Question

A cuboid has a total surface area of $50{{m}^{2}}$and lateral surface area is $30{{m}^{2}}$ . Find the area of its base.

Answer 1

Hint: We have a cuboid whose total surface area is $50{{m}^{2}}$ and the lateral surface area is $30{{m}^{2}}$. We need to find the area of its base. Let us assume that the length, breadth, and height of the cuboid are l, b, and h respectively. As we know that the total surface area of cuboid is given by: $TSA=2\left( lb+bh+lh \right)$ and lateral surface area of cuboid is given by: $LSA=2\left( bh+lh \right)$ . Now, put the values of total surface area and lateral surface area in these formulas and get the value of $\left( l\times b \right)$ . The value of $\left( l\times b \right)$ is the area of its base.

Complete step by step answer:
We have a cuboid as shown in the figure:

We have:
Total surface area of cuboid:$TSA=50{{m}^{2}}$
Lateral surface area of cuboid: $LSA=30{{m}^{2}}$
As we know that:
$TSA=2\left( lb+bh+lh \right)$
$LSA=2\left( bh+lh \right)$
So, by putting the value of TSA and LSA, we get:
$\begin{align}
  & \Rightarrow 2\left( lb+bh+lh \right)=50{{m}^{2}}......(1) \\
 & \Rightarrow 2\left( bh+lh \right)=30{{m}^{2}}......(2) \\
\end{align}$
We can write equation (1) as:
$\Rightarrow 2\left( lb \right)+2\left( bh+lh \right)=50{{m}^{2}}......(3)$
Now, put the value of $2\left( bh+lh \right)$ from equation (2) in equation (3), we get:
$\begin{align}
  & \Rightarrow 2\left( lb \right)+30{{m}^{2}}=50{{m}^{2}} \\
 & \Rightarrow 2\left( lb \right)=20{{m}^{2}} \\
 & \Rightarrow \left( l\times b \right)=10{{m}^{2}} \\
\end{align}$

Hence, the area of the base is $10{{m}^{2}}$.

Note: The lateral surface of an object covers all of the sides of the object, excluding its base and top (when they exist). The lateral surface area is the area of the lateral surface. Whereas, the total surface area of an object is the area of all the faces of an object, which means that the total surface area includes the lateral surface area as well as the area of top and base of the object.