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 :
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
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 cm2 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 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 cm2 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 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
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 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 cm2 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 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 cm2 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) 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 cm2 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 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
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)3 = 8 x 8 x 8 = 512 m3
Hence, the volume of the cube is 512 m3.
Option (a) 512 m3 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 :–
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