Billabong High International School,
Bhoomi Park, Bafhira Nagar, Malad West,Mumbai –400095
Email –info.malad@billabonghigh.com Contact : 02229673217 /18/19
Subject: Mathematics Chapter 14: Mensuration Grade: 8
Topical Helpdesk -12
OBJECTIVE TYPE QUESTIONS (1 MARK EACH)
1. iii. 10 cm
2. iii. 17.5 cm
3. i. 2640 m3
4. iv. 180 cm2
5. iii. 24 cm2
VERY SHORT ANSWER QUESTIONS (2 MARKS EACH)
6.
7.
8.
9. Solution:
Let length of parallel sides be (5x) cm and (3x) cm.
1
Area of trapezium = × (��� �� �������� �����) × ℎ���ℎ�
2
1
450 = 2 × (5� + 3�) × 12.5
450 = 50 x
X=9
5x = 45 cm; 3x = 27 cm
Billabong High International School,
Bhoomi Park, Bafhira Nagar, Malad West,Mumbai –400095
Email –info.malad@billabonghigh.com Contact : 02229673217 /18/19
Subject: Mathematics Chapter 14: Mensuration Grade: 8
Topical Helpdesk -12
10.
SHORT ANSWER QUESTIONS (3 MARKS EACH)
11. Solution:
Let ABCD be the given rhombus in which AB = 17 cm and
diagonal AC = 30 cm.
Let AC and BD intersect at O.
We know that the diagonal of a rhombus bisects each other at right angles.
12. Solution:
Total surface area of 1 cubical stool = 6a2 = 6 × 60 × 60 = 21,600 cm2
Total surface area of 6 cubes = 21600 × 6 = 129600 cm2
Quantity of paint required to paint 1200 cm2 = 1 can
1
Quantity of paint required to paint 1 cm2 =1200
129600
Quantity of paint required to paint 129600cm2 = 1200
= 108 cans
13.
14. Solution:
Let h₁ = height of cylinder 1
Let h₂ = height of cylinder 2
Billabong High International School,
Bhoomi Park, Bafhira Nagar, Malad West,Mumbai –400095
Email –info.malad@billabonghigh.com Contact : 02229673217 /18/19
Subject: Mathematics Chapter 14: Mensuration Grade: 8
Topical Helpdesk -12
Let r₁ = radius of first cylinder
Let r₂ = radius of second cylinder
r₁/r₂ = 1:6
r₁²/r₂² = (1/6)² = 1/36
If the volume of cylinders 1 and 2 are same then we can write:
πr₁²h₁ = πr₂²h₂
r₁²/r₂² = h₂/h₁
h₂/h₁ = r₁²/r₂² = 1/36
Therefore
h₁/h₂ = 36:1
The ratio of the heights of the two cylinders will be 36:1.
15.
LONG ANSWER QUESTIONS (4 MARKS EACH)
16. Solution:
Area of shaded region = Area of circle - Area of four triangles - Area of square
199.5 sq. cm
17. Solution:
r+h =37 m
2πr (r + h) = 1628
r= 7 m, h=30m
Volume = 4620 m3
Billabong High International School,
Bhoomi Park, Bafhira Nagar, Malad West,Mumbai –400095
Email –info.malad@billabonghigh.com Contact : 02229673217 /18/19
Subject: Mathematics Chapter 14: Mensuration Grade: 8
Topical Helpdesk -12
18. Solution:
19.
20. Solution:
Ratio between the dimensions of a cuboid = 1:2:3
Billabong High International School,
Bhoomi Park, Bafhira Nagar, Malad West,Mumbai –400095
Email –info.malad@billabonghigh.com Contact : 02229673217 /18/19
Subject: Mathematics Chapter 14: Mensuration Grade: 8
Topical Helpdesk -12
Total surface area of the cuboid = 88 m2
Let the dimensions of the cuboid be x , 2x and 3x respectively .
Then,
----------
2 × [(x × 2x) + (x × 3x) + (2x × 3x)] = 88
∴2x2 + 3x2 + 6x2 = 88/2
∴ 11x2 = 44
∴ x2 = 44/11
∴ x2 = 4
∴ x = √4
∴x=2m
∴ Height = x = 2m
∴ Breadth = 2x = ( 2 × 2) m = 4m
∴ Length = 3x = (3 × 2) m = 6m
