
If $l,b,h$are the length, breadth and height of a room, then area of 4 walls will be,
A. $2\left( l+b \right)h$
B. $\left( l+b \right)h$
C. $lbh$
D. $2\left( lb+bh+hl \right)h$
Answer
588.9k+ views
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,
$l,b\ and\ h$ 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 $=AB\times AE$
$=\left( l\times h \right)$………….(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
$\begin{align}
& =BC\times BF \\
& =\left( b\times h \right)...........\left( 2 \right) \\
\end{align}$
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
$\begin{align}
& =CD\times CG \\
& =lh...........\left( 3 \right) \\
\end{align}$
For the rectangular wall DAEH,
Length of AD $=b$
Length of AE $=h$
So, Area of rectangular wall DAEH
$\begin{align}
& =AD\times AE \\
& =b\times h.........\left( 4 \right) \\
\end{align}$
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 $=\left( l\times h+b\times h+l\times h+b\times h \right)$
$\begin{align}
& =\left( lh+bh+lh+bh \right) \\
& =2lh+2bh \\
\end{align}$
Area of four walls $=2lh+2bh$
Taking $2h$ common from both terms, we get,
Area of four walls $=2\left( l+b \right)h$
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 $=2\left( lh \right)+2\left( bh \right)$
$=2\left( l+b \right)h$
(Taking $2h$ common from both terms).
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,
$l,b\ and\ h$ 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 $=AB\times AE$
$=\left( l\times h \right)$………….(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
$\begin{align}
& =BC\times BF \\
& =\left( b\times h \right)...........\left( 2 \right) \\
\end{align}$
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
$\begin{align}
& =CD\times CG \\
& =lh...........\left( 3 \right) \\
\end{align}$
For the rectangular wall DAEH,
Length of AD $=b$
Length of AE $=h$
So, Area of rectangular wall DAEH
$\begin{align}
& =AD\times AE \\
& =b\times h.........\left( 4 \right) \\
\end{align}$
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 $=\left( l\times h+b\times h+l\times h+b\times h \right)$
$\begin{align}
& =\left( lh+bh+lh+bh \right) \\
& =2lh+2bh \\
\end{align}$
Area of four walls $=2lh+2bh$
Taking $2h$ common from both terms, we get,
Area of four walls $=2\left( l+b \right)h$
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 $=2\left( lh \right)+2\left( bh \right)$
$=2\left( l+b \right)h$
(Taking $2h$ common from both terms).
Recently Updated Pages
The height of a solid metal cylinder is 20cm Its r-class-10-maths-ICSE

If a train crossed a pole at a speed of 60kmhr in 30 class 10 physics CBSE

Name the Writs that the High Courts are empowered to class 10 social science CBSE

A tower is 5sqrt 3 meter high Find the angle of el-class-10-maths-CBSE

Immediate cause of variations of A Mutations B Environmental class 10 biology CBSE

A rectangular container whose base is a square of side class 10 maths CBSE

Trending doubts
Why is there a time difference of about 5 hours between class 10 social science CBSE

Why is Sardar Vallabhbhai Patel called the Iron man class 10 social science CBSE

Tropical deciduous trees shed their leaves in the dry class 10 social science CBSE

Write a letter to the principal requesting him to grant class 10 english CBSE

The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths

Write an application to the principal requesting five class 10 english CBSE

