Given a polygonal region $R$
in a plane $E$
and a point $V$
not in $E$, the **pyramid **with base $R$
and vertex $V$
is the union of all line segments $\overline{VN}$
such that $N$
is a point of $R$.

Most of the pyramids that are studied in high school are **regular pyramids**. These pyramids have the following characteristics:

- $1)$ The base is a regular polygon.
- $2)$ All lateral edges are congruent.
- $3)$ All lateral faces are congruent isosceles triangles.
- $4)$ The altitude meets the base at its center.

The altitude of a lateral face of a regular pyramid is the **slant height.** In a non-regular pyramid, slant height is not defined.

The **lateral **surface area of a regular pyramid is the sum of the areas of its lateral faces.

The general formula for the **lateral surface area** of a regular pyramid is $L.S.A.=\frac{1}{2}pl$
where $p$
represents the perimeter of the base and $l$
the slant height.

**Example 1:**

Find the lateral surface area of a regular pyramid with a triangular base if each edge of the base measures $8$ inches and the slant height is $5$ inches.

The perimeter of the base is the sum of the sides.

$p=3\left(8\right)=24$ inches

$L.S.A.=\frac{1}{2}\left(24\right)\left(5\right)=60\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\text{inches}}^{2}$

The **total surface area of a regular pyramid **is the sum of the areas of its lateral faces and its base.
The general formula for the **total surface area **of a regular pyramid is $T.S.A.=\frac{1}{2}pl+B$
where $p$
represents the perimeter of the base, $l$
the slant height and $B$
the area of the base.

**Example 2:**

Find the total surface area of a regular pyramid with a square base if each edge of the base measures $16$ inches, the slant height of a side is $17$ inches and the altitude is $15$ inches.

The perimeter of the base is $4s$ since it is a square.

$p=4\left(16\right)=64$ inches

The area of the base is ${s}^{2}$.

$B={16}^{2}=256\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\text{inches}}^{2}$

$\begin{array}{l}T.S.A.=\frac{1}{2}\left(64\right)\left(17\right)+256\\ =544+256\\ =800\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\text{inches}}^{2}\end{array}$

There is no formula for a surface area of a non-regular pyramid since slant height is not defined. To find the area, find the area of each face and the area of the base and add them.

The **volume** of a pyramid equals one-third the area of the base times the altitude (height) of the pyramid. $\left(V=\frac{1}{3}Bh\right)$
.

**Example 3:**

Find the volume of a regular square pyramid with base sides $10$ cm and altitude $12$ cm.

$\begin{array}{l}V=\frac{1}{3}Bh\\ V=\frac{1}{3}{\left(10\right)}^{2}\left(12\right)=400\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\text{cm}}^{2}\end{array}$