# 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

(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

(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

#### Answer :

Option (c) is correct

(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

(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

(a) doubled
(b) tripled
(c) four times
(d) remains same

#### Answer :

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

(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

#### (a) 11 cm(b) 18 cm(c) 25 cm(d) 36 cmAnswer :

OACB is a quadrant of a circle of radius 7 cm

Option (c) 25 cm is correct

(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

Option (a) 10.5 cm2 is correct

(a) 70 cm
(b) 56 cm
(c) 78 cm
(d) 46 cm

#### 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

(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 cm2Answer :

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

(a) 44 cm
(b) 88 cm
(c) 66 cm
(d) 132 cm

#### Answer :

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

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

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

(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

(a) 15 m
(b) 16 m
(c) 10 m
(d) 12 m

#### Answer 20

Option (a) 15  cm is correct

(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

(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

(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

(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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