Question

If $l,b,h$are the length, breadth and height of a room, then area of 4 walls will be,
A. $2(l+b)h$
B. $(l+b)h$
C. $lbh$
D. $2(lb+bh+hl)h$

Answer 1

Hint:

If we look around and observe the walls of a room, we find that generally the walls are in the shape of a rectangle. The floor and the ceiling of the room are also of rectangular shape. We have a formula for finding the area of a rectangle.
Area of rectangle $=length\times breadth$
Apply this formula for the four walls separately and then add them to get the area of four walls.


Complete step-by-step solution -

Suppose ABFE, BCGF, CDHG and DAEH are the four walls whose area we need to find.
Given,
$l,bandh$ are the length, breadth and height of a room.
So,
Length of AB $=l$
Length of AD $=b$
Length of AE $=h$
For the rectangle ABFE,
Length of AB $=l$
Length of AE $=h$
So, Area of rectangle ABFE $=AB\times AE$
$=(l\times h)$………….(1)
For the rectangle BCGF,
Length of BC = length of AD $=b$
Length of BF = length of AE $=h$
So, Area of rectangular wall BCGF
$\begin{array}{rl}& =BC\times BF\\ & =(b\times h)...........\left(2\right)\end{array}$
For the rectangular wall CDHG,
Length of CD = length of AB $=l$
Length of CG = length of AE $=h$
So, Area of rectangular wall CDHG
$\begin{array}{rl}& =CD\times CG\\ & =lh...........\left(3\right)\end{array}$
For the rectangular wall DAEH,
Length of AD $=b$
Length of AE $=h$
So, Area of rectangular wall DAEH
 $\begin{array}{rl}& =AD\times AE\\ & =b\times h.........\left(4\right)\end{array}$
Now, the total area of four walls = Area of wall ABFE + Area of wall BCGF + Area of wall CDHG +Area of wall DAEH
So, adding equations (1), (2), (3) and (4), we get,
Total Area of four walls $=(l\times h+b\times h+l\times h+b\times h)$
$\begin{array}{rl}& =(lh+bh+lh+bh)\\ & =2lh+2bh\end{array}$
Area of four walls $=2lh+2bh$
Taking $2h$ common from both terms, we get,
Area of four walls $=2(l+b)h$
Hence, option (A) is correct.

Note: You can also do it quickly.
Look around in the room you are sitting to understand better. Observe that all four walls whose area we have to find have the $h$ (height of the room) as one side and other side will be length $\left(l\right)$ for two and breadth $\left(b\right)$ for the other two walls.
Also, observe the opposite walls being of the same size and shape and hence, have the same area too.
So,
Area of four walls $=2\left(lh\right)+2\left(bh\right)$
$=2(l+b)h$
(Taking $2h$ common from both terms).