Question

The base and height of a square-based pyramid is $10$cm and $12$cm respectively. Find its total surface area.

Answer 1

We first calculate slant height $L$ of the pyramid with base $s=10$ cm and height $12$ cm:

$L^2=H^2+{\left(\frac{s}{2}\right)}^2={12}^2+{\left(\frac{10}{2}\right)}^2={12}^2+5^2=144+25=169\Rightarrow L=\sqrt{169}=13$

The perimeter of the base is $P=4s$, since it is a square, therefore,

$P=4\times 10=40$ cm

The general formula for the lateral surface area of a regular pyramid is $LSA=\frac{1}{2}Pl$ where $P$ represents the perimeter of the base and $l$ is the slant height.

Since the perimeter of the pyramid is $P=40$ cm and the slant height is $l=13$ cm, therefore, the lateral surface area is:

$LSA=\frac{1}{2}Pl=\frac{1}{2}\times 40\times 13=260$ cm${}^2$

Now, the area of the base $B=s^2$ with $s=10$ cm is:

$B=s^2={10}^2=100$ cm${}^2$

The general formula for the total surface area of a regular pyramid is $TSA=\frac{1}{2}Pl+B$ where $P$ represents the perimeter of the base, $l$ is the slant height and $B$ is the area of the base.

Since $LSA=\frac{1}{2}Pl=260$ cm${}^2$ and area of the base is $B=100$ cm${}^2$, therefore, the total surface area is:

$TSA=\frac{1}{2}Pl+B=260+100=360$ cm${}^2$

Hence, total surface area of the pyramid is $360$ cm${}^2$.