# Parveen 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 2.5 m, with base dimensions 4 m × 3 m?

**Solution:**

Since the shelter is a box-like structure (cuboid) with a tarpaulin that covers all four sides and the top of the car, the surface area of the shelter is the sum of the lateral surface area of the cuboid and area of the top.

Lateral surface area of cuboid = 2(l + b)h

Then the area of the tarpaulin required to make the shelter = lb + 2(l + b)h

Let the length, breadth and height of the shelter be l, b and h respectively.

Length, l = 4 m

Breadth, b = 3 m

Height, h = 2.5 m

The area of the tarpaulin required to make the shelter = lb + 2(l + b)h

= (4 m × 3 m) + 2 × (4 m + 3 m) × 2.5 m

= 12 m² + 2 × 7 m × 2.5 m

= 12 m² + 35 m²

= 47 m²

Hence, 47 m² of tarpaulin will be required.

**Video Solution:**

## Parveen 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 2.5 m, with base dimensions 4 m × 3 m?

### Class 9 Maths NCERT Solutions - Chapter 13 Exercise 13.1 Question 8:

**Summary:**

It is given that Parveen 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. We have found that 47 m² of tarpaulin will be required to make the shelter of height 2.5 m, with base dimensions 4 m × 3 m.