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 cm**^{2}. If its base is 10 cm, then its height is

^{2}. 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 cm^{2}

(b) 600 cm^{2}

(c) 400 cm^{2}

(d) 200 cm^{2}

**Answer :**

**Question 3. ****Area of a parallelogram is 48 cm**^{2}. If its height is 6 cm then its base is

^{2}. 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**

**Answer :**

**Option (c) is correct**

**Question 5. ****If the area of a trapezium is 64 cm**^{2} and the distance between parallel sides is 8 cm, then sum of its parallel sides is

^{2}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. b****Area of a rhombus whose diagonals are 8 cm and 6 cm is**

(a) 48 cm^{2}

(b) 24 cm^{2}

(c) 12 cm^{2}

(d) 96 cm^{2}

**Answer :**

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

**= 24 cm ^{2}**

**Option (b) 24 cm ^{2 }is 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 :**

**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 cm^{2}

(b) 200 cm^{2}

(c) 50 cm^{2}

(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 cm ^{2 }is 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**

**Answer :**

**OACB is a quadrant of a circle of radius 7 cm**

**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**

(a) 10.5 cm^{2}

(b) 38.5 cm

(c) 49 cm^{2}

(d) 11.5 cm^{2}

**Answer :**

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

**Option (a) 10.5 cm ^{2}^{ }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

**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 cm^{2}

(b) 77 cm^{2}

(c) 294 cm^{2}

(d) 217 cm^{2}

**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 cm ^{2 }is 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 cm^{2}

(b) 385 cm^{2}

(c) 231 cm^{2}

(d) 308 cm^{2}

**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 cm ^{2 }is 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 :**

**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) cm^{2}

(b) (30 + 36π) cm^{2}

(c) (30+18π) cm^{2}

(d) (30 + 9π) cm^{2}

**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) cm ^{2}^{ }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 m**^{3}, then its surface area is

^{3}, then its surface area is

(a) 486 cm^{2}

(b) 324 cm^{2}

(c) 162 cm^{2}

(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 cm ^{2 }cm^{ }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 cm**^{2}, then the volume of the cube is

^{2}, then the volume of the cube is

(a) 8 cm^{3}

(b) 512 cm^{3}

(c) 64 cm^{3}

(d) 27 cm^{3}

**Answer :**

Surface area of a cube = 96 cm^{2}

Surface area of a cube = 6 (Side)^{2} = 96 ⇒ (Side)^{2} = 16

(Side) = 4 cm

Volume of cube = (Side)^{3} = (4)3 = 64 cm^{3}

**Option (c) 64 cm ^{3}^{ }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**

**Option (a) 15 cm ^{ }is correct**

**Question 21. ****The lateral surface area of a cube is 256 m**^{2}. The volume of the cube is

^{2}. The volume of the cube is

(a) 512 m^{3}

(b) 64 m^{3}

(c) 216 m^{3}

(d) 256 m^{3}

**Answer 21. **

Given, lateral surface area of a cube = 256 m^{2}

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)^{3 }= 8 x 8 x 8 = 512 m^{3}

Hence, the volume of the cube is 512 m^{3}.

**Option (a) 512 m ^{3 }^{ }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 :**

**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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