The dimensions of a rectangular room (cuboidal) are \[l,{\text{ }}b{\text{ }}and{\text{ }}h\]. What is the area of its four walls?
Answer
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Hint: The cuboid is the box-like three dimensional shaped can be either solid or hollow and has six rectangular faces. The faces opposite to each other are equal. First visualize the rectangular room and accordingly as per the required area of four walls find it using the formula of the lateral surface area$2h(l + b)$. We can also subtract the area of the roof and floor from the total surface area of the cuboid room.
Complete step-by-step solution:
Given: Dimensions of the cuboidal room are-
Length $ = l$
Breadth $ = b$ and
Height $ = h$
Area of four walls = Lateral surface area of cuboid =$2h(l + b)$
Where h is height of the cuboid,
l is length of the cuboid and
b is the breadth.
Also, we can find\[area{\text{ }}of{\text{ }}four{\text{ }}walls{\text{ }} = {\text{ }}total{\text{ }}surface{\text{ }}area{\text{ }}of{\text{ }}the{\text{ }}cuboid{\text{ }} - 2 \times {\text{ }}area{\text{ }}of{\text{ }}floor\]
As [Area \[\]of floor and roof is the same from the properties of cuboid].
$ \Rightarrow 2 \times (lb + bh + lh) - 2 \times (lb)$
\[\left[ {area{\text{ }}of{\text{ }}rectangle = {\text{ }}l \times b{\text{ }}as,{\text{ }}room{\text{ }}has{\text{ }}one{\text{ }}rectangular{\text{ }}floor{\text{ }}and{\text{ }}one{\text{ }}roof{\text{ }}hence{\text{ }}we{\text{ }}will{\text{ }}subtract{\text{ }}2 \times l \times b} \right]\]
\[
\Rightarrow 2lb + 2bh + 2lh - 2lb \\
\Rightarrow 2bh + 2lh \\
\Rightarrow 2h(l + b) \\
\]
Therefore, the required answer – the area of the four wall of the room is $2h(l + b)$
Note:In such types of problems we should first understand the exact shape then we can use properties to solve the problem. It is not needed to memorise each formula to find the required value rather we can split the whole 3d shape in 2d shapes and relate them to find the required solution. Like in this case, we have calculated the area of four walls by using total surface area so we split the whole cuboid into four walls and two rectangular surfaces (roof and floor).
Complete step-by-step solution:
Given: Dimensions of the cuboidal room are-
Length $ = l$
Breadth $ = b$ and
Height $ = h$
Area of four walls = Lateral surface area of cuboid =$2h(l + b)$
Where h is height of the cuboid,
l is length of the cuboid and
b is the breadth.
Also, we can find\[area{\text{ }}of{\text{ }}four{\text{ }}walls{\text{ }} = {\text{ }}total{\text{ }}surface{\text{ }}area{\text{ }}of{\text{ }}the{\text{ }}cuboid{\text{ }} - 2 \times {\text{ }}area{\text{ }}of{\text{ }}floor\]
As [Area \[\]of floor and roof is the same from the properties of cuboid].
$ \Rightarrow 2 \times (lb + bh + lh) - 2 \times (lb)$
\[\left[ {area{\text{ }}of{\text{ }}rectangle = {\text{ }}l \times b{\text{ }}as,{\text{ }}room{\text{ }}has{\text{ }}one{\text{ }}rectangular{\text{ }}floor{\text{ }}and{\text{ }}one{\text{ }}roof{\text{ }}hence{\text{ }}we{\text{ }}will{\text{ }}subtract{\text{ }}2 \times l \times b} \right]\]
\[
\Rightarrow 2lb + 2bh + 2lh - 2lb \\
\Rightarrow 2bh + 2lh \\
\Rightarrow 2h(l + b) \\
\]
Therefore, the required answer – the area of the four wall of the room is $2h(l + b)$
Note:In such types of problems we should first understand the exact shape then we can use properties to solve the problem. It is not needed to memorise each formula to find the required value rather we can split the whole 3d shape in 2d shapes and relate them to find the required solution. Like in this case, we have calculated the area of four walls by using total surface area so we split the whole cuboid into four walls and two rectangular surfaces (roof and floor).
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