ML Aggarwal Mensuration MCQs Class 9 ICSE Maths Solutions

ML Aggarwal Mensuration MCQs Class 9 ICSE Maths Solutions Ch-16. Step by Step Solutions of MCQs Questions on Mensuration of ML Aggarwal for ICSE Class 9th Mathematics. Visit official website CISCE for detail information about ICSE Board Class-9.

ML Aggarwal Mensuration MCQs Class 9 ICSE Maths Solutions

Board ICSE
Subject Maths
Class 9th
Chapter-16 Mensuration
Topics Solution of MCQs Questions
Academic Session 2024-2025

MCQs Solutions of ML Aggarwal for ICSE Class-9 Ch-16, Mensuration

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

Answer :

let h be the height of the triangle.

given

area of triangle =30cm²

base of the triangle =10cm

we know

area of triangle =1/2*b*h

or, 30cm²=1/2*10cm*h

or, 30cm²=5cm*h

or, h=30cm²/5cm

or, h=6cm

Option (b) 6 cm is correct

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

Answer :

Perimeter of a  square = 4 * length of each side
80 = 4 * length
length = 20 cm.
Area of a square = length * length
Area = 20 * 20 = 400 sq.cm
Option (c) 400 cm is  correct

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

Answer :

Area of Parallelogram = Base ×Height

48= Base × 6

Base = 48 ÷ 6

= 8cm

Option (a) 8 cm is  correct

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

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

Answer :

Option (c) is correct

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

Answer :

Area of trapezium = 1/2 (a+b) * h
64 cm² = 1/2 (a+b) * 8
64*(2/8) = a. + b
16 = a+b

Option (a) 8 cm is correct

Question 6. bArea of a rhombus whose diagonals are 8 cm and 6 cm is

(a) 48 cm2
(b) 24 cm2
(c) 12 cm2
(d) 96 cm2

Answer :

Area of a rhombus = (8 x 6)/2

= 24 cm2

Option (b) 24 cmis correct

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

Answer :

ml class 9 Mensuration img 90

Option (c) four times is correct

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

Answer :

area of quadrilateral = 1/2(sum of perpendiculars dropped from opposite sides)diagonal
=1/2 × (4+6) × 10
5 × 10 cm square
= 50 cm²

Option (c) 50 cmis correct

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

Answer :

Area of rhombus = 1/2 * d1* d2 (here d1

and d2 are diagonals)

90 =1/2*10* d2

90=5*d2

d2=90/5

d2=18 cm

Option (a) 18 cm  is correct


Mensuration Exercise-MCQs

ML Aggarwal Class 9 ICSE Maths Solutions

Page 396

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
In the given figure, OACB is a quadrant of a circle of radius 7 cm. The perimeter of the quadrant is
Answer :

OACB is a quadrant of a circle of radius 7 cm

ml class 9 Mensuration img 91

Option (c) 25 cm is correct

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

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

Answer :

OABC is a square of side 7 cm. OAC is a quadrant

ml class 9 Mensuration img 92

Option (a) 10.5 cm2 is correct

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
The given figure shows a rectangle and a semicircle. The perimeter of the shaded region is

Answer :

perimeter of shaded region =circumference of semicircle + perimeter of rectangle – length of rectangle

2×(22/7) × (14/2) + (2×10) +14

22+20+14

56

Option (b) 56 cm is correct

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

Answer :

The area of the shaded region shown in Q. 12

= Area of rectangle + area of semicircle

l x b + (1/2)π²

14 x 10 + 1/2 x 22/7 x 7 x 7

= 140 + 77 

= 217 cm²

Option (d) 217 cmis correct

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

Answer :

area of the shaded region

Area of semicircle of 14 cm radius – area of semicircle of 7 cm radius + area of semicircle of 7 cm radius

= area of semicircle of radius 14 cm

= 1/2πr²

= 1/2 x 22/7 x `14 x 14

= 308 cm²

Option (d) 308 cmis correct

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

Answer :

ml class 9 Mensuration img 93

Option (b) 88 cm is correct

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

Answer :

In the given figure, ABC is a right angled triangle right angled at B,AB is diameter = 12 cm, BC = 5 cm

Area of shaded portion = area of semicircle + area of right triangle ABC

ml class 9 Mensuration img 94

Option (c) (30+18π) cm2 is correct

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

Answer :

perimeter of the shaded region shown in Q. 16

ml class 9 Mensuration img 95

Option (d) (18 + 6π) cm is correct

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

Answer :

edge of the cube = a cm

volume(v)= a³

a³ = 729

a³ = (9cm)³

a = 9cm

lateral surface area = 4 a²
= 4* 9²
= 4* 81
=324cm²

total surface area = 6a²
= 6* 9²
= 6*81
= 486 cm²

Option (a) 486 cmcm is correct


Mensuration Exercise-MCQs

ML Aggarwal Class 9 ICSE Maths Solutions

Page 395

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

Answer :

Surface area of a cube = 96 cm2
Surface area of a cube = 6 (Side)2 = 96 ⇒  (Side)2 = 16
(Side) = 4 cm
Volume of cube = (Side)3 = (4)3 = 64 cm3

Option (c) 64 cm3 cm is correct

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

Answer 20

ml class 9 Mensuration img 96

Option (a) 15  cm is correct

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

Answer 21. 

Given, lateral surface area of a cube = 256 m2

We know that, lateral surface area of a cube = 4 x (Side)2
⇒ 256 = 4 x (Side)2
⇒ (Side)2 = 256/4 = 64
⇒ Side = √64 = 8 m

Now, volume of a cube = (Side)3 = (8)= 8 x 8 x 8 = 512 m3
Hence, the volume of the cube is 512 m3.

Option (a) 512 m is correct

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

Answer 22

One side of cube = 40 cm.

We need to find the sum of length of its edges?

Follow the below steps to find the solution

As we know the edge of one side =4

Perimeter of one side of cube=40 cm

Length of one edge = 40/ 4

Length of one edge = 10 cm.

As we know, the cube has 12 edges

Sum of length of edges = Total number of edge x length of one edge.

Sum of the length of its edges = 12 x 10

Multiply the two values which gives 120.

we got the sum of length of edges of cube is 120 cm.

Option (b) 120 cm  is correct

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

Answer :

let the length, breadth and height of the container be x, y and z cm respectively.
So the volume of the container will be xyz cm³
So xyz cm³ = 50 boxes
If the dimensions of the container are doubled the the length, breadth and height of the container will be 2x, 2y and 2z
So the volume will change to 8xyz cm³
Since xyz cm³ can contain 50 boxes therefore 8xyz cm^3 can contain 50 * 8 boxes = 400 boxes
Option (c) 400 boxes is correct

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

Answer :

Length of the plank = 4m

= 400cm
Breadth = 50cm
Height = 20cm
Volume of the plank = L*B*H
=400*50*20
=400000cm³

Length of the pit = 16m

= 1600cm
Breadth = 12m = 1200cm
Height = 4m

= 400cm
Volume of the pit = L*B*H
= 1600*1200*400
= 768000000 cm³

Number of planks that can be fitted=
768000000/400000
=1920 planks is the answer.

Option (b) 1920 is correct

—  : End of ML Aggarwal Mensuration MCQs Class 9 ICSE Maths Solutions :–

Return to :-   ML Aggarawal Maths Solutions for ICSE  Class-9

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