Exercise 13.1 question 8

CBSE, JEE, NEET, CUET
Question Bank, Mock Tests, Exam Papers
NCERT Solutions, Sample Papers, Notes, Videos
Posted by Rajat Agarwal Agarwal 5 years, 1 month ago
- 1 answers
Related Questions
Posted by Alvin Thomas 3 months, 2 weeks ago
- 0 answers
Posted by Akhilesh Patidar 1 year, 5 months ago
- 0 answers
Posted by Savitha Savitha 1 year, 5 months ago
- 0 answers
Posted by Duruvan Sivan 6 months, 2 weeks ago
- 0 answers
Posted by Yash Pandey 6 months, 2 weeks ago
- 0 answers
Posted by Sheikh Alfaz 2 months ago
- 0 answers

myCBSEguide
Trusted by 1 Crore+ Students

Test Generator
Create papers online. It's FREE.

CUET Mock Tests
75,000+ questions to practice only on myCBSEguide app
myCBSEguide
Gaurav Seth 5 years, 1 month ago
8. Praveen wanted to make a temporary shelter for her car, by making a box – like structure with tarpaulin that covers all the four sides and the top of the car (with the front face as a flap which can be rolled up). Assuming that the stitching margins are very small, and therefore negligible, how much tarpaulin would be required to make the shelter of height 2.5m, with base dimensions 4m×3m?
Solution:
Let l, b and h be the length, breadth and height of the shelter.
Given:
l = 4m
b = 3m
h = 2.5m
Tarpaulin will be required for the top and four wall sides of the shelter.
Using formula, Area of tarpaulin required = 2(lh+bh)+lb
On putting the values of l, b and h, we get
= [2(4×2.5+3×2.5)+4×3] m2
= [2(10+7.5)+12]m2
= 47m2
Therefore, 47 m2 tarpaulin will be required.
1Thank You