Question

The surface area of the four walls of a hall having length 15m, width 10m and and height 6m is​

Answer 1

Given :

• Length of cuboid = 15 m

• Breadth of cuboid = 10 m

• Height of cuboid = 6 m

To Find :

• TSA of the four walls of the hall .

Solution:-

As we know that,

$\sf \: TSA \: of \: cuboid \: = 2(lb + bh + lh)$

$\sf \: TSA \: of \: cuboid \: = 2(15 \times 10 + 10 \times 6 + 15 \times 6)$

$\sf \: TSA \: of \: cuboid \: = 2(150 + 60 + 90)$

$\sf \: TSA \: of \: cuboid \: = 2(300)$

$\sf \: TSA\: of \: cuboid \: = 600 \:m^3$

Let's find TSA of 4 walls ,

$\sf \: TSA \: of \: 4 \: walls \: = 600 \times 4$

$\sf \: TSA \: of \: 4 \: walls \: = 2400 \:m^3$

Answer 2

Answer:

the surface area of the four for of a hall having length 15 M with 10 M height 6 m