Question

The roof of a temple is in the shape of a square based pyramid. The edge of the base is $7.5m$and the slant height is $7m$. Find the area of metal sheet required to cover the top of the roof.

Answer 1

Hint: First, we will see what is perimeter.
Perimeter is the total length of the sides in a given two-dimensional or a three-dimensional shape. Where the area of the triangle is the total region enclosed by three sides of any triangle.
In this problem, we are going to use the formula of the perimeter to find the area of the metal sheet which is required to cover the top of the given roof.
Formula used: Perimeter of the given as $P = 4s$where P is the given perimeter and s is the edge of the given base of the temple.
The general formula for the lateral surface is $LSA = \dfrac{1}{2}Pl$ the area of the given pyramid.

Complete step-by-step solution:
First we will see the given things from the question are, the roof of the temple is in the shape of the square-based pyramid where the edge of the base is $7.5m$and the slant height is$7m$.
Let us know the formula for the perimeter $P = 4s$, where s is the base of the edge which is$7.5m$.
Hence, we get the perimeter as $P = 4 \times 7.5$(multiple of four times into the base)
Further solving we get the perimeter $P = 30m$.
Also, the formula for the surface area of the given pyramid is given by $LSA = \dfrac{1}{2}Pl$ where P is the perimeter and l is the height of the given slant.
Hence substituting the know values, we get $LSA = \dfrac{1}{2}Pl \Rightarrow \dfrac{1}{2} \times 30 \times 7$(where P is the perimeter and the values are thirty, l is the height and the value is seven)
Thus, solving this we get $LSA = \dfrac{1}{2} \times 30 \times 7 \Rightarrow 105{m^2}$ (meter of the perimeter and meter of the height also multiplied)
Therefore, the area of the required metal sheet to cover the distance of the top of the roof is $105{m^2}$.

Note: Since $LSA = \dfrac{1}{2}Pl$ which is the lateral surface area formula equals half of the given perimeter and slant height.
The area of the semi perimeter can also able to calculated using the same formula $\dfrac{1}{2}P = 4s$;
Which is the half of the given perimeter being semi perimeter.