Question

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 $4m\times 3m$?

Answer 1

Step-1: Find the area of the lateral surface area of the shelter:

Given,

Length $\left(l\right)=4m$

Breadth $\left(b\right)=3m$

Height $\left(h\right)=2.5m$

Area of the lateral sides$=$( Perimeter of the base)$\times$Height

Perimeter of the base $=2\times (l+b)$

$$\begin{gathered} =2\times \left(4+3\right) \\ =14m \end{gathered}$$

$\therefore$Area of the lateral sides$=$$14\times \left(2.5\right)$

$=35m^2$

Step-2: Find the area of the top surface of the shelter :

Area of the top $=l\times b$

$$\begin{gathered} =4\times 3 \\ =12m^2 \end{gathered}$$

Step-3: Find the area the tarpaulin required to make the shelter:

Area of tarpaulin required $=$ Area of the lateral surface$+$ Area of the top

$$\begin{gathered} =35+12 \\ =47m^2 \end{gathered}$$

Hence, $47m^2$ tarpaulin is required to make the shelter