -1\(\frac { 2 }{ 3 }\) and ordinate -2 \(\frac { 1 }{ 4 }\) .
(iv) whose ordinate is 5 and abscissa is -2
(v) whose abscissa is -2 and lies on x-axis.
(vi) whose ordinate is \(\frac { 3 }{ 2 }\) and lies on y-axis.
Solution:

Question 2.
In which quadrant or on which axis each of the following points lie?
(-3, 5), (4, -1) (2, 0), (2, 2), (-3, -6)
Solution:

Question 3.
Which of the following points lie on
(i) x-axis? (ii) y-axis?
A (0, 2), B (5, 6), C (23, 0), D (0, 23), E (0, -4), F (-6, 0), G (√3,0)
Solution:

Question 4.
Plot the following points on the same graph paper :
A (3, 4), B (-3, 1), C (1, -2), D (-2, -3), E (0, 5), F (5, 0), G (0, -3), H (-3, 0).
Solution:

Question 5.
Write the co-ordinates of the points A, B, C, D, E, F, G and H shown in the adjacent figure.

Solution:

Question 6.
In which quadrants are the points A, B, C and D of problem 3 located ?
Solution:
A Lies in the first quadrant, B lies on x-axis C lies in the third quadrant and D lies in the fourth quadrant.

Question 7.
Plot the following points on the same graph paper :

Solution:

Question 8.
Plot the following points on the same graph paper.

Solution:

Question 9.
Plot the following points and check whether they are collinear or not:
(i) (1,3), (-1,-1) and (-2,-3)
(ii) (1,2), (2,-1) and (-1, 4)
(iii) (0,1), (2, -2) and (\(\frac { 2 }{ 3 }\) ,0)
Solution:

Question 10.
Plot the point P(-3, 4). Draw PM and PN perpendiculars to x-axis and y-axis respectively. State the co-ordinates of the points M and N.
Solution:

Question 11.
Plot the points A (1,2), B (-4,2), C (-4, -1) and D (1, -1). What kind of quadrilateral is ABCD ? Also find the area of the quadrilateral ABCD.
Solution:

Question 12.
Plot the points (0,2), (3,0), (0, -2) and (-3,0) on a graph paper. Join these points (in order). Name the figure so obtained and find the area of the figure obtained.
Solution:

Question 13.
Three vertices of a square are A (2,3), B(-3, 3) and C (-3, -2). Plot these points on a graph paper and hence use it to find the co-ordinates of the fourth vertex. Also find the area of the square.
Solution:

Question 14.
Write the co-ordinates of the vertices of a rectangle which is 6 units long and 4 units wide if the rectangle is in the first quadrant, its longer side lies on the x-axis and one vertex is at the origin.
Solution:

Question 15.
Repeat problem 12 assuming that the rectangle is in the third quadrant with all other conditions remaining the same.
Solution:
A rectangle which is 6 unit long and 4 units wide and this rectangle is in the third quadrant.

Question 16.
The adjoining figure shows an equilateral triangle OAB with each side = 2a units. Find the coordinates of the vertices.
Solution:

Question 17.
In the given figure, APQR is equilateral. If the coordinates of the points Q and R are (0, 2) and (0, -2) respectively, find the coordinates of the point P.

Solution:

EXERCISE 19.2

Question 1.
Draw the graphs of the following linear equations :
(i) 2x + + 3 = 0
(ii) x- 5y- 4 = 0
Solution:

Question 2.
Draw the graph of 3y= 12 – 2x. Take 2cm = 1 unit on both axes.
Solution:

Question 3.
Draw the graph of 5x + 6y – 30 = 0 and use it to find the area of the triangle formed by the line and the co-ordinate axes.
Solution:

Question 4.
Draw the graph of 4x- 3y + 12 = 0 and use it to find the area of the triangle formd by the line and the co-ordinate axes. Take 2 cm = 1 unit on both axes.
Solution:

Question 5.
Draw the graph of the equation y = 3x – 4. Find graphically.
(i) the value of y when x = -1
(ii) the value of x when y = 5.
Solution:

Question 6.
The graph of a linear equation in x and y passes through (4, 0) and (0, 3). Find the value of k if the graph passes through (A, 1.5).
Solution:

Question 7.
Use the table given alongside to draw the graph of a straight line. Find, graphically the values of a and b.

Solution:

EXERCISE 19.3

Question 1.
Solve the following equations graphically: 3x – 2y = 4, 5x – 2y = 0
Solution:

Question 2.
Solve the following pair of equations graphically. Plot at least 3 points for each straight line 2x – 7y = 6, 5x – 8y = – 4
Solution:

Question 3.
Using the same axes of co-ordinates and the same unit, solve graphically.
x+y = 0, 3x – 2y = 10
Solution:

Question 4.
Take 1 cm to represent 1 unit on each axis to draw the graphs of the equations 4x- 5y = -4 and 3x = 2y – 3 on the same graph sheet (same axes). Use your graph to find the solution of the above simultaneous equations.
Solution:

Question 5.
Solve the following simultaneous equations graphically, x + 3y = 8, 3x = 2 + 2y
Solution:

Question 6.
Solve graphically the simultaneous equations 3y = 5 – x, 2x = y + 3 (Take 2cm = 1 unit on both axes).
Solution:

Question 7.
Use graph paper for this question.
Take 2 cm = 1 unit on both axes.
(i) Draw the graphs of x +y + 3 = 0 and 3x-2y + 4 = 0. Plot only three points per line.
(ii) Write down the co-ordinates of the point of intersection of the lines.
(iii) Measure and record the distance of the point of intersection of the lines from the origin in cm.

Solution:

Question 8.
Solve the following simultaneous equations graphically :
2x-3y + 2 = 4x+ 1 = 3x – y + 2
Solution:

Question 9.
Use graph paper for this question.
(i) Draw the graphs of 3x -y – 2 = 0 and 2x + y – 8 = 0. Take 1 cm = 1 unit on both axes and plot three points per line.
(ii) Write down the co-ordinates of the point of intersection and the area of the traingle formed by the lines and the x-axis.
Solution:

Question 10.
Solve the following system of linear equations graphically : 2x -y – 4 = 0, x + y + 1 = 0. Hence, find the area of the triangle formed by these lines and the y-axis.
Solution:

Question 11.
Solve graphically the following equations: x + 2y = 4, 3x – 2y = 4
Take 2 cm = 1 unit on each axis. Write down the area of the triangle formed by the lines and the x-axis.
Solution:

Question 12.
On graph paper, take 2 cm to represent one unit on both the axes, draw the lines : x + 3 = 0, y –  2 = 0, 2x + 3y = 12 .
Write down the co-ordinates of the vertices of the triangle formed by these lines.
Solution:

Question 13.
Find graphically the co-ordinates of the vertices of the triangle formed by the lines y = 0, y – x and 2x + 3y= 10. Hence find the area of the triangle formed by these lines.
Solution:

EXERCISE 19.4

Question 1.
Find the distance between the following pairs of points :
(i) (2, 3), (4, 1)
(ii) (0, 0), (36, 15)
(iii) (a, b), (-a, -b)
Solution:

Question 2.
A is a point on y-axis whose ordinate is 4 and B is a point on x-axis whose abscissa is -3. Find the length of the line segment AB.
Solution:

Question 3.
Find the value of a, if the distance between the points A (-3, -14) and B (a, -5) is 9 units.
Solution:

Question 4.
(i) Find points on the x-axis which are at a distance of 5 units from the point (5, -4).
(ii) Find points on the y-axis are at a distance of 10 units from the point (8, 8) ?
(iii) Find points (or points) which are at a distance of √10 from the point (4, 3) given that the ordinate of the point or points is twice the abscissa.
Solution:

Question 5.
Find the point on the x-axis which, is equidistant from the points (2, -5) and (-2, 9).
Solution:

Question 6.
Find the value of x such that PQ = QR where the coordinates of P, Q and R are (6, -1), (1, 3) and (x, 8) respectively.
Solution:

Question 7.
If Q (0, 1) is equidistant from P (5, -3) and R (x, 6) find the values of x.
Solution:

Question 8.
Find a relation between x and y such that the point (x, y) is equidistant from the points (7, 1) and (3, 5).
Solution:

Question 9.
The x-coordinate of a point P is twice its y-coordinate. If P is equidistant from the points Q (2, -5) and U (-3, 6), then find the coordinates of P.
Solution:

Question 10.
If the points A (4,3) and B (x, 5) are on a circle with centre C (2, 3), find the value of x.
Solution:

Question 11.
If a point A (0, 2) is equidistant from the points B (3, p) and C (p, 5), then find the value of p.
Solution:

Question 12.
Using distance formula, show that (3, 3) is the centre of the circle passing through the points (6, 2), (0, 4) and (4, 6).
Solution:

Question 13.
The centre of a circle is C (2α – 1, 3α + 1) and it passes through the point A (-3, -1). If a diameter of the circle is of length 20 units, find the value(s) of α.
Solution:

Question 14.
Using distance formula, show that the points A (3, 1), B (6, 4) and C (8, 6) are coliinear.
Solution:

Question 15.
Check whether the points (5, -2), (6, 4) and (7, -2) are the vertices of an isosceles triangle.
Solution:

Question 16.
Name the type of triangle formed by the points A (-5, 6), B (-4, -2) and (7, 5).
Solution:

Question 17.
Show that the points (1, 1), (- 1, – 1) and (-√3,√3) form an equilateral triangle.
Solution:

Question 18.
Show that the points (7, 10), (-2, 5) and (3, -4) are the vertices of an isosceles right triangle.
Solution:

Question 19.
The points A (0, 3), B (- 2, a) and C (- 1, 4) are the vertices of a right angled triangle at A, find the value of a.
Solution:

Question 20.
Show that the points (0, – 1), (- 2, 3), (6, 7) and (8, 3), taken in order, are the vertices of a rectangle. Also find its area.
Solution:

Question 21.
If P (2, -1), Q (3, 4), R (-2, 3) and S (-3, -2) be four points in a plane, show that PQRS is a rhombus but not a square. Find the area of the rhombus.
Solution:

Question 22.
Prove that the points A (2, 3), B {-2, 2), C (-1, -2) aqd D (3, -1) are the vertices of a square ABCD.
Solution:

Question 23.
Name the type of quadrilateral formedby the following points and give reasons for your answer :
(i) (-1, -2), (1, 0), (-1, 2), (-3, 0)
(ii) (4, 5), (7, 6), (4, 3), (1, 2)
Solution:

Question 24.
Find the coordinates of the circumcentre of the triangle whose vertices are (8, 6), (8, -2) and (2, -2). Also, find its circumradius.
Solution:

Question 25.
If two opposite vertices of a square are (3, 4) and (1, -1), find the coordinates of the other two vertices.
Solution:

Multiple Choice Questions

Choose the correct answer from the given four options (1 to 16):
Question 1.
Point (-3, 5) lies in the
(a) first quadrant
(b) second quadrant
(c) third quadrant
(d) fourth quadrant
Solution:
Point (-3, 5) lies in second quadrant, (b)

Question 2.
Point (0, -7) lies
(a) on the x-axis
(b) in the second quadrant
(c) on the y-axis
(d) the fourth quadrant
Solution:
Point (0, -7) lies on y-axis (as x = 0) (c)

Question 3.
Abscissa of a point is positive in
I and II quadrants
I and IV quadrants
I quadrant only
II quadrant only
Solution:
Abscissa of a point is positive in first and fourth quadrants. (b)

Question 4.
The point which lies ony-axis at a distance of 5 units in the negative direction of y- axis is
(a) (0, 5)
(b) (5, 0)
(c) (0, -5)
(d) (-5, 0)
Solution:
(0, -5) is the required point. (c)

Question 5.
If the perpendicular distance of a point P from the x-axis is 5 units and the foot of perpendicular lies on the negative direction of x-axis, then the point P has
(a) x-coordinate = -5
(b) y-coordinate = 5 only
(c) y-coordinate = -5 only
(d) y-coordinate = 5 or -5
Solution:
Perpendicular distance for a point P on x- axis in negative direction.
It will has y = 5 and x = -5 (d)

Question 6.
The points whose abscissa and ordinate have different signs will lie in
(a) I and II quadrants
(b) II and III quadrants
(c) I and III quadrants
(d) II and IV quadrants
Solution:
Point which has abscissa and ordinate having different signs will lie in second and fourth quadrants. (d)

Question 7.
The points (-5, 2) and (2, -5) lie in
(a) same quadrant
(b) II and III quadrants respectively
(c) II and IV quadrants respectively
(d) IV and II quadrants respectively
Solution:
Points (-5, 2) and (2, -5) lie in second and fourth quadrants respectively. (b)

Question 8.
If P (-1,1), Q (3, -4), R (1, -1), S (-2, -3) and T (-4, 4) are plotted on the graph paper, then point(s) in the fourth quadrant are
(a) P and T
(b) Q and R
(c) S only
(d) P and R
Solution:
Points P (-1, 1), Q (3, -4), R (1, -1), S (-2, -3) and T (-4, 4) are plotted on graph The points in 4th quadrant are Q and R (b)

Question 9.
On plotting the points O (0, 0), A (3, 0), B (3, 4), C (0, 4) and joining OA, AB, BC and CO which of the following figure is obtained?
(a) Square
(b) Rectangle
(c) Trapezium
(d) Rhombus
Solution:
On plotting the points O (0, 0), A (3, 0), B (3, 4), C (0, 4)
OA, AB, BC and CO are joined
The figure so formed will a rectangle (b)

Question 10.
Which of the following points lie on the graph of the equation :
3x-5y + 7 = 0?
(a) (1, -2)
(b) (2, 1)
(c) (-1, 2)
(d) (1, 2)
Solution:

Question 11.
The pair of equation x – a and y = b graphically represents lines which are
(a) parallel
(b) intersecting at (b, a)
(c) coincident
(d) intersecting at (a, b)
Solution:
x = a, y = 6
Which are intersecting at (a, b) (d)

Question 12.
The distance of the point P (2, 3) from the x>axis is
(a) 2 units
(b) 3 units
(c) 1 unit
(d) 5 units
Solution:
The distance of the point P (2, 3) from x- axis is 3 units (as y = 3). (b)

Question 13.
The distance of the point P (-4, 3) from the y-axis is
(a) 5 units
(b) -4 units
(c) 4 units
(d) 3 units
Solution:
The distance of the point P (-4, 3) from y- axis will be 4 units. (c)

Question 14.
The distance of the point P (-6, 8) from the origin is
(a) 8 units
(b) 2\(\sqrt { 7 }\) units
(c) 10 units
(d) 6 units
Solution:

Question 15.
The distance between the points A (0, 6) and B (0, -2) is
(a) 6 units
(b) 8 units
(c) 4 units
(d) 2 units
Solution:

Question 16.
The distance between the points (0, 5) and (-5, 0) is
(a) 5 units
(b) 5\(\sqrt { 2 }\)units
(c) 2 \(\sqrt { 7 }\) units
(d) 10 units
Solution:
The distance between the points (0, 5) and (-5, 0) is

Question 17.
AOBC is a rectangle whose three vertices are A (0, 3), O (0, 0) and B (5, 0). The length of its diagonal is
(a) 5 units
(b) 3 units
(c) \(\sqrt { 34 }\) units
(d) 4 units
Solution:

Question 18.
If the distance between the points (2, -2) and (-1, x) is S units, then one of the value of x is
(a) -2
(b) 2
(c) -1
(d) 1
Solution:

Question 19.
The distance between the points (4, p) and (1, 0) is 5 units, then the value of p is
(a) 4 only
(b) -4 only
(c) ±4
(d) 0
Solution:

Question 20.
The points (-4, 0), (4, 0) and (0, 3) are the vertices of a
(a) right triangle
(b) isosceles triangle
(c) equilateral triangle
(d) scalene triangle
Solution:

Question 21.
The area of a square whose vertices are A (0, -2), B (3, 1), C (0, 4) and D (-3, 1) is
(a) 18 sq. units
(b) 15 sq. units
(c) \(\sqrt { 18 }\) sq. units
(d) \(\sqrt { 15 }\) sq. units
Solution:

Question 22.
In the given figure, the area of the triangle ABC is
(a) 15 sq. units
(b) 10 sq. units
(c) 7.5 sq. units
(d) 2.5 sq. units
Solution:

Question 23.
The perimeter of a triangle with vertices (0, 4), (0, 0) and (3, 0) is
(a) 5 units
(b) 12 units
(c) 11 units
(d) 7 + \(\sqrt { 5 }\) units
Solution:

Question 24.
If A is a point on the .y-axis whose ordinate is 5 and B is the point (-3, 1), then the length of AB is
(a) 8 units
(b) 5 units
(c) 3 units
(d) 25 units
Solution:

Question 25.
The point A (9, 0), B (9, 6), C (-9, 6) and D (-9, 0) are the vertices of a
(a) rectangle
(b) square
(c) rhombus
(d) trapezium
Solution:

Chapter Test

Question 1.
Three vertices of a rectangle are A (2, -1), B (2, 7) and C(4, 7). Plot these points on a graph and hence use it to find the co-ordinates of the fourth vertex D Also find the co-ordinates of
(i) the mid-point of BC
(ii) the mid point of CD
(iii) the point of intersection of the diagonals. What is the area of the rectangle ?
Solution:

Question 2.
Three vertices of a parallelogram are A (3, 5), B (3, -1) and C (-1, -3). Plot these points on a graph paper and hence use it to find the coordinates of the fourth vertex D. Also find the coordinates of the mid-point of the side CD. What is the area of the parallelogram?
Solution:

Question 3.
Draw the graphs of the following linear equations.
(i) y = 2x – 1
(ii) 2x + 3y = 6
(iii) 2x – 3y = 4.
Also find slope and y-intercept of these lines.
Solution:

Question 4.
Draw the graph of the equation 3x – 4y = 12. From the graph, find :
(i) the value of y when x = -4
(ii) the value of x when y = 3.
Solution:

Question 5.
Solve graphically, the simultaneous equations: 2x – 3y = 7; x + 6y = 11.
Solution:

Question 6.
Solve the following system of equations graphically: x – 2y – 4 = 0, 2x + .y – 3 = 0.
Solution:

Question 7.
Using a scale of l cm to 1 unit for both the axes, draw the graphs of the following equations : 6y = 5x:+ 10,y = 5;c-15. From the graph, find
(i) the coordinates of the point where the two lines intersect.
(ii) the area of the triangle between the lines and the x-axis.
Solution:

Question 8.
Find, graphically, the coordinates of the vertices of the triangle formed by the lines : 8y – 3x + 7 = 0, 2x-y + 4 = 0 and 5x + 4y = 29.
Solution:

Question 9.
Find graphically the coordinates of the vertices of the triangle formed by the lines y-2 = 0,2y + x = 0 and y + 1 = 3(x – 2). Hence, find the area of the triangle formed by these lines.
Solution:

Question 10.
A line segment is of length 10 units and one of its end is (-2,3). If the ordinate of the other end is 9, find the abscissa of the other end.
Solution:

Question 11.
A (-4, -1), B (-1, 2) and C (a, 5) are the vertices of an isosceles triangle. Find the value of a, given that AB is the unequal side.
Solution:

Question 12.
If A (-3, 2), B (a, p) and C (-1, 4) are the vertices of an isosceles triangle, prove that α + β = 1, given AB = BC.
Solution:

Question 13.
Prove that the points (3, 0), (6, 4) and (-1, 3) are the vertices of a right angled isosceles triangle.
Solution:

Question 14.
(i) Show that the points (2, 1), (0,3), (-2, 1) and (0, -1), taken in order, are the vertices of a square. Also find the area of the square.
(ii) Show that the points (-3, 2), (-5, -5), (2, -3) and (4, 4), taken in order, are the vertices of rhombus. Also find its area. Do the given points form a square?
Solution:

Question 15.
The ends of a diagonal of a square have co-ordinates (-2, p) and (p, 2). Find p if the area of the square is 40 sq. units.
Solution:

Question 16.
What type of quadrilateral do the points A (2, -2), B (7, 3), C (11, -1) and D (6, -6), taken in the order, form?
Solution:

Question 17.
Find the coordinates of the centre of the circle passing through the three given points A (5, 1), B (-3, -7) and C (7, -1).
Solution:

ML Aggarwal Class 9 Solutions for ICSE Maths Chapter 16 Mensuration

EXERCISE 16.1

Question 1.
Find the area of a triangle whose base is 6 cm and corresponding height is 4 cm.
Solution:

Question 2.
Find the area of a triangle whose sides are
(i) 3 cm, 4 cm and 5 cm
(ii) 29 cm, 20 cm and 21 cm
(iii) 12 cm, 9.6 cm and 7.2 cm
Solution:

Question 3.
Find the area of a triangle whose sides are 34 cm, 20 cm and 42 cm. Hence, find the length of the altitude corresponding to the shortest side.
Solution:

Question 4.
The sides of a triangular field are 975m, 1050 m and 1125 m. If this field is sold at the rate of Rs. 1000 per hectare, find its selling price. [1 hectare = 10000 m²].
Solution:

Question 5.
The base of a right angled triangle is 12 cm and its hypotenuse is 13 cm long. Find its area and the perimeter.

Solution:

Question 6.
Find the area of an equilateral triangle whose side is 8 m. Given your answer correct to two decimal places.
Solution:

Question 7.
If the area of an equilateral triangle is 81√3 cm² find its. perimeter.
Solution:

Question 8.
If the perimeter of an equilateral triangle is 36 cm, calculate its area and height.
Solution:

Question 9.
(i) If the length of the sides of a triangle are in the ratio 3:4:5 and its perimeter is 48 cm, find its area.
(ii) The sides of a triangular plot are in the ratio 3: 5:7 and its perimeter is 300 m. Find its area.
Solution:

Question 10.
ABC is a triangle in which AB = AC = 4 cm and ∠ A = 90°. Calculate the area of ∆ABC. Also find the length of perpendicular from A to BC.
Solution:

Question 11.
Find the area of an isosceles triangle whose equal sides are 12 cm each and the perimeter is 30 cm.
Solution:

Question 12.
Find the area of an isosceles triangle whose base is 6 cm and perimeter is 16 cm.
Solution:

Question 13.
The sides of a right angled triangle containing the right angle are 5x cm and (3x – 1) cm. Calculate the length of the hypotenuse of the triangle if its area is 60 cm².
Solution:

Question 14.
In ∆ ABC, ∠B = 90°, AB = (2A + 1) cm and BC = (A + 1). cm. If the area of the ∆ ABC is 60 cm², find its perimeter.
Solution:

Question 15.
If the perimeter of a right angled triangle is 60 cm and its hypotenuse is 25 cm, find its area.
Solution:

Question P.Q.
In ∆ ABC, ∠B = 90° and D is mid-point of AC. If AB = 20 cm and BD = 14.5 cm, find the area and the perimeter of ∆ ABC.
Solution:

Question 16.
The perimeter of an isosceles triangle is 40 cm. The base is two third of the sum of equal sides. Find the length of each side.
Solution:

Question 17.
If the area of an isosceles triangle is 60 cm2 and the length of each of its equal sides is 13 cm, find its base.
Solution:

Question 18.
The base of a triangular field is 3 times its height If the cost of cultivating the field at the rate of ₹25 per 100m2 is ₹60000, find its base and height.
Solution:

Question 19.
A triangular park ABC? has sides 120 m, 80 m and 50 m (as shown in the given figure). A gardner Dhania has to put a fence around it and also plant grass inside. How much area does she need to plant? Find the cost of fencing it with barbed wire at the rate of ₹20 per metre leaving a space 3 m wide for a gate on one side.
Solution:

Question 20.
An umbrella is made by stitching 10 triangular pieces of cloth of two different colours (shown in the given figure), each piece measuring 20 cm, 50 cm and 50 cm. How much cloth of each colour is required for the umbrella?

Solution:

Question 21.
(a) In the figure (1) given below, ABC is an equilateral triangle with each side of length 10 cm. In ∆ BCD, ∠D = 90° and CD = 6 cm.
Find the area of the shaded region. Give your answer correct to one decimal place.

(b) In the figure given, ABC is an isosceles right angled triangle and DEFG is a rectangle. If AD = AE = 3 cm and DB = EC = 4 cm, find the area of the shaded region.

Solution:

EXERCISE 16.2

Question 1.
(i) Find the area of quadrilateral whose one diagonal is 20 cm long and the perpendiculars to this diagonal from other vertices are of length 9 cm and 15 cm.
Solution:

Question 2.
Find the area of the quadrilateral field ABCD whose sides AB = 40 m, BC = 28 m, CD = 15 m, AD = 9 m and ∠A = 90°
Solution:

Question 3.
Find the area of the quadrilateral ABCD in which ∠BCA= 90°, AB = 13 cm and ACD is an equilateral triangle of side 12 cm.

Solution:

Question 4.
Find the area of quadrilateral ABCD in which ∠B = 90°, AB = 6 cm, BC = 8 cm 13 and CD = AD = 13 cm.

Solution:

Question 5.
The perimeter of a rectangular cardboard is 96 cm ; If its breadth is 18 cm, find the length and the area of the cardboard.
Solution:

Question 6.
The length of a rectangular hall is 5 m more than its breadth, If the area of the hall is 594 m2, find its perimeter.

Solution:
Let ABCD be rectangular field.

Question 7.
(a) The diagram (i) given below shows two paths drawn inside a rectangular field 50 m long and 35 m wide. The width of each path is 5 metres. Find the area of the shaded portion.
(b) In the diagram (ii) given below, calculate the area of the shaded portion. All measurements are in centimetres.
Solution:

Question 8.
A rectangular plot 20 m long and 14 m wide is to be covered with grass leaving 2 m all around. Find the area to be laid with grass.
Solution:

Question 9.
The shaded region of the given diagram represents the lawn in front of a house. On three sides of the lawn there are flower beds of width 2 m.
(i) Find the length and the breadth of the lawn.
(ii) Hence, or otherwise, find the area of the flower – beds.

Solution:
BCDE is the lawn

Question 10.
A foot path of uniform width runs all around the inside of a rectangular field 50 m long and 38m wide. If the area of the path is 492 m². Find its width.
Solution:

Question 11.
The cost of enclosing a rectangular garden with a fence all around at the rate of Rs. 15 per metre is Rs. 5400. If the length of the garden is 100 m And the area of the garden.
Solution:

Question 12.
A rectangular floor which measures 15 m x 8 m is to be laid with tiles measuring 50 cm x 25 cm find the number of tiles required further, if a carpet is laid on the floor so that a space of 1 m exists between its edges and the edges of the floor, what fraction of the floor is uncovered?
Solution:

Question 13.
The width of a rectangular room is \(\frac { 3 }{ 5 }\) of its length x metres. If its perimeter isy metres, write an equation connecting.vandy. Find the floor area of the room if its perimeter is 32 m.
Solution:

Question 14.
A rectangular garden 10 m by 16 m is to be surrounded by a concrete walk of uniform width. Given that the area of the walk is 120 square metres, assuming the width of the walk to be x, form an equation in x and solve it to find the value of x.
Solution:

Question 15.
A rectangular room is 6 m long, 4.8 m wide and 3.5 m high. Find the inner surface area of the four walls.
Solution:

Question 16.
A rectangular plot of land measures 41 metres in length and 22.5 metres in width. A boundary wall 2 metres high is built all around the plot at a distance of 1.5 m from the plot. Find the inner surface area of the boundary wall.
Solution:

Question 17.
(a) Find the perimeter and area of the figure
(i) given below in which all corners are right angled.
(b) Find the perimeter and area of the figure
(ii) given below in which all corners are right angles.
(c) Find the area and perimeter of the figure
(iii) given below in which all corners are right angles and all measurement in centimetres.

Solution:

Question 18.
The length and the breadth of a rectangle are 12 cm and 9 cm respectively. Find the height of a triangle whose base is 9 cm and whose area is one third that of rectangle.
Solution:

Question 19.
The area of a square plot is 484 mV Find the length of its one side and the length of its one diagonal.

Solution:

Question 20.
A square has the perimeter 56 m. Find its area and the length of one diagonal correct upto two decimal places.
Solution:

Question 21.
A wire when bent in the form of an equilateral triangle encloses an area of 36√3 cm2. Find the area enclosed by the same wire when bent to form:
(i) a square, and
(ii) a rectangle whose length is 2 cm more than its width.
Solution:

Question 22.
Two adjacent sides of a parallelogram are: 15 cm and 10 cm. If the distance between the longer sides is 8 cm, find the area of the parallelogram. Also find the distance between shorter sides.
Solution:

Question 23.
ABCD is a parallelogram with sides AB = 12 cm, BC = 10 cm and diagonal AC = 16 cm. Find the area of the parallelogram. Also find the distance between its shorter sides.
Solution:

Question 24.
Diagonals AC and BD of a parallelogram ABCD intersect at O. Given that AB = 12 cm and perpendicular distance between AB and DC is 6 cm. Calculate the area of the triangle AOD.
Solution:

Question 25.
ABCD is a parallelogram with side AB = 10 cm. Its diagonals AC and BD are of length 12 cm and 16 cm respectively. Find the area of the parallelogram ABCD.
Solution:

Question 26.
The area of a parallelogram is p cm2 and its height is q cm. A second parallelogram has equal area but its base is ‘r’ cm more than that of the first. Obtain an expression in terms of p, q and r for the height h of the second parallelogram.
Solution:

Question 27.
What is the area of a rhombus whose diagonals are 12 cm and 16 cm ?
Solution:

Question 28.
The area of a rhombus is 98 cm². If one of its diagonal is 14 cm, what is the length of the other diagonal?
Solution:

Question 29.
The perimeter of a rhombus is 45 cm. If its height is 8 cm, calculate its area.
Solution:

Question 30.
PQRS is a rhombus. If it is given that PQ = 3 cm and the height of the rhombus is 2.5 cm, calculate its area.
Solution:

Question 31.
If the diagonals of a rhombus are 8 cm and 6 cm, find its perimeter.
Solution:

Question 32.
If the sides of a rhombus are 5 cm each and one diagonal is 8 cm, calculate
(i) the length of the other diagonal, and
(ii) the area of the rhombus.
Solution:

Question 33.
(a) The diagram (t) given below is a trapezium. Find the length of BC and the area of the trapezium Assume AB = 5 cm, AD = 4 cm, CD = 8 cm
(b) The diagram (ii) given below is a trapezium Find (i) AB (ii) area of trapezium ABCD.
(c) The cross-section of a canal is shown in figure (iii) given below. If the canal is 8 m wide at the top and 6 m wide at the bottom and the area of the cross-section is 16.8 m², calculate its depth

Solution:

Question 34.
The distance between parallel sides of a trapezium is 12 cm and the distance between mid-points of other sides is 18 cm. Find the area of the trapezium.
Solution:

Question 35.
The area of a trapezium is 540 cm². If the ratio of parallel sides is 7 : 5 and the distance between them is 18 cm, find the length of parallel sides.
Solution:

Question 36.
The parallel sides of an isosceles trapezium are in the ratio 2 : 3. If its height is 4 cm and area is 60 cm2, find the perimeter.
Solution:

Question 37.
The area of a parallelogram is 98 cm². If one altitude is half the corresponding base, determine the base and the altitude of the parallelogram.
Solution:

Question 38.
The length of a rectangular garden is 12m more than its breadth. The numerical value of its area is equal to 4 times the numerical value of its perimeter. Find the dimensions of the garden
Solution:

Question 39.
If the perimeter of a rectangular plot is 68 m and length of its diagonal is 26 m, find its area.
Solution:

Question 40.
A rectangle has twice the area of a square. The length of the rectangle is 12 cm greater and the width is 8 cm greater than 2 side of a square. Find the perimeter of the square.
Solution:

Question 41.
The perimeter of a square is 48 cm. The area of a rectangle is 4 cm2 less than the area of the square. If the length of the rectangle is 4 cm greater than its breadth, find the perimeter of the rectangle.
Solution:

Question 42.
In the adjoining figure, ABCD is a rectangle with sides AB = 10 cm and BC = 8 cm. HAD and BFC are equilateral triangles; AEB and DCG are right angled isosceles triangles. Find the area of the shaded region and the perimeter of the figure.
Solution:

Question 43.
(a) Find the area enclosed by the figure (i) given below, where ABC is an equilateral triangle and DGFG is an isosceles trapezium.
All measurements are in centmetces.
(b) Find the area enclosed by the figure (ii) given below. AH measurements are in centimetres.
(c) In the figure (iii) given below, from a 24. cm x 24 cm piece of cardboard, a block in the shape of letter M is cut off. Find the area of the cardboard left over, all measurements are in centimetres.

Solution:

Question 44.
(a) The figure (i) given below shows the cross-section of the concrete structure with the measurements as given. Calculate the area of cross-section.
(b) The figure (ii) given below shows a field with the measurements given in metres. Find the area of the field.
(c) Calculate the area of the pentagon ABCDE shown in fig. (iii) below, given that AX = BX = 6 cm, EY = CY = 4 cm, DE = DC = 5cm,DX = 9cmand DX is perpendicular to EC and AB.

Solution:

Question 45.
If the length and the breadth of a room are increased by 1 metre the area is increased by 21 square metres. If the length is increased by 1 metre and breadth is decreased by 1 metre the, area is decreased by 5 square metres. Find the perimeter of the room.
Solution:

Question 46.
A triangle and a parallelogram have the same base and same area. If the sides of the triangle are 26 cm, 28 cm and 30 cm and the parallelogram stands on the base 28 cm, find the height of the parallelogram.
Solution:

Question 47.
A rectangle of area 105 cm² has its length equal to x cm. Write down its breadth in terms of x. Given that its perimeter is 44 cm, write down an equation in x and solve it to determine the dimensions of the rectangle.
Solution:

Question 48.
The perimeter of a rectangular plot is 180 m and its area is 1800 m². Take the length of plot as x m. Use the perimeter 180 m to write the value of the breadth in terms of x. Use the value of the length, breadth and the area to,write an equation in x. Solve the equation to calculate the length and breadth of the plot.
Solution:

EXERCISE 16.3

Question 1.
Find the length of the diameter of a circle whose circumference is 44 cm.
Solution:

Question 2.
Find the radius and area of a circle if its circumference is 18π cm.
Solution:

Question 3.
Find the perimeter of a semicircular plate of radius 3.85 cm.
Solution:

Question 4.
Find the radius and circumference of a circle whose area is 144π cm².
Solution:

Question 5.
A sheet is 11 cm long and 2 cm wide. Circular pieces 0.5 cm in diameter are cut from it to prepare discs. Calculate the number of discs that can be prepared.
Solution:

Question 6.
If the area of a semicircular region is 77cm², find its perimeter.
Solution:

Question 7.
(a) In the figure (i) given below, AC and BD are two perpendicular diameters of a circle ABCD. Given that the ara of the shaded portion is 308 cm2, calculate
(i) the length of AC and
(ii) the circumference of the circle.
(b) In the figure (ii) given below, AC and BD are two perpendicular diameters of a circle with centre O. If AC = 16 cm, calculate the area and perimeter of the shaded part. (Take π = 3.14)

Solution:

Question 8.
A bucket is raised from a well by means of a rope which is wound round a wheel of diameter 77 cm. Given that the bucket ascends in 1 minute 28 seconds with a uniform speed of 1.1 m/sec, calculate the number of complete revolutions the wheel makes in raising the bucket.
Solution:

Question 9.
The wheel of a cart is making 5 revolutions per second. If the diameter of the wheel is 84 cm, find its speed in km/hr. Give your answer correct to the nearest km.
Solution:

Question 10.
The circumference of a circle is 123.2 cm. Calculate :
(i) the radius of the circle in cm.
(ii) the area of the circle in cm², correct to the nearest cm².
(iii) the effect on the area of the circle if the radius is doubled.
Solution:

Question 11.
(a) In the figure (i) given below, the area enclosed between the concentric circles is 770 cm². Given that the radius of the outer circle is 21 cm, calculate the radius of the inner circle.
(b) In the figure (ii) given below, the area enclosed between the circumferences of two concentric circles is 346.5 cm². The circumference of the inner circle is 88 cm. Calculate the radius of the outer circle.

Solution:

Question 12.
A road 3.5 m wide surrounds a circular plot whose circumference is 44 m. Find the cost of paving the road at ₹50 per m².
Solution:

Question 13.
The sum of diameters of two circles is 14 cm and the difference of their circumferences is 8 cm. Find the circumference of the two circles.
Solution:

Question 14.
Find the circumference of the circle whose area is equal to the sum of the areas of three circles with radius 2 cm, 3 cm and 6 cm.
Solution:

Question 15.
A copper wire when bent in the form of a square encloses an area of 121 cm2. If the same wire is bent into the form of a circle, find the area of the circle.
Solution:

Question 16.
A copper wire when bent into an equilateral triangle has area 121√3 cm2. If the same wire is bent into the form of a circle, find the area enclosed by the wire.
Solution:

Question 17.
(a) Find the circumference of the circle whose area is 16 times the area of the circle with diameter 7 cm.
(b) In the given figure, find the area of the unshaded portion within the rectangle. (Take π = 3.14)

Solution:

Question 18.
In the adjoining figure, A6CD is a square of side 21 cm. AC and BD are two diagonals of the square. Two semicircle are drawn with AD and BC as diameters. Find the area of the shaded region. Take π = \(\frac { 22 }{ 7 }\).
Solution:

Question 19.
(a) In the figure (i) given below, ABCD is a square of side 14 cm and APD and BPC are semicircles. Find the area and the perimeter of the shaded region.
(b) In the figure (ii) given below, ABCD is a square of side 14 cm. Find the area of the shaded region.
(c) In the figure (iii) given below, the diameter of the semicircle is equal to 14 cm. Calculate the area of the shaded region. Take π = \(\frac { 22 }{ 7 }\).


Solution:

Question 20.
(a) Find the area and the perimeter of the shaded region in figure (i) given below. The dimensions are in centimetres.
(b) In the figure (ii) given below, area of ∆ABC = 35 cm2. Find the area of the shaded region.

Solution:

Question 21.
(a) In the figure (i) given below, AOBC is a quadrant of a circle of radius 10 m. Calculate the area of the shaded portion. Take π = 3.14 and give your answer correct to two significant figures.
(b) In the figure, (ii) given below, OAB is a quadrant of a cirlce. The radius OA = 3.5 cm and OD = 2 cm. Calculate the area of the shaded portion.

Solution:

Question 22.
A student takes a rectangular piece of paper 30 cm long and 21 cm wide. Find the area of the biggest circle that can be cut out from the paper. Also find the area of the paper left after cutting out the circle. (Take π = \(\frac { 22 }{ 7 }\) )
Solution:

Question 23.
A rectangle with one side 4 cm is inscribed in a circle of radius 2.5 cm. Find the area of the rectangle.
Solution:

Question 24.
(a) In the figure (i) given below, calculate the area of the shaded region correct to two decimal places. (Take π = 3. 142).
(b) In the figure (ii) given below, ABC is an isosceles right angled triangle with ∠ABC = 90°. A semicircle is drawn with AC as diameter. If AB = BC = 7 cm, find the area of the shaded region. Take π = \(\frac { 22 }{ 7 }\) .

Solution:

Question 25.
A circular field has perimeter 660 m. A plot in the shape of a square having its vertices on the circumference is marked in the field. Calculate the area of the square field.
Solution:

Question 26.
In the adjoining figure, ABCD is a square. Find the ratio between

(i) the circumferences
(ii) the areas of the incircle and the circumcircle of the square.
Solution:

Question 27.
(a) The figure (i) given below shows a running track surrounding a grassed enclosure PQRSTU. The enclosure consists of a rectangle PQST with a semicircular region at each end.
PQ = 200 m ; PT = 70 m.

(i) Calculate the area of the grassed enclosure in m2.
(ii) Given that the track is of constant width 7 m, calculate the outer perimeter ABCDEF of the track.
(b) In the figure (ii) given below, the inside perimeter of a practice running track with semi-circular ends and straight parallel sides is 312 m. The length of the straight portion of the track is 90 m. If the track has a uniform width of 2 m throughout, find its area.
Solution:

Question 28.
(a) In the figure (i) given below, two circles with centres A and B touch each other at the point C. If AC = 8 cm and AB = 3 cm, find the area of the shaded region.
(b) The quadrants shown in the figure (ii) given below are each of radius 7 cm. Calculate the area of the shaded portion.

Solution:

Question 29.
(a) In the figure (i) given below, two circular flower beds have been shown on the two sides of a square lawn ABCD of side 56 m. If the centre of each circular flower bed is the point of intersection O of the diagonals of the square lawn, find the sum of the areas of the lawn and the flower beds.

(b) In the figure (ii) given below, a square OABC is inscribed in a quadrant OPBQ of a circle. If OA = 20 cm, find the area of the shaded region. (Use π = 3.14)

Solution:

Question 30.
(a) In the figure (i) given below, ABCD is a rectangle, AB = 14 cm and BC = 7 cm. Taking DC, BC and AD as diameters, three semicircles are drawn as shown in the figure. Find the area of the shaded portion.

(b) In the figure (ii) given below, O is the centre of a circle with AC = 24 cm, AB = 7 cm and ∠BOD = 90°. Find the area of the shaded region. (Use π = 3.14).

Solution:

Question 31.
(a) In the figure given below ABCD is a square of side 14 cm. A, B, C and D are centres of the equal circle which touch externally in pairs. Find the area of the shaded region.

(b) In the figure (ii) given below, the boundary of the shaded region in the given diagram consists of three semi circular arcs, the smaller being equal. If the diameter of the larger one is 10 cm, calculate.
(i) the length of the boundary.
(ii) the area of the shaded region. (Take π to be 3.14)

Solution:

Question 32.
(a) In the figure (i) given below, the points A, B and C are centres of arcs of circles of radii 5 cm, 3 cni and 2 cm respectively. Find the perimeter and the area of the shaded region. (Take π = 3.14).
(b) In the figure (ii) given below, ABCD is a square of side 4 cm. At each corner of the square a quarter circle of radius 1 cm, and at the centre a circle of diameter 2 cm are drawn. Find the perimeter and the area of the shaded region. Take π = 3.14.

Solution:

Question 33.
(a) In the figure given below, ABCD is a rectangle. AB = 14 cm, BC = 7 cm. From the rectangle, a quarter circle BFEC and a semicircle DGE are removed. Calculate the area of the remaining piece of the rectangle. (Take π = 22/7)
(b) The figure (ii) given below shows a kite, in which BCD is in the shape of a quadrant of circle of radius 42 cm. ABCD is a square and ∆ CEF is an isosceles right angled triangle whose equal sides are 6 cm long. Find the area of the shaded region.

Solution:

Question 34.
(a) In the figure (i) given below, the boundary of the shaded region in the given diagram consists of four semi circular arcs, the smallest two being equal. If the diameter of the largest is 14 cm and of the smallest is 3.5 cm, calculate
(i) the length of the boundary.
(ii) the area of the shaded region.

(b) In the figure (ii) given below, a piece of cardboard, in the shape of a trapezium ABCD, and AB || DC and ∠BCD = 90°, quarter circle BFEC is removed. Given AB = BC = 3.5 cm and DE = 2 cm. Calculate the area of the remaining piece of the cardboard.
Solution:

Question 35.
(a) In the figure (i) given below, ABC is a right angled triangle, ∠B = 90°, AB = 28 cm and BC = 21 cm. With AC as diameter a semi-circle is drawn and with BC as radius a quarter circle is drawn. Find the area of the shaded region correct to two decimal places.
(b) In the figure (ii) given below, ABC is an equilateral triangle of side 8 cm. A, B and C are the centres of circular arcs of equal radius. Find the area of the shaded region correct upto 2 decimal places.    (Take π = 3.142 and √3 = 1.732).

Solution:

Question 36.
A circle is inscribed in a regular hexagon of side 2√3 cm. Find
(i) the circumference of the inscribed circle
(ii) the area of the inscribed circle
Solution:
ABCDEF is a regular hexagon of side 2√3 cm. and a circle is inscribed in it with centre O.

Question 37.
In the figure (i) given below, a chord AB of a circle of radius 10 cm subtends a right angle at the centre O. Find the area of the sector OACB and of the major segment. Take π = 3.14.
Solution:

EXERCISE 16.4

Question 1.
Find the surface area and volume of a cube whose one edge is 7 cm.
Solution:

Question 2.
Find the surface area and the volume of a rectangular solid measuring 5 m by 4 m by 3 m. Also find the length of a diagonal.
Solution:

Question 3.
The length and breadth of a rectangular solid are respectively 25 cm and 20 cm. If the volume is 7000 cm3, find its height.
Solution:

Question 4.
A class room is 10 m long, 6 m broad and 4 m high. How many students can it accommodate if one student needs 1.5 m2 of floor area ? How many cubic metres of air will each student have ?
Solution:

Question 5.
(a) The volume of a cuboid is 1440 cm³. Its height is 10 cm and the cross-section is a square. Find the side of the square.
(b) The perimeter of one face of a cube is 20 cm. Find the surface area and the volume of the cube.
Solution:

Question 6.
Mary wants to decorate her Christmas tree. She wants to place the tree on a wooden box covered with coloured papers with pictures of Santa Claus. She must know the exact quantity of paper to buy for this purpose. If the box has length 80 cm, breadth 40 cm and height 20 cm respectively, then how many square sheets of paper of side 40 cm would she require ?
Solution:

Question 7.
The volume of a cuboid is 3600 cm³ and its height is 12 cm. The cross-section is a rectangle whose length and breadth are in the ratio 4 :3. Find the perimeter of the cross-section.
Solution:

Question 8.
The volume of a cube is 729 cm³. Find its surface area and the length of a diagonal.
Solution:

Question 9.
The length of the longest rod which can be kept inside a rectangular box is 17 cm. If the inner length and breadth of the box are 12 cm and 8 cm respectively, find its inner height.
Solution:

Question 10.
A closed rectangular box has inner dimensions 90 cm by 80 cm by 70 cm. Calculate its capacity and the area of tin-foil needed to line its inner surface.
Solution:

Question 11.
The internal measurements of a box are 20 cm long, 16 cm wide and 24 cm high. How many 4 cm cubes could be put into the box ?
Solution:

Question 12.
The internal measurements of a box are 10 cm long, 8 cm wide and 7 cm high. How many cubes of side 2 cm can be put into the box ?
Solution:

Question 13.
A certain quantity of wood costs Rs. 250 per m3. A solid cubical block of such wood is bought for Rs. 182.25. Calculate the volume of the block and use the method of factors to find the length of one edge of the block.
Solution:

Question 14.
A cube of 11 cm edge is immersed completely in a rectangular vessel containing water. If the dimensions of the base of the vessel are 15 cm x 12 cm, find the rise in the water level in centimetres correct to 2 decimal places, assuming that no water over flows.
Solution:

Question 15.
A rectangular container, whose base is a square of side 6 cm, stands on a horizontal table and holds water upto 1 cm from the top. When a cube is placed in the water and is completely submerged, the water rises to the top and 2 cm3 of water over flows.. Calculate the volume of the cube.
Solution:

Question 16.
(a) Two cubes, each with 12 cm edge, are joined end to end. Find the surface area of the resulting cuboid,
(b) A solid cube of side 12 cm is cut into eight cubes of equal volume. What will be the side of the new cube ? Also, find the ratio between the surface area of the original cube and the sum of the surface areas of the new cubes.
Solution:

Question 17.
A cube of a metal of 6 cm edge is melted and cast into a cuboid whose base is 9 cm x g cm. Find the height of the cuboid.
Solution:

Question 18.
The area of a playground is 4800 m2. Find the cost of covering it with gravel 1 cm deep, if the gravel costs Rs. 260 per cubic metre.
Solution:

Question 19.
A field is 30 m long and 18 m broad. A pit 6 m long, 4m wide and 3 m deep is dug out from the middle of the field and the earth removed is evenly spread over the remaining area of the field. Find the rise in the level of the remaining part of the field in centimetres correct to two decimal places.
Solution:

Question 20.
A rectangular plot is 24 m long and 20 m wide. A cubical pit of edge 4 m is dug at each of the four corners of the field and the soil removed is evenly spread over the remaining part of the plot. By what height does the remaining plot get raised?
Solution:

Question 21.
The inner dimensions of a closed wooden box are 2 m, 1.2 m and .75 m. The thickness of the wood is 2.5 cm. Find the cost of wood required to make the box if 1 m³ of wood costs Rs. 5400.
Solution:

Question 22.
A cubical wooden box of internal edge 1 mis made of 5 cm thick wood. The box is open at the top. If the wood costs Rs. 9600 per cubic metre, find the cost of the wood required to make the box.
Solution:

Question 23.
A square brass plate of side x cm is 1mm thick and weighs 4725 g. If one. cc of brass weighs 8.4 gm, find the value of x.
Solution:

Question 24.
Three cubes whose edges are x cm, 8 cm and 10 cm respectively are melted and recast into a single cube of edge 12 cm. Find x.
Solution:

Question 25.
The area of cross-section of a pipe is 3.5 cm2 and water is flowing out of pipe at the rate of 40 cm/s. How much water is delivered by the pipe in one minute ?
Solution:

Question 26.
(a) The figure (i) given below shows a solid of uniform cross-section. Find the volume of the solid. All measurements are in cm and all angles in the figure are right angles.
(b) The figure (ii) given below shows the cross section of a concrete wall to be constructed. It is 2 m wide at the top, 3.5 m wide at the bottom and its
height is 6 m, and its length is 400 m. Calculate (i) The cross-sectional area, and (ii) volume of concrete in the wall.
(c) The figure (iii) given below show the cross section of a swimming pool 10 m broad, 2 m deep at one end and 3 m deep at the other end. Calculate the volume of water it will hold when full, given that its length is 40 m.

Solution:

Question 27.
A swimming pool is 50 metres long and 15 metres wide. Its shallow and deep ends arc 1\(\frac { 1 }{ 2 }\) metres and 14\(\frac { 1 }{ 2 }\) metres deep respectively. If the bottom of the pool slopes uniformly, find the amount of water required to fill the pool.
Solution:

Multiple Choice Questions

Choose the correct answer from the given four options (1 to 24):
Question 1.
Area of a triangle is 30 cm². If its base is 10 cm, then its height is
(a) 5 cm
(b) 6 cm
(c) 7 cm
(d) 8 cm
Solution:

Question 2.
If the perimeter of a square is 80 cm, then its area is
(a) 800 cm²
(b) 600 cm²
(c) 400 cm²
(d) 200 cm²
Solution:

Question 3.
Area of a parallelogram is 48 cm². If its height is 6 cm then its base is
(a) 8 cm
(b) 4 cm
(c) 16 cm
(d) None of these
Solution:

Question 4.
If d is the diameter of a circle, then its area is

Solution:

Question 5.
If the area of a trapezium is 64 cm² and the distance between parallel sides is 8 cm, then sum of its parallel sides is
(a) 8 cm
(b) 4 cm
(c) 32 cm
(d) 16 cm
Solution:

Question 6.
Area of a rhombus whose diagonals are 8 cm and 6 cm is
(a) 48 cm²
(b) 24 cm²
(c) 12 cm²
(d) 96 cm²
Solution:

Question 7.
If the lengths of diagonals of a rhombus is doubled, then area of rhombus will be
(a) doubled
(b) tripled
(c) four times
(d) remains same
Solution:

Question 8.
If the length of a diagonal of a quadrilateral is 10 cm and lengths of the perpendiculars on it from opposite vertices are 4 cm and 6 cm, then area of quadrilateral is
(a) 100 cm²
(b) 200 cm²
(c) 50 cm²
(d) None of these
Solution:

Question 9.
Area of a rhombus is 90 cm2. If the length of one diagonal is 10 cm then the length of other diagonal is
(a) 18 cm
(b) 9 cm
(c) 36 cm
(d) 4.5 cm
Solution:

Question 10.
In the given figure, OACB is a quadrant of a circle of radius 7 cm. The perimeter of the quadrant is
(a) 11 cm
(b) 18 cm
(c) 25 cm
(d) 36 cm

Solution:

Question 11.
In the given figure, OABC is a square of side 7 cm. OAC is a quadrant of a circle with O as centre. The area of the shaded region is

(a) 10.5 cm²
(b) 38.5 cm
(c) 49 cm²
(d) 11.5 cm²
Solution:

Question 12.
The given figure shows a rectangle and a semicircle. The perimeter of the shaded region is
(a) 70 cm
(b) 56 cm
(c) 78 cm
(d) 46 cm

Solution:

Question 13.
The area of the shaded region shown in Q. 12 (above is
(a) 140 cm²
(b) 77 cm²
(c) 294 cm²
(d) 217 cm²
Solution:

Question 14.
In the given figure, the boundary of the shaded region consists of semicircular arcs. The area of the shaded region is equal to
(a) 616 cm²
(b) 385 cm²
(c) 231 cm²
(d) 308 cm²

Solution:

Question 15.
The perimeter of the shaded region shown in Q. 14 (above) is
(a) 44 cm
(b) 88 cm
(c) 66 cm
(d) 132 cm
Solution:

Question 16.
In the given figure, ABC is a right angled triangle at B. A semicircle is drawn on AB as diameter. If AB = 12 cm and BC = 5 cm, then the area of the shaded region is
(a) (60 + 18π) cm²
(b) (30 + 36π) cm²
(c) (30+18π) cm²
(d) (30 + 9π) cm²

Solution:

Question 17.
The perimeter of the shaded region shown in Q. 16 (above) is
(a) (30 + 6π) cm
(b) (30 + 12π) cm
(c) (18 + 12π) cm
(d) (18 + 6π) cm
Solution:

Question 18.
If the volume of a cube is 729 m³, then its surface area is
(a) 486 cm²
(b) 324 cm²
(c) 162 cm²
(d) None of these
Solution:

Question 19.
If the total surface area of a cube is 96 cm², then the volume of the cube is
(a) 8 cm³
(b) 512 cm³
(c) 64 cm³
(d) 27 cm³
Solution:

Question 20.
The length of the longest pole that can be put in a room of dimensions (10 m x 10 m x 5 m) is
(a) 15 m
(b) 16 m
(c) 10 m
(d) 12 m
Solution:

Question 21.
The lateral surface area of a cube is 256 m². The volume of the cube is
(a) 512 m³
(b) 64 m³
(c) 216 m³
(d) 256 m³
Solution:

Question 22.
If the perimeter of one face of a cube is 40 cm, then the sum of lengths of its edge is
(a) 80 cm
(b) 120 cm
(c) 160 cm
(d) 240 cm
Solution:

Question 23.
A cuboid container has the capacity to hold 50 small boxes. If all the dimensions of the container are doubled, then it can hold (small boxes of same size)
(a) 100 boxes
(b) 200 boxes
(c) 400 boxes
(d) 800 boxes
Solution:

Question 24.
The number of planks of dimensions (4 m x 50 cm x 20 cm) that can be stored in a pit which is 16 m long, 12 m wide and 4 m deep is
(a) 1900
(b) 1920
(c) 1800
(d) 1840
Solution:

Chapter Test

Question 1.
(a) Calculate the area of the shaded region.

(b) If the sides of a square are lengthened by 3 cm, the area becomes 121 cm2. Find the perimeter of the original square.
Solution:

Question P.Q.
The given figure shows a kite in the shape of a square with a diagonal 32 cm and an isosceles triangle of base 8 cm and side 6cm each. How much paper is used in making the kite ? Ignore the wastage of the paper is making the kite.

Solution:

Question 2.
(a) Find the area enclosed by the figure (i) given below. All measurements are in centimetres:
(b) Find the area of the quadrilateral ABCD shown in figure (ii) given below. All measurements are in centimetres.
(c) Calculate the area of the shaded region shown in figure (iii) given below. All measurements are in metres.

Solution:

Question 3.
Asifa cut an aeroplane from a coloured chart paper (as shown in the adjoining figure). Find the total area of the chart paper used, correct to 1 decimal place.
Solution:

Question 4.
If the area of a circle is 78.5 cm², find its circumference. (Take π = 3.14)
Solution:

Question 5.
From a square cardboard, a circle of biggest area was cut out. If the area of the circle is 154 cm2, calculate the original area of the cardboard.
Solution:

Question 6.
(a) From a sheet of paper of dimensions = 2m x 1.5m, how many circles can you cut of radius 5cm. Also find the area of the paper wasted. Take π = 3.14.
(b) If the diameter of a semicircular protractor is 14cm, then find its perimeter.
Solution:

Question 7.
A road 3.5 m wide surrounds a circular park whose circumference is 88 m. Find the cost of paving the road at the rate of Rs. 60 per square metre.
Solution:

Question 8.
The adjoining sketch shows a running tract 3.5 m wide all around which consists of two straight paths and two semicircular rings. Find the area of the track.

Solution:

Question 9.
In the adjoining figure, O is the centre of a circular arc and AOB is a line segment.Find the perimeter and the area of the shaded region correct to one decimal place. (Take π = 3.142)

Solution:

Question 10.
(a) In the figure (1) given below, the radius is 3.5 cm. Find the perimeter of the quarter of the circle.
(b) In the figure (ii) given below, there are five squares each of side 2 cm.
(i) Find the radius of the circle.
(ii) Find the area of the shaded region. (Take π= 3.14).
Solution:

Question 11.
(a) In the figure (i) given below, a piece of cardboard in the shape of a quadrant of a circle of radius 7 cm is bounded by perpendicular radii OX and OY. Points A and B lie on OX and OY respectively such that OA = 3 cm and OB = 4 cm. The triangular part OAB is removed. Calculate the area and the perimeter of the remaining piece.

(b) In the figure (ii) given below, ABCD is a square. Points A, B, C and D are centres of quadrants of circles of the same radius. If the area of the shaded portion is 21\(\frac { 3 }{ 7 }\)
cm2, find the radius of the quadrants. Take π = \(\frac { 22 }{ 7 }\).
Solution:

Question 12.
In the adjoining figure, ABC is a right angled triangle right angled at B. Semicircle are drawn on AB, BC and CA as diameter. Show that the sum of areas of semi circles drawn on AB and BC as diameter is equal to the area of the semicircle drawn on CA as diameter.

Solution:

Question 13.
The length of minute hand of a clock is 14 cm. Find the area swept by the minute hand in 15 minutes.
Solution:

Question 14.
Find the radius of a circle if a 90° arc has a length of 3.5 n cm. Hence, find the area of sector formed by this arc.
Solution:

Question 15.
A cube whose each edge is 28 cm long has a circle of maximum radius on each of its face painted red. Find the total area of the unpainted surface of the cube.
Solution:

Question 16.
Can a pole 6.5 m long fit into the body of a truck with internal dimensions of 3.5m, 3 m and 4m?
Solution:

Question 17.
A car has a petrol tank 40 cm long, 28 cm wide and 25 cm deep. If the fuel consumption of the car averages 13.5 km per litre, how far can the car travel with a full tank of petrol ?
Solution:

Question 18.
An aquarium took 96 minutes to completely fill with water. Water was filling the aquarium at a rate of 25 litres every 2 minutes. Given that the aquarium was 2 m long and 80 cm wide, compute the height of the aquarium.
Solution:

Question 19.
The lateral surface area of a cubiod is 224 cm2. Its height is 7 cm and the base is a square. Find :
(i) a side of the square, and
(ii) the volume of the cubiod.
Solution:

Question 20.
If the volume of a cube is V m³, its surface area is S m2 and the length of a diagonal is d metres, prove that 6√3 V = S d.
Solution:

Question 21.
The adjoining figure shows a victory stand, each face is rectangular. All measurement are in centimetres. Find its volume and surface area (the bottom of the stand is open).
Solution:

Question 22.
The external dimensions of an open rectangular wooden box are 98 cm by 84 cm by 77 cm. If the wood is 2 cm thick all around, find :
(i) the capacity of the box
(ii) the volume of the wood used in making the box, and
(iii) the weight of the box in kilograms correct to one decimal place, given that 1 cm3 of wood weighs 0.8 gm.
Solution:

Question 23.
A cuboidal block of metal has dimensions 36 cm by 32 cm by 0.25 m. It is melted and recast into cubes with an edge of 4 cm.
(i) How many such cubes can be made ?
(ii) What is the cost of silver coating the surfaces of the cubes at the rate of Rs. 1.25 per square centimetre ?
Solution:

Question 24.
Three cubes of silver with edges 3 cm, 4 cm and 5 cm are melted and recast into a single cube. Find the cost of coating the surface of the new cube with gold at the rate of Rs. 3.50 per square centimetre.
Solution:

ML Aggarwal Class 9 Solutions for ICSE Maths Chapter 15 Circle

EXERCISE 15.1

Question 1.
Calculate the length of a chord which is at a distance of 12 cm from the centre of a circle of radius 13 cm.
Solution:

Question 2.
A chord of length 48 cm is drawn in a circle of radius 25 cm. Calculate its distance from the centre of the circle.
Solution:

Question 3.
A chord of length 8 cm is at a distance of 3 cm from the centre of the circle. Calculate the radius of the circle.
Solution:

Question 4.
Calculate the length of the chord which is at a distance of 6 cm from the centre of a circle of diameter 20 cm.
Solution:

Question 5.
A chord of length 16 cm is at a distance of 6 cm from the centre of the circle. Find the length of the chord of the same circle which is at a distance of 8 cm from the centre.
Solution:

Question 6.
In a circle of radius 5 cm, AB and CD are two parallel chords of length 8 cm and 6 cm respectively. Calculate the distance between the chords if they are on :
(i) the same side of the centre.
(ii) the opposite sides of the centre.
Solution:

Question 7.
(a) In the figure given below, O is the centre of the circle. AB and CD are two chords of the circle, OM is perpendicular to AB and ON is perpendicular to CD. AB = 24 cm, OM = 5 cm, ON = 12 cm. Find the:
(i) radius of the circle.
(ii) length of chord CD.

(b) In the figure (ii) given below, CD is the diameter which meets the chord AB in E such that AE = BE = 4 cm. If CE = 3 cm, find the radius of the circle.

Solution:

Question 8.
In the adjoining figure, AB and CD ate two parallel chords and O is the centre. If the radius of the circle is 15 cm, find the distance MN between the two chords of length 24 cm and 18 cm respectively.

Solution:

Question 9.
AB and CD are two parallel chords of a circle of lengths 10 cm and 4 cm respectively. If the chords lie on the same side of the centre and the distance between them is 3 cm, find the diameter of the circle.
Solution:

Question 10.
ABC is an isosceles triangle inscribed in a circle. If AB = AC = 12√5 cm and BC = 24 cm, find the radius of the circle.
Solution:

Question 11.
An equilateral triangle of side 6 cm is inscribed in a circle. Find the radius of the circle.
Solution:

Question 12.
AB is a diameter of a circle. M is a point in AB such that AM = 18 cm and MB = 8 cm. Find the length of the shortest chord through M.
Solution:

Question 13.
A rectangle with one side of length 4 cm is inscribed in a circle of diameter 5 cm. Find the area of the rectangle.
Solution:

Question 14.
The length of the common chord of two intersecting circles is 30 cm. If the radii of the two circles are 25 cm and 17 cm, find the distance between their centres.
Solution:

Question 15.
The line joining the mid-points of two chords of a circle passes through its centre. Prove that the chords are parallel.
Solution:

Question 16.
If a diameter of a circle is perpendicular to one of two parallel chords of the circle, prove that it is perpendicular to the other and bisects it.
Solution:

Question 17.
In an equilateral triangle, prove that the centroid and the circumcentre of the triangle coincide.
Solution:

Question 18.
(a) In the figure (i) given below, OD is perpendicular to the chord AB of a circle whose centre is O. If BC is a diameter, show that CA = 2 OD.
(b) In the figure (ii) given below, O is the centre of a circle. If AB and AC are chords of the circle such that AB = AC and OP ⊥ AB, OQ ⊥ AC, Prove that PB = QC.

Solution:

Question 19.
(a) In the figure (i) given below, a line l intersects two concentric circles at the points A, B, C and D. Prove that AB = CD.
(b) In the figure (it) given below, chords AB and CD of a circle with centre O intersect at E. If OE bisects ∠AED, Prove that AB = CD.

Solution:

Question 20.
(a) In the figure (i) given below, AD is a diameter of a circle with centre O.
If AB || CD, prove that AB = CD.
(b) In the figure (ii) given below, AB and CD are equal chords of a circle with centre O. If AB and CD meet at E (outside the circle) Prove that :
(i) AE = CE (ii) BE = DE.

Solution:

EXERCISE 15.2

Question 1.
If arcs APB and CQD of a circle are congruent, then find the ratio of AB: CD.
Solution:

Question 2.
A and B are points on a circle with centre O. C is a point on the circle such that OC bisects ∠AOB, prove that OC bisects the arc AB.
Solution:

Question 3.
Prove that the angle subtended at the centre of a circle is bisected by the radius passing through the mid-point of the arc.
Solution:

Question 4.
In the given figure, two chords AB and CD of a circle intersect at P. If AB = CD, prove that arc AD = arc CB.
Solution:

Multiple Choice Questions

Choose the correct answer from the given four options (1 to 6) :
Question 1.
If P and Q are any two points on a circle, then the line segment PQ is called a
(a) radius of the circle
(b) diameter of the circle
(c) chord of the circle
(d) secant of the circle
Solution:
chord of the circle (c)

Question 2.
If P is a point in the interior of a circle with centre O and radius r, then
(a) OP = r
(b) OP > r
(c) OP ≥ r
(d) OP < r
Solution:
OP > r (b)

Question 3.
The circumference of a circle must be
(a) a positive real number
(b) a whole number
(c) a natural number
(d) an integer
Solution:
a positive real number (a)

Question 4.
AD is a diameter of a circle and AB is a chord. If AD = 34 cm and AB = 30 cm, then the distance of AB from the centre of circle is
(a) 17 cm
(b) 15 cm
(c) 4 cm
(d) 8 cm
Solution:

Question 5.
If AB = 12 cm, BC = 16 cm and AB is perpendicular to BC, then the radius of the circle passing through the points A, B and C is
(a) 6 cm
(b) 8 cm
(c) 10 cm
(d) 12 cm
Solution:

Question 6.
In the given figure, O is the centre of the circle. If OA = 5 cm, AB = 8 cm and OD ⊥ AB, then length of CD is equal to
(a) 2 cm
(b) 3 cm
(c) 4 cm
(d) 5 cm
Solution:

Chapter Test

Question 1.
In the given figure, a chord PQ of a circle with centre O and radius 15 cm is bisected at M by a diameter AB. If OM = 9 cm, find the lengths of :
(i) PQ
(ii) AP
(iii) BP

Solution:

Question 2.
The radii of two concentric circles are 17 cm and 10 cm ; a line PQRS cuts the larger circle at P and S and the smaller circle at Q and R. If QR = 12 cm, calculate PQ.
Solution:

Question 3.
A chord of length 48 cm is at a distance of 10 cm from the centre of a circle. If another chord of length 20 cm is drawn in the same circle, find its distance from the centre of the circle.
Solution:

Question 4.
(a) In the figure (i) given below, two circles with centres C, D intersect in points P, Q. If length of common chord is 6 cm and CP = 5 cm, DP = 4 cm, calculate the distance CD correct to two decimal places.

(b) In the figure (ii) given below, P is a point of intersection of two circles with centres C and D. If the st. line APB is parallel to CD, Prove that AB = 2 CD.
Solution:

Question 5.
(a) In the figure (i) given below, C and D are centres of two intersecting circles. The line APQB is perpendicular to the line of centres CD.Provethat:
(i) AP=QB
(ii) AQ = BP.
(b) In the figure (ii) given below, two equal chords AB and CD of a circle with centre O intersect at right angles at P. If M and N are mid-points of the chords AB and CD respectively, Prove that NOMP is a square.

Solution:

Question 6.
In the given figure, AD is diameter of a circle. If the chord AB and AC are equidistant from its centre O, prove that AD bisects ∠BAC and ∠BDC.
Solution:

ML Aggarwal Class 9 Solutions for ICSE Maths Chapter 14 Theorems on Area

Question 1.
Prove that the line segment joining the mid-points of a pair of opposite sides of a parallelogram divides it into two equal parallelograms.
Solution:

Question 2.
Prove that the diagonals of a parallelogram divide it into four triangles of equal area.
Solution:

Question 3.
(a) In the figure (1) given below, AD is median of ∆ABC and P is any point on AD. Prove that
(i) area of ∆PBD = area of ∆PDC.
(ii) area of ∆ABP = area of ∆ACP.
(b) In the figure (2) given below, DE || BC. prove that (i) area of ∆ACD = area of ∆ ABE.
(ii) area of ∆OBD = area of ∆OCE.

Solution:

Question 4.
(a) In the figure (1) given below, ABCD is a parallelogram and P is any point in BC. Prove that: Area of ∆ABP + area of ∆DPC = Area of ∆APD.
(b) In the figure (2) given below, O is any point inside a parallelogram ABCD. Prove that:
(i) area of ∆OAB + area of ∆OCD = \(\frac { 1 }{ 2 }\) area of || gm ABCD.
(ii) area of ∆ OBC + area of ∆ OAD = \(\frac { 1 }{ 2 }\) area of ||gmABCD

Solution:

Question 5.
If E, F, G and H are mid-points of the sides AB, BC, CD and DA respectively of a parallelogram ABCD, prove that area of quad. EFGH = 1/2 area of || gm ABCD.
Solution:

Question 6.
(a) In the figure (1) given below, ABCD is a parallelogram. P, Q are any two points on the sides AB and BC respectively. Prove that, area of ∆ CPD = area of ∆ AQD.

(b) In the figure (2) given below, PQRS and ABRS are parallelograms and X is any point on the side BR. Show that area of ∆ AXS = \(\frac { 1 }{ 2 }\) area of ||gm PQRS
Solution:

Question 7.
D,EandF are mid-point of the sides BC, CA and AB respectively of a ∆ ABC. Prove that
(i) FDCE is a parallelogram
(ii) area of ADEF = \(\frac { 1 }{ 4 }\) area of A ABC
(iii) area of || gm FDCE = \(\frac { 1 }{ 2 }\) area of ∆ ABC.
Solution:

Question 8.
In the given figure, D, E and F are mid points of the sides BC, CA and AB respectively of AABC. Prove that BCEF is a trapezium and area of trap. BCEF = \(\frac { 3 }{ 4 }\) area of ∆ ABC.

Solution:

Question P.Q.
Prove that two triangles having equal areas and having one side of one of the triangles equal to one side of the other, have their corresponding altitudes equal.
Solution:

Question 9.
(a) In the figure (1) given below, the point D divides the side BC of ∆ABC in the ratio m : n. Prove that area of ∆ ABD: area of ∆ ADC = m : n.
(b) In the figure (2) given below, P is a point on the sidoBC of ∆ABC such that PC = 2BP, and Q is a point on AP such that QA = 5 PQ, find area of ∆AQC : area of ∆ABC.
(c) In the figure (3) given below, AD is a median of ∆ABC and P is a point in AC such that area of ∆ADP : area of AABD = 2:3. Find
(i) AP : PC (ii) area of ∆PDC : area of ∆ABC.

Solution:

Question 10.
(a) In the figure (1) given below, area of parallelogram ABCD is 29 cm2. Calculate the height of parallelogram ABEF if AB = 5.8 cm
(b) In the figure (2) given below, area of ∆ABD is 24 sq. units. If AB = 8 units, find the height of ABC.
(c) In the figure (3) given below, E and F are mid points of sides AB and CD respectively of parallelogram ABCD. If the area of parallelogram ABCip is 36 cm2.
(i) State the area of ∆ APD.
(ii) Name the parallelogram whose area is equal to the area of ∆ APD.

Solution:

Question 11.
(a) In the figure (1) given below, ABCD is a parallelogram. Points P and Q on BC trisect BC into three equal parts. Prove that :
area of ∆APQ = area of ∆DPQ = \(\frac { 1 }{ 6 }\) (area of ||gm ABCD)
(b) In the figure (2) given below, DE is drawn parallel to the diagonal AC of the quadrilateral ABCD to meet BC produced at the point E. Prove that area of quad. ABCD = area of ∆ABE.
(c) In the figure (3) given below, ABCD is a parallelogram. O is any point on the diagonal AC of the parallelogram. Show that the area of ∆AOB is equal to the area of ∆AOD.

Solution:

Question P.Q.
(a) In the figure (1) given below, two parallelograms ABCD and AEFB are drawn on opposite sides of AB, prove that: area of || gm ABCD + area of || gm AEFB = area of || gm EFCD.
(b) In the figure (2) given below, D is mid-point of the side AB of ∆ABC. P is any point on BC, CQ is drawn parallel to PD to meet AB in Q. Show that area of ∆BPQ = \(\frac { 1 }{ 2 }\) area of ∆ABC.
(c) In the figure (3) given below, DE is drawn parallel to the diagonal AC of the quadrilateral ABCD to meet BC produced at the point E. Prove that area of quad. ABCD = area of ∆ABE.


Solution:

Question 12.
(a) In the figure given, ABCD and AEFG are two parallelograms.
Prove that area of || gm ABCD = area of || gm AEFG.
(b) In the fig. (2) given below, the side AB of the parallelogram ABCD is produced to E. A st. line At through A is drawn parallel to CE to meet CB produced at F and parallelogram BFGE is Completed prove that area of || gm BFGE=Area of || gmABCD.

(c) In the figure (3) given below AB || DC || EF, AD || BEandDE || AF. Prove the area ofDEFH is equal to the area of ABCD.

Solution:

Question 13.
Any point D is taken on the side BC of, a ∆ ABC and AD is produced to E such that AD=DE, prove that area of ∆ BCE = area of ∆ ABC.
Solution:

Question 14.
ABCD is a rectangle and P is mid-point of AB. DP is produced to meet CB at Q. Prove that area of rectangle ∆BCD = area of ∆ DQC.
Solution:

Question P.Q.
ABCD is a square, E and F are mid-points of the sides AB and AD respectively Prove that area of ∆CEF = \(\frac { 3 }{ 8 }\) (area of square ABCD).
Solution:

Question P.Q.
A line PQ is drawn parallel to the side BC of ∆ABC. BE is drawn parallel to CA to meet QP (produced) at E and CF is drawn parallel to BA to meet PQ (produced) at F. Prove that
area of ∆ABE=area of ∆ACF.
Solution:

Question 15.
(a) In the figure (1) given below, the perimeter of parallelogram is 42 cm. Calculate the lengths of the sides of the parallelogram.
(b) In the figure (2) given below, the perimeter of ∆ ABC is 37 cm. If the lengths of the altitudes AM, BN and CL are 5x, 6x, and 4x respectively, Calculate the lengths of the sides of ∆ABC.
(c) In the fig. (3) given below, ABCD is a parallelogram. P is a point on DC such that area of ∆DAP = 25 cm² and area of ∆BCP = 15 cm². Find
(i) area of || gm ABCD
(ii) DP : PC.

Solution:

Question 16.
In the adjoining figure, E is mid-point of the side AB of a triangle ABC and EBCF is a parallelogram. If the area of ∆ ABC is 25 sq. units, find the area of || gm EBCF.
Solution:

Question 17.
(a) In the figure (1) given below, BC || AE and CD || BE. Prove that: area of ∆ABC= area of ∆EBD.
(b) In the llgure (2) given below, ABC is right angled triangle at A. AGFB is a square on the side AB and BCDE is a square on the hypotenuse BC. If AN ⊥ ED, prove that:
(i) ∆BCF ≅ ∆ ABE.
(ii)arca of square ABFG = area of rectangle BENM.


Solution:

Multiple Choice Questions

Choose the correct answer from the given four options (1 to 8):
Question 1.
In the given figure, if l || m, AF || BE, FC ⊥ m and ED ⊥ m , then the correct statement is
(a) area of ||ABEF = area of rect. CDEF
(b) area of ||ABEF = area of quad. CBEF
(c) area of ||ABEF = 2 area of ∆ACF
(d) area of ||ABEF = 2 area of ∆EBD

Solution:
In the given figure,
l ||m, AF || BE, FC ⊥ m and ED ⊥ m
∵ ||gm ABEF and rectangle CDEF are on the same base EF and between the same parallel
∴ area ||gm ABEF = area rect. CDEF (a)

Question 2.
Two parallelograms are on equal bases and between the same parallels. The ratio of their areas is
(a) 1 : 2
(b) 1 : 1
(c) 2 : 1
(d) 3 : 1
Solution:
A triangle and a parallelogram are on the same base and between same parallel, then
∴ They are equal in area
∴ Their ratio 1:1 (b)

Question 3.
If a triangle and a parallelogram are on the same base and between same parallels, then the ratio of area of the triangle to the area of parallelogram is
(a) 1 : 3
(b) 1 : 2
(c) 3 : 1
(d) 1 : 4
Solution:
A triangle and a parallelogram are on the same base and between same parallel, then area of
triangle = \(\frac { 1 }{ 2 }\) area ||gm
∴ Their ratio 1 : 2 (b)

Question 4.
A median of a triangle divides it into two
(a) triangles of equal area
(b) congruent triangles
(c) right triangles
(d) isosceles triangles
Solution:
A median of a triangle divides it into two triangle equal in area. (a)

Question 5.
In the given figure, area of parallelogram ABCD is
(a) AB x BM
(b) BC x BN
(c) DC x DL
(d) AD x DL

Solution:
In the given figure,
Area of ||gm ABCD = AB x DL or DC x DL (∵ AB = DC) (c)

Question 6.
The mid-points of the sides of a triangle along with any of the vertices as the fourth point make a parallelogram of area equal to
(a) \(\frac { 1 }{ 2 }\) area of ∆ABC
(b) \(\frac { 1 }{ 3 }\) area of ∆ABC
(c) \(\frac { 1 }{ 4 }\) area of ∆ABC
(d) area of ∆ABC
Solution:

Question 7.
In the given figure, ABCD is a trapezium with parallel sides AB = a cm and DC = b cm. E and F are mid-points of the non parallel sides. The ratio of area of ABEF and area of EFCD is
(a) a : b
(b) (3a + b) : (a + 3b)
(c) (a + 3b) : (3a + b)
(d) (2a + b) : (3a + b)
Solution:

Question 8.
In the given figure, AB || DC and AB ≠ DC. If the diagonals AC and BD of the trapezium ABCD intersect at O, then which of the following statements is not true?
(a) area of ∆ABC = area of ∆ABD
(b) area of ∆ACD = area of ∆BCD
(c) area of ∆OAB = area of ∆OCD
(d) area of ∆OAD = area of ∆OBC

Solution:

Chapter test

Question 1.
(a) In the figure (1) given below, ABCD is a rectangle (not drawn to scale ) with side AB = 4 cm and AD = 6 cm. Find :
(i) the area of parallelogram DEFC
(ii) area of ∆EFG.
(b) In the figure (2) given below, PQRS is a parallelogram formed by drawing lines parallel to the diagonals of a quadrilateral ABCD through its corners. Prove that area of || gm PQRS = 2 x area of quad. ABCD.

Solution:

Question P.Q.
In the adjoining figure, ABCD and ABEF are parallelogram and P is any point on DC. If area of || gm ABCD = 90 cm2, find:
(i) area of || gm ABEF
(ii) area of ∆ABP.
(iii) area of ∆BEF.

Solution:

Question 2.
In the parallelogram ABCD, P is a point on the side AB and Q is a point on the side BC. Prove that
(i) area of ∆CPD = area of ∆AQD
(ii)area of ∆ADQ = area of ∆APD + area of ∆CPB.

Solution:

Question 3.
In the adjoining figure, X and Y are points on the side LN of triangle LMN. Through X, a line is drawn parallel to LM to meet MN at Z. Prove that area of ∆LZY = area of quad. MZYX.

Solution:

Question P.Q.
If D is a point on the base BC of a triangle ABC such that 2BD = DC, prove that area of ∆ABD= \(\frac { 1 }{ 3 }\) area of ∆ ABC.
Solution:

Question 4.
Perpendiculars are drawn from a point within an equilateral triangle to the three sides. Prove that the sum of the three perpendiculars is equal to the altitude of the triangle.
Solution:

Question 5.
If each diagonal of a quadrilateral’ divides it into two triangles of equal areas, then prove that the quadrilateral is a parallelogram.
Solution:

Question 6.
In the given figure, ABCD is a parallelogram in which BC is produced to E such that CE = BC. AE intersects CD at F. If area of ∆DFB = 3 cm², find the area of parallelogram ABCD.

Solution:

Question 7.
In the given figure, ABCD is a square. E and F are mid-points of sides BC and CD respectively. If R is mid-point of EF, prove that: area of ∆AER = area of ∆AFR.
Solution:

Question 8.
In the given figure, X and Y are mid-points of the sides AC and AB respectively of ∆ABC. QP || BC and CYQ and BXP are straight lines. Prove that area of ∆ABP = area of ∆ACQ.

Solution:

 

ML Aggarwal Class 9 Solutions for ICSE Maths Chapter 12 Pythagoras Theorem

Question 1.
Lengths of sides of triangles are given below. Determine which of them are right triangles. In case of a right triangle, write the length of its hypotenuse:
(i) 3 cm, 8 cm, 6 cm
(ii) 13 cm, .12 cm, 5 cm
(iii) 1.4 cm, 4.8 cm, 5 cm
Solution:

Question 2.
Foot of a 10 m long ladder leaning against a vertical well is 6 m away from the base of the wail. Find the height of the point on the wall where the top of the ladder reaches.
Solution:

Question 3.
A guy attached a wire 24 m long to a vertical pole of height 18 m and has a stake attached to the other end. How far from the base of the pole should the stake be driven so that the wire will be taught?
Solution:

Question 4.
Two poles of heights 6 m and 11 m stand on a plane ground. If the distance between their feet is 12 m, find the distance between their tops.
Solution:

Question 5.
In a right-angled triangle, if hypotenuse is 20 cm and the ratio of the other two sides is 4:3, find the sides.
Solution:

Question 6.
If the sides of a triangle are in the ratio 3:4:5, prove that it is right-angled triangle.
Solution:

Question 7.
For going to a city B from city A, there is route via city C such that AC ⊥ CB, AC = 2x km and CB=2(x+ 7) km. It is proposed to construct a 26 km highway which directly connects the two cities A and B. Find how much distance will be saved in reaching city B from city A after the construction of highway.
Solution:

Question 8.
The hypotenuse of right triangle is 6m more than twice the shortest side. If the third side is 2m less than the hypotenuse, find the sides of the triangle.
Solution:

Question 9.
ABC is an isosceles triangle right angled at C. Prove that AB² = 2AC².
Solution:

Question 10.
In a triangle ABC, AD is perpendicular to BC. Prove that AB² + CD² = AC² + BD².
Solution:

Question 11.
In ∆PQR, PD ⊥ QR, such that D lies on QR. If PQ = a, PR = b, QD = c and DR = d, prove that (a + b) (a – b) = (c + d) (c – d).
Solution:

Question 12.
ABC is an isosceles triangle with AB = AC = 12 cm and BC = 8 cm. Find the altitude on BC and Hence, calculate its area.
Solution:

Question 13.
Find the area and the perimeter of a square whose diagonal is 10 cm long.
Solution:

Question 14.
(a) In fig. (i) given below, ABCD is a quadrilateral in which AD = 13 cm, DC = 12 cm, BC = 3 cm, ∠ ABD = ∠BCD = 90°. Calculate the length of AB.
(b) In fig. (ii) given below, ABCD is a quadrilateral in which AB = AD, ∠A = 90° =∠C, BC = 8 cm and CD = 6 cm. Find AB and calculate the area of ∆ ABD.

Solution:

Question 15.
(a) In figure (i) given below, AB = 12 cm, AC = 13 cm, CE = 10 cm and DE = 6 cm.Calculate the length of BD.
(b) In figure (ii) given below, ∠PSR = 90°, PQ = 10 cm, QS = 6 cm and RQ = 9 cm. Calculate the length of PR.
(c) In figure (iii) given below, ∠ D = 90°, AB = 16 cm, BC = 12 cm and CA = 6 cm. Find CD.

Solution:

Question 16.
(a) In figure (i) given below, BC = 5 cm,
∠B =90°, AB = 5AE, CD = 2AE and AC = ED. Calculate the lengths of EA, CD, AB and AC.
(b) In the figure (ii) given below, ABC is a right triangle right angled at C. If D is mid-point of BC, prove that AB2 = 4AD² – 3AC².

Solution:

Question 17.
In ∆ ABC, AB = AC = x, BC = 10 cm and the area of ∆ ABC is 60 cm². Find x.
Solution:

Question 18.
In a rhombus, If diagonals are 30 cm and 40 cm, find its perimeter.
Solution:

Question 19.
(a) In figure (i) given below, AB || DC, BC = AD = 13 cm. AB = 22 cm and DC = 12cm. Calculate the height of the trapezium ABCD.
(b) In figure (ii) given below, AB || DC, ∠ A = 90°, DC = 7 cm, AB = 17 cm and AC = 25 cm. Calculate BC.
(c) In figure (iii) given below, ABCD is a square of side 7 cm. if
AE = FC = CG = HA = 3 cm,
(i) prove that EFGH is a rectangle.
(ii) find the area and perimeter of EFGH.

Solution:

Question 20.
AD is perpendicular to the side BC of an equilateral Δ ABC. Prove that 4AD² = 3AB².
Solution:

Question 21.
In figure (i) given below, D and E are mid-points of the sides BC and CA respectively of a ΔABC, right angled at C.

Solution:

Question 22.
If AD, BE and CF are medians of ΕABC, prove that 3(AB² + BC² + CA²) = 4(AD² + BE² + CF²).
Solution:

Question 23.
(a) In fig. (i) given below, the diagonals AC and BD of a quadrilateral ABCD intersect at O, at right angles. Prove that
AB² + CD² = AD² + BC².
(b) In figure (ii) given below, OD⊥BC, OE ⊥CA and OF ⊥ AB. Prove that :
(i) OA² + OB² + OC² = AF² + BD² + CE² + OD² + OE² + OF².
(ii) OAF² + BD² + CE² = FB² + DC² + EA².

Solution:

Question 24.
In a quadrilateral, ABCD∠B = 90° = ∠D. Prove that 2 AC² – BC2 = AB² + AD² + DC².
Solution:

Question 25.
In a ∆ ABC, ∠ A = 90°, CA = AB and D is a point on AB produced. Prove that :
DC² – BD² = 2AB. AD.
Solution:

Question 26.
In an isosceles triangle ABC, AB = AC and D is a point on BC produced. Prove that AD² = AC² + BD.CD.
Solution:

Question P.Q.
(a) In figure (i) given below, PQR is a right angled triangle, right angled at Q. XY is parallel to QR. PQ = 6 cm, PY = 4 cm and PX : OX = 1:2. Calculate the length of PR and QR.
(b) In figure (ii) given below, ABC is a right angled triangle, right angled at B.DE || BC.AB = 12 cm, AE = 5 cm and AD : DB = 1: 2. Calculate the perimeter of A ABC.
(c)In figure (iii) given below. ABCD is a rectangle, AB = 12 cm, BC – 8 cm and E is a point on BC such that CE = 5 cm. DE when produced meets AB produced at F.
(i) Calculate the length DE.
(ii) Prove that ∆ DEC ~ AEBF and Hence, compute EF and BF.

Solution:

Multiple Choice Questions

Choose the correct answer from the given four options (1 to 7):
Question 1.
In a ∆ABC, if AB = 6√3 cm, BC = 6 cm and AC = 12 cm, then ∠B is
(a) 120°
(b) 90°
(c) 60°
(d) 45°
Solution:

Question 2.
If the sides of a rectangular plot are 15 m and 8 m, then the length of its diagonal is
(a) 17 m
(b) 23 m
(c) 21 m
(d) 17 cm
Solution:

Question 3.
The lengths of the diagonals of a rhombus are 16 cm and 12 cm. The length of the side of the rhombus is
(a) 9 cm
(b) 10 cm
(c) 8 cm
(d) 20 cm
Solution:

Question 4.
If a side of a rhombus is 10 cm and one of the diagonals is 16 cm, then the length of the other diagonals is
(a) 6 cm
(b) 12 cm
(c) 20 cm
(d) 12 cm
Solution:

Question 5.
If a ladder 10 m long reaches a window 8 m above the ground, then the distance of the foot of the ladder from the base of the wall is
(a) 18 m
(b) 8 m
(c) 6 m
(d) 4 m
Solution:

Question 6.
A girl walks 200 m towards East and then she walks ISO m towards North. The distance of the girl from the starting point is
(a) 350 m
(b) 250 m
(c) 300 m
(d) 225 m
Solution:

Question 7.
A ladder reaches a window 12 m above the ground on one side of the street. Keeping its foot at the same point, the ladder is turned to the other side of the street to reach a window 9 m high. If the length of the ladder is 15 m, then the width of the street is
(a) 30 m
(b) 24 m
(c) 21 m
(d) 18 m
Solution:

Chapter Test

Question 1.
(a) In fig. (i) given below, AD ⊥ BC, AB = 25 cm, AC = 17 cm and AD = 15 cm. Find the length of BC.
(b) In figure (ii) given below, ∠BAC = 90°, ∠ADC = 90°, AD = 6 cm, CD = 8 cm and BC = 26 cm. Find :
(i) AC (ii) AB (iii) area of the shaded region.
(c) In figure (iii) given below, triangle ABC is right angled at B. Given that AB = 9 cm, AC = 15 cm and D, E are mid-points of the sides AB and AC respectively, calculate
(i) the length of BC (ii) the area of ∆ ADE.

Solution:

Question 2.
If in ∆ ABC, AB > AC and ADI BC, prove that AB² – AC² = BD² – CD².
Solution:

Question 3.
In a right angled triangle ABC, right angled at C, P and Q are the points on the sides CA and CB respectively which divide these sides in the ratio 2:1. Prove that
(i) 9AQ² = 9AC² + 4BC²
(ii) 9BP² = 9BC² + 4AC²
(iii) 9(AQ² + BP²) = 13AB².
Solution:
A right angled ∆ ABC in which ∠ C

Question 4.
In the given figure, ∆PQR is right angled at Q and points S and T trisect side QR. Prove that 8PT² – 3PR² + 5PS².
Solution:

Question 5.
In a quadrilateral ABCD, ∠B = 90°. If AD² = AB² + BC² + CD², prove that ∠ACD = 90°.
Solution:

Question 6.
In the given figure, find the length of AD in terms of b and c.
Solution:

Question 7.
ABCD is a square, F is mid-point of AB and BE is one-third of BC. If area of ∆FBE is 108 cm², find the length of AC.
Solution:

Question 8.
In a triangle ABC, AB = AC and D is a point on side AC such that BC² = AC x CD, Prove that BD = BC.
Solution:

ML Aggarwal Class 9 Solutions for ICSE Maths Chapter 11 Mid Point Theorem

Question 1.
(a) In the figure (1) given below, D, E and F are mid-points of the sides BC, CA and AB respectively of ∆ ABC. If AB = 6 cm, BC = 4.8 cm and CA= 5.6 cm, find the perimeter of (i) the trapezium FBCE (ii) the triangle DEF.
(b) In the figure (2) given below, D and E are mid-points of the sides AB and AC respectively. If BC =
5.6 cm and∠B = 72°, compute (i) DE (ii)∠ADE.
(c) In the figure (3) given below, D and E are mid-points of AB, BC respectively and DF || BC. Prove that DBEF is a parallelogram. Calculate AC if AF = 2.6 cm.

Solution:
(a) (i) Given : AB = 6 cm, BC = 4.8 cm, and CA = 5.6 cm
Required : The perimeter of trapezium FBCA.

Question 2.
Prove that the four triangles formed by joining in pairs the mid-points of the sides c of a triangle are congruent to each other.
Solution:
Given: In ∆ ABC, D, E and r,
F are mid-points of AB, BC and CA respectively. Join DE, EF and FD.

Question 3.
If D, E and F are mid-points of sides AB, BC and CA respectively of an isosceles triangle ABC, prove that ∆DEF is also F, isosceles.
Solution:
Given : ABC is an isosceles triangle in which AB = AC

Question 4.
The diagonals AC and BD of a parallelogram ABCD intersect at O. If P is the mid-point of AD, prove that
(i) PQ || AB
(ii) PO=\(\frac { 1 }{ 2 }\)CD.
Solution:
(i) Given : ABCD is a parallelogram in which diagonals AC and BD intersect each other. At point O, P is the mid-point of AD. Join OP.

Question 5.
In the adjoining figure, ABCD is a quadrilateral in which P, Q, R and S are mid-points of AB, BC, CD and DA respectively. AC is its diagonal. Show that
(i) SR || AC and SR =\(\frac { 1 }{ 2 }\)AC
(ii) PQ = SR
(iii) PQRS is a parallelogram.

Solution:

Question 6.
Show that the quadrilateral formed by joining the mid-points of the adjacent sides of a square, is also a square,
Solution:

Question 7.
In the adjoining figure, AD and BE are medians of ∆ABC. If DF U BE, prove that CF = \(\frac { 1 }{ 4 }\) AC.

Solution:

Question 8.
(a) In the figure (1) given below, ABCD is a parallelogram. E and F are mid-points of the sides AB and CO respectively. The straight lines AF and BF meet the straight lines ED and EC in points G and H respectively. Prove that
(i) ∆HEB = ∆HCF
(ii) GEHF is a parallelogram.
(b) In the diagram (2) given below, ABCD is a parallelogram. E is mid-point of CD and P is a point on AC such that PC = \(\frac { 1 }{ 4 }\) AC. EP produced meets BC at F. Prove that
(i) F is mid-point of BC (ii) 2EF = BD

Solution:

Question 9.
ABC is an isosceles triangle with AB = AC. D, E and F are mid-points of the sides BC, AB and AC respectively. Prove that the line segment AD is perpendicular to EF and is bisected by it.
Solution:

Question 10.
(a) In the quadrilateral (1) given below, AB || DC, E and F are mid-points of AD and BD respectively. Prove that:

Solution:

Question 11.
(a) In the quadrilateral (1) given below, AD = BC, P, Q, R and S are mid-points of AB, BD, CD and AC respectively. Prove that PQRS is a rhombus.
(b) In the figure (2) given below, ABCD is a kite in which BC = CD, AB = AD, E, F, G are mid-points of CD, BC and AB respectively. Prove that:
(i) ∠EFG = 90
(ii) The line drawn through G and parallel to FE bisects DA.

Solution:

Question 12.
In the adjoining figure, the lines l, m and n are parallel to each other, and G is mid-point of CD. Calculate:
(i) BG if AD = 6 cm
(ii) CF if GE = 2.3 cm
(iii) AB if BC = 2.4 cm
(iv) ED if FD = 4.4 cm.

Solution:

Multiple Choice Questions

Choose the correct answer from the given four options (1 to 6):
Question 1.
In a ∆ABC, AB = 3 cm, BC = 4 cm and CA = 5 cm. IfD and E are mid-points of AB and BC respectively, then the length of DE is
(a) 1.5 cm
(b) 2 cm
(c) 2.5 cm
(d) 3.5 cm
Solution:
In ∆ABC, D and E are the mid-points of sides AB and BC

Question 2.
In the given figure, ABCD is a rectangle in which AB = 6 cm and AD = 8 cm. If P and Q are mid-points of the sides BC and CD respectively, then the length of PQ is
(a) 7 cm
(b) 5 cm
(c) 4 cm
(d) 3 cm
Solution:

Question 3.
D and E are mid-points of the sides AB and AC of ∆ABC and O is any point on the side BC. O is joined to A. If P and Q are mid-points of OB and OC respectively, then DEQP is
(a) a square
(b) a rectangle
(c) a rhombus
(d) a parallelogram
Solution:
D and E are mid-points of sides AB and AC respectively of AABC O is any point on BC and AO is joined P and Q are mid-points of OB and OC, EQ and DP are joined

Question 4.
The quadrilateral formed by joining the mid-points of the sides of a quadrilateral PQRS, taken in order, is a rectanlge if
(a) PQRS is a parallelogram
(b) PQRS is a rectangle
(c) the diagonals of PQRS are perpendicular to each other
(d) the diagonals of PQRS are equal.
Solution:
A, B, C and D are the mid-points of the sides PQ, QR, RS and SP respectively

Question 5.
The quadrilateral formed by joining the mid-points of the sides of a quadrilateral ABCD, taken in order, is a rhombus if
(a) ABCD is a parallelogram
(b) ABCD is a rhombus
(c) the diagonals of ABCD are equal
(d) the diagonals of ABCD are perpendicular to each other.
Solution:
P, Q, R and S are the mid-points of the quadrilateral ABCD and a quadrilateral is formed by joining the mid-points in order

Question 6.
The figure formed by joining the mid points of the sides of a quadrilateral
ABCD, taken in order, is a square only if
(a) ABCD is a rhombus r
(b) diagonals of ABCD are equal
(c) diagonals of ABCD are perpendicular to each other
(d) diagonals of ABCD are equal and perpendicular to each other.
Solution:

Chapter Test

Question 1.
ABCD is a rhombus with P, Q and R as midpoints of AB, BC and CD respectively. Prove that PQ ⊥ QR.
Solution:

Question 2.
The diagonals of a quadrilateral ABCD are perpendicular. Show that the quadrilateral formed by joining the mid-points of its adjacent sides is a rectangle.
Solution:

Question 3.
If D, E, F are mid-points of the sides BC, CA and AB respectively of a ∆ ABC, Prove that AD and FE bisect each other.
Solution:

Question 4.
In ∆ABC, D and E are mid-points of the sides AB and AC respectively. Through E, a straight line is drawn parallel to AB to meet BC at F. Prove that BDEF is a parallelogram. If AB = 8 cm and BC = 9 cm, find the perimeter of the parallelogram BDEF.
Solution:
In ∆ABC, D and E are the mid-points of

Question 5.
In the given figure, ABCD is a parallelogram and E is mid-point of AD. DL EB meets AB produced at F. Prove that B is mid-point of AF and EB = LF.

Solution:
Given In the figure

Question 6.
In the given figure, ABCD is a parallelogram. If P and Q are mid-points of sides CD and BC respectively. Show that CR = \(\frac { 1 }{ 2 }\) AC.

Solution:

ML Aggarwal Class 9 Solutions for ICSE Maths Chapter 10 Triangles

Exercise 10.1

Question 1.
It is given that ∆ABC ≅ ∆RPQ. Is it true to say that BC = QR ? Why?
Solution:

Question 2.
“If two sides and an angle of one triangle are equal to two sides and an angle of another triangle, then the two triangles must be congruent.” Is the statement true? Why?
Solution:
No, it is not true statement as the angles should be included angle of there two given sides.

Question 3.
In the given figure, AB=AC and AP=AQ. Prove that
(i) ∆APC ≅ ∆AQB
(ii) CP = BQ
(iii) ∠APC = ∠AQB.
Solution:

Question 4.
In the given figure, AB = AC, P and Q are points on BA and CA respectively such that AP = AQ. Prove that
(i) ∆APC ≅ ∆AQB
(ii) CP = BQ
(iii) ∠ACP = ∠ABQ.
Solution:

Question 5.
In the given figure, AD = BC and BD = AC. Prove that :
∠ADB = ∠BCA and ∠DAB = ∠CBA.

Solution:

Question 6.
In the given figure, ABCD is a quadrilateral in which AD = BC and ∠DAB = ∠CBA. Prove that
(i) ∆ABD ≅ ∆BAC
(ii) BD = AC
(iii) ∠ABD = ∠BAC.
Solution:

Question 7.
In the given figure, AB = DC and AB || DC. Prove that AD = BC.
Solution:

Question 8.
In the given figure. AC = AE, AB = AD and ∠BAD = ∠CAE. Show that BC = DE.

Solution:

Question 9.
In the adjoining figure, AB = CD, CE = BF and ∠ACE = ∠DBF. Prove that
(i) ∆ACE ≅ ∆DBF
(ii) AE = DF.
Solution:

Question 10.
In the given figure, AB = AC and D is mid-point of BC. Use SSS rule of congruency to show that
(i) ∆ABD ≅ ∆ACD
(ii) AD is bisector of ∠A
(iii) AD is perpendicular to BC.
Solution:

Question 11.
Two line segments AB and CD bisect each other at O. Prove that :
(i) AC = BD
(ii) ∠CAB = ∠ABD
(iii) AD || CB
(iv) AD = CB.

Solution:

Question 12.
In each of the following diagrams, find the values of x and y.

Solution:

Exercise 10.2

Question 1.
In triangles ABC and PQR, ∠A= ∠Q and ∠B = ∠R. Which side of APQR should be equal to side AB of AABC so that the two triangles are congruent? Give reason for your answer.
Solution:

Question 2.
In triangles ABC and PQR, ∠A = ∠Q and ∠B = ∠R. Which side of APQR should be equal to side BC of AABC so that the two triangles are congruent? Give reason for your answer.
Solution:

Question 3.
“If two angles and a side of one triangle are equal to two angles and a side of another triangle, then the two triangles must be congruent”. Is the statement true? Why?
Solution:
The given statement can be true only if the corresponding (included) sides are equal otherwise not.

Question 4.
In the given figure, AD is median of ∆ABC, BM and CN are perpendiculars drawn from B and C respectively on AD and AD produced. Prove that BM = CN.
Solution:

Question 5.
In the given figure, BM and DN are perpendiculars to the line segment AC. If BM = DN, prove that AC bisects BD.

Solution:

Question 6.
In the given figure, l and m are two parallel lines intersected by another pair of parallel lines p and q. Show that ∆ABC ≅ ∆CDA.

Solution:

Question 7.
In the given figure, two lines AB and CD intersect each other at the point O such that BC || DA and BC = DA. Show that O is the mid-point of both the line segments AB and CD.
Solution:

Question 8.
In the given figure, ∠BCD = ∠ADC and ∠BCA = ∠ADB. Show that
(i) ∆ACD ≅ ∆BDC
(ii) BC = AD
(iii) ∠A = ∠B.
Solution:

Question 9.
In the given figure, ∠ABC = ∠ACB, D and E are points on the sides AC and AB respectively such that BE = CD. Prove that
(i) ∆EBC ≅ ∆DCB
(ii) ∆OEB ≅ ∆ODC
(iii) OB = OC.
Solution:

Question 10.
ABC is an isosceles triangle with AB=AC. Draw AP ⊥ BC to show that ∠B = ∠C.
Solution:

Question 11.
In the given figure, BA ⊥ AC, DE⊥ DF such that BA = DE and BF = EC.
Solution:

Question 12.
ABCD is a rectanige. X and Y are points on sides AD and BC respectively such that AY = BX. Prove that BY = AX and ∠BAY = ∠ABX.
Solution:

Question 13.
(a) In the figure (1) given below, QX, RX are bisectors of angles PQR and PRQ respectively of A PQR. If XS⊥ QR and XT ⊥ PQ, prove that
(i) ∆XTQ ≅ ∆XSQ
(ii) PX bisects the angle P.
(b) In the figure (2) given below, AB || DC and ∠C = ∠D. Prove that
(i) AD = BC
(ii) AC = BD.
(c) In the figure (3) given below, BA || DF and CA II EG and BD = EC . Prove that, .
(i) BG = DF
(ii) EG = CF.


Solution:

Question 14.
In each of the following diagrams, find the values of x and y.

Solution:

Exercise 10.3

Question 1.
ABC is a right angled triangle in which ∠A = 90° and AB = AC. Find ∠B and ∠C.
Solution:

Question 2.
Show that the angles of an equilateral triangle are 60° each.
Solution:

Question 3.
Show that every equiangular triangle is equilateral.
Solution:

Question 4.
In the following diagrams, find the value of x:

Solution:

Question 5.
In the following diagrams, find the value of x:

Solution:

Question 6.
(a) In the figure (1) given below, AB = AD, BC = DC. Find ∠ ABC.
(b)In the figure (2) given below, BC = CD. Find ∠ACB.
(c) In the figure (3) given below, AB || CD and CA = CE. If ∠ACE = 74° and ∠BAE =15°, find the values of x and y.
Solution:

Question 7.
In ∆ABC, AB = AC, ∠A = (5x + 20)° and each of the base angle is \(\frac { 2 }{ 5 }\) th of ∠A. Find the measure of ∠A.
Solution:

Question 8.
(a) In the figure (1) given below, ABC is an equilateral triangle. Base BC is produced to E, such that BC’= CE. Calculate ∠ACE and ∠AEC.
(b) In the figure (2) given below, prove that ∠ BAD : ∠ ADB = 3 : 1.
(c) In the figure (3) given below, AB || CD. Find the values of x, y and ∠.

Solution:

Question 9.
In the given figure, D is mid-point of BC, DE and DF are perpendiculars to AB and AC respectively such that DE = DF. Prove that ABC is an isosceles triangle.

Solution:

Question 10.
In the given figure, AD, BE and CF arc altitudes of ∆ABC. If AD = BE = CF, prove that ABC is an equilateral triangle.
Solution:

Question 11.
In a triangle ABC, AB = AC, D and E are points on the sides AB and AC respectively such that BD = CE. Show that:
(i) ∆DBC ≅ ∆ECB
(ii) ∠DCB = ∠EBC
(iii) OB = OC,where O is the point of intersection of BE and CD.
Solution:

Question 12.
ABC is an isosceles triangle in which AB = AC. P is any point in the interior of ∆ABC such that ∠ABP = ∠ACP. Prove that
(a) BP = CP
(b) AP bisects ∠BAC.
Solution:

Question 13.
In the adjoining figure, D and E are points on the side BC of ∆ABC such that BD = EC and AD = AE. Show that ∆ABD ≅ ∆ACE.
Solution:

Question 14.
(a) In the figure (i) given below, CDE is an equilateral triangle formed on a side CD of a square ABCD. Show that ∆ADE ≅ ∆BCE and hence, AEB is an isosceles triangle.

(b) In the figure (ii) given below, O is a point in the interior of a square ABCD such that OAB is an equilateral trianlge. Show that OCD is an isosceles triangle.

Question 15.
In the given figure, ABC is a right triangle with AB = AC. Bisector of ∠A meets BC at D. Prove that BC = 2AD.
Solution:

Exercise 10.4

Question 1.
In ∆PQR, ∠P = 70° and ∠R = 30°. Which side of this triangle is longest? Give reason for your answer.
Solution:

Question 2.
Show that in a right angled triangle, the hypotenuse is the longest side.
Solution:

Question 3.
PQR is a right angle triangle at Q and PQ : QR = 3:2. Which is the least angle.
Solution:

Question 4.
In ∆ ABC, AB = 8 cm, BC = 5.6 cm and CA = 6.5 cm. Which is (i) the greatest angle ?
(ii) the smallest angle ?
Solution:

Question 5.
In ∆ABC, ∠A = 50°, ∠B= 60°, Arrange the sides of the triangle in ascending order.
Solution:

Question 6.
In figure given alongside, ∠B = 30°, ∠C = 40° and the bisector of ∠A meets BC at D. Show
(i) BD > AD
(ii) DC > AD
(iii) AC > DC
(iv) AB > BD

Solution:

Question 7.
(a) In the figure (1) given below, AD bisects ∠A. Arrange AB, BD and DC in the descending order of their lengths.
(b) In the figure (2) given below, ∠ ABD = 65°, ∠DAC = 22° and AD = BD. Calculate ∠ ACD and state (giving reasons) which is greater : BD or DC ?

Solution:

Question 8.
(a) In the figure (1) given below, prove that (i) CF> AF (ii) DC>DF.
(b) In the figure (2) given below, AB = AC.
Prove that AB > CD.
(c) In the figure (3) given below, AC = CD. Prove that BC < CD.

Solution:

Question 9.
(a) In the figure (i) given below, ∠B < ∠A and ∠C < ∠D. Show that AD < BC. (b) In the figure (ii) given below, D is any point on the side BC of ∆ABC. If AB > AC, show that AB > AD.
Solution:

Question 10.
(i) Is it possible to construct a triangle with lengths of its sides as 4 cm, 3 cm and 7 cm? Give reason for your answer,
(ii) Is it possible to construct a triangle with lengths of its sides as 9 cm, 7 cm and 17 cm? Give reason for your answer.
(iii) Is it possible to construct a triangle with lengths of its sides as 8 cm, 7 cm and 4 cm? Give reason for your answer.
Solution:
(i) Length of sides of a triangle are 4 cm, 3 cm and 7 cm
We know that sum of any two sides of a triangle is greatar than its third side But 4 + 3 = 7 cm
Which is not possible
Hence to construction of a triangle with sides 4 cm, 3 cm and 7 cm is not possible.
(ii) Length of sides of a triangle are 9 cm, 7 cm and 17 cm
We know that sum of any two sides of a triangle is greater than its third side Now 9 + 7 = 16 < 17 ∴ It is not possible to construct a triangle with these sides.
(iii) Length of sides of a triangle are 8 cm, 7 cm and 4 cm We know that sum of any two sides of a triangle is greater than its third side Now 7 + 4 = 11 > 8
Yes, It is possible to construct a triangle with these sides.

Multiple Choice Questions

Choose the correct answer from the given four options (1 to 18):
Question 1.
Which of the following is not a criterion for congruency of triangles?
(a) SAS
(b) ASA
(c) SSA
(d) SSS
Solution:
Criteria of congruency of two triangles ‘SSA’ is not the criterion. (c)

Question 2.
In the adjoining figure, AB = FC, EF=BD and ∠AFE = ∠CBD. Then the rule by which ∆AFE = ∆CBD is
(a) SAS
(b) ASA
(c) SSS
(d) AAS

Solution:

Question 3.
In the adjoining figure, AB ⊥ BE and FE ⊥ BE. If AB = FE and BC = DE, then
(a) ∆ABD ≅ ∆EFC
(b) ∆ABD ≅ ∆FEC
(c) ∆ABD ≅ ∆ECF
(d) ∆ABD ≅ ∆CEF
Solution:
In the figure given,

Question 4.
In the adjoining figure, AB=AC and AD is median of ∆ABC, then AADC is equal to
(a) 60°
(b) 120°
(c) 90°
(d) 75°
Solution:

Question 5.
In the adjoining figure, O is mid point of AB. If ∠ACO = ∠BDO, then ∠OAC is equal to
(a) ∠OCA
(b) ∠ODB
(c) ∠OBD
(d) ∠BOD
Solution:

Question 6.
In the adjoining figure, AC = BD. If ∠CAB = ∠DBA, then ∠ACB is equal to
(a) ∠BAD
(b) ∠ABC
(c) ∠ABD
(d) ∠BDA

Solution:

Question 7.
In the adjoining figure, ABCD is a quadrilateral in which BN and DM are drawn perpendiculars to AC such that BN = DM. If OB = 4 cm, then BD is
(a) 6 cm
(b) 8 cm
(c) 10 cm
(d) 12 cm
Solution:

Question 8.
In ∆ABC, AB = AC and ∠B = 50°. Then ∠C is equal to
(a) 40°
(b) 50°
(c) 80°
(d) 130°
Solution:

Question 9.
In ∆ABC, BC = AB and ∠B = 80°. Then ∠A is equal to
(a) 80°
(b) 40°
(c) 50°
(d) 100°
Solution:

Question 10.
In ∆PQR, ∠R = ∠P, QR = 4 cm and PR = 5 cm. Then the length of PQ is
(a) 4 cm
(b) 5 cm
(c) 2 cm
(d) 2.5 cm
Solution:

Question 11.
In ∆ABC and APQR, AB = AC, ∠C = ∠P and ∠B = ∠Q. The two triangles are
(a) isosceles but not congruent
(b) isosceles and congruent
(c) congruent but isosceles
(d) neither congruent nor isosceles
Solution:

Question 12.
Two sides of a triangle are of lenghts 5 cm and 1.5 cm. The length of the third side of the triangle can not be
(a) 3.6 cm
(b) 4.1 cm
(c) 3.8 cm
(d) 3.4 cm
Solution:

Question 13.
If a, b, c are the lengths of the sides of a trianlge, then
(a) a – b > c
(b) c > a + b
(c) c = a + b
(d) c < A + B
Solution:
a, b, c are the lengths of the sides of a trianlge than a + b> c or c < a + b
(Sum of any two sides is greater than its third side) (d)

Question 14.
It is not possible to construct a triangle when the lengths of its sides are
(a) 6 cm, 7 cm, 8 cm
(b) 4 cm, 6 cm, 6 cm
(c) 5.3 cm, 2.2 cm, 3.1 cm
(d) 9.3 cm, 5.2 cm, 7.4 cm
Solution:
We know that sum of any two sides of a triangle is greater than its third side 2.2 + 3.1 = 5.3 ⇒ 5.3 = 5.3 is not possible (c)

Question 15.
In ∆PQR, if ∠R> ∠Q, then
(a) QR > PR
(b) PQ > PR
(c) PQ < PR
(d) QR < PR
Solution:
In ∆PQR, ∠R> ∠Q
∴ PQ > PR (b)

Question 16.
If triangle PQR is right angled at Q, then
(a) PR = PQ
(b) PR < PQ
(c) PR < QR
(d) PR > PQ
Solution:

Question 17.
If triangle ABC is obtuse angled and ∠C is obtuse, then
(a) AB > BC
(b) AB = BC
(c) AB < BC
(d) AC > AB
Solution:

Question P.Q.
A triangle can be constructed when the lengths of its three sides are
(a) 7 cm, 3 cm, 4 cm
(b) 3.6 cm, 11.5 cm, 6.9 cm
(c) 5.2 cm, 7.6 cm, 4.7 cm
(d) 33 mm, 8.5 cm, 49 mm
Solution:
We know that in a triangle, if sum of any two sides is greater than its third side, it is possible to construct it 5.2 cm, 7.6 cm, 4.7 cm is only possible. (c)

Question P.Q.
A unique triangle cannot be constructed if its
(a) three angles are given
(b) two angles and one side is given
(c) three sides are given
(d) two sides and the included angle is given
Solution:
A unique triangle cannot be constructed if its three angle are given, (a)

Question 18.
If the lengths of two sides of an isosceles are 4 cm and 10 cm, then the length of the third side is
(a) 4 cm
(b) 10 cm
(c) 7 cm
(d) 14 cm
Solution:
Lengths of two sides of an isosceles triangle are 4 cm and 10 cm, then length of the third side is 10 cm
(Sum of any two sides of a triangle is greater than its third side and 4 cm is not possible as 4 + 4 > 10 cm.

Chapter Test

Question 1.
In triangles ABC and DEF, ∠A = ∠D, ∠B = ∠E and AB = EF. Will the two triangles be congruent? Give reasons for your answer.
Solution:

Question 2.
In the given figure, ABCD is a square. P, Q and R are points on the sides AB, BC and CD respectively such that AP= BQ = CR and ∠PQR = 90°. Prove that
(a) ∆PBQ ≅ ∆QCR
(b) PQ = QR
(c) ∠PRQ = 45°

Solution:

Question 3.
In the given figure, AD = BC and BD = AC. Prove that ∠ADB = ∠BCA.
Solution:

Question 4.
In the given figure, OA ⊥ OD, OC X OB, OD = OA and OB = OC. Prove that AB = CD.

Solution:

Question 5.
In the given figure, PQ || BA and RS CA. If BP = RC, prove that:
(i) ∆BSR ≅ ∆PQC
(ii) BS = PQ
(iii) RS = CQ.

Solution:

Question 6.
In the given figure, AB = AC, D is a point in the interior of ∆ABC such that ∠DBC = ∠DCB. Prove that AD bisects ∠BAC of ∆ABC.
Solution:

Question 7.
In the adjoining figure, AB || DC. CE and DE bisects ∠BCD and ∠ADC respectively. Prove that AB = AD + BC.
Solution:

Question 8.
In ∆ABC, D is a point on BC such that AD is the bisector of ∠BAC. CE is drawn parallel to DA to meet BD produced at E. Prove that ∆CAE is isosceles
Solution:

Question 9.
In the figure (ii) given below, ABC is a right angled triangle at B, ADEC and BCFG are squares. Prove that AF = BE.

Solution:

Question 10.
In the given figure, BD = AD = AC. If ∠ABD = 36°, find the value of x .

Solution:

Question 11.
In the adjoining figure, TR = TS, ∠1 = 2∠2 and ∠4 = 2∠3. Prove that RB = SA.

Solution:
Given: In the figure , RST is a triangle

Question 12.
(a) In the figure (1) given below, find the value of x.
(b) In the figure (2) given below, AB = AC and DE || BC. Calculate
(i)x
(ii) y
(iii) ∠BAC
(c) In the figure (1) given below, calculate the size of each lettered angle.

Solution:

Question 13.
(a) In the figure (1) given below, AD = BD = DC and ∠ACD = 35°. Show that
(i) AC > DC (ii) AB > AD.
(b) In the figure (2) given below, prove that
(i) x + y = 90° (ii) z = 90° (iii) AB = BC

Solution:

Question 14.
In the given figure, ABC and DBC are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC. If AD is extended to intersect BC at P, show that
(i) ∆ABD ≅ ∆ACD
(ii) ∆ABP ≅ ∆ACP
(iii) AP bisects ∠A as well as ∠D
(iv) AP is the perpendicular bisector of BC.
Solution:

Question 15.
In the given figure, AP ⊥ l and PR > PQ. Show that AR > AQ.
Solution:

Question 16.
If O is any point in the interior of a triangle ABC, show that
OA + OB + OC > \(\frac { 1 }{ 2 }\)
(AB + BC + CA).

Solution:

Question P.Q.
Construct a triangle ABC given that base BC = 5.5 cm, ∠ B = 75° and height = 4.2 cm.
Solution:

Question P.Q.
Construct a triangle ABC in which BC = 6.5 cm, ∠ B = 75° and ∠ A = 45°. Also construct median of A ABC passing through B.
Solution:

Question P.Q.
Construct triangle ABC given that AB – AC = 2.4 cm, BC = 6.5 cm. and ∠ B = 45°.
Solution: