Question

Find the lateral and total surface area of the following pyramids.
(a) Square-based pyramid with base $6$cm and slant height $14$cm;
(b) Triangular-based pyramid with base $12$cm and slant height $20$cm.

Answer 1

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

$P=4\times 6=24$ 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 $P=24$ cm and the slant height is $l=14$ cm, therefore, the lateral surface area is:

$LSA=\frac{1}{2}Pl=\frac{1}{2}\times 24\times 14=168$ cm${}^2$

Now, the area of the base $B=s^2$ with $s=6$ cm is:

$B=s^2=6^2=36$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 $LSA=\frac{1}{2}Pl=168$cm${}^2$ and area of the base is $B=36$ cm${}^2$, therefore, the total surface area is:

$TSA=\frac{1}{2}Pl+B=168+36=204$cm${}^2$

Hence, lateral surface area of the pyramid is $168$cm${}^2$ and total surface area is $204$cm${}^2$.

(b) The perimeter of the base is $P=4s$, since it is a triangle, therefore,

$P=3\times 12=36$ 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 $P=36$ cm and the slant height is $l=20$ cm, therefore, the lateral surface area is:

$LSA=\frac{1}{2}Pl=\frac{1}{2}\times 36\times 20=360$ cm${}^2$

Now, the area of the base $B=\frac{\sqrt{3}}{4}{s}^2$ with $s=12$ cm is:

$B=\frac{\sqrt{3}}{4}{s}^2=\frac{\sqrt{3}}{4}\times 12\times 12=36\sqrt{3}$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 $LSA=\frac{1}{2}Pl=360$cm${}^2$ and area of the base is $B=36\sqrt{3}$ cm${}^2$, therefore, the total surface area is:

$TSA=\frac{1}{2}Pl+B=360+36\sqrt{3}=36(10+\sqrt{3})$cm${}^2$

Hence, lateral surface area of the pyramid is $360$cm${}^2$ and total surface area is $36(10+\sqrt{3})$cm${}^2$.