Find the following for the regular triangular pyramid.

If AB = 12 and VA = 10, find the lateral area and the volume.

The length of VA is equal to10 and length of ABis equal to12 in the given pyramid:

The length of each base edge of pyramid is equal. So, the length of the sideBC is equal to the length of the sideAB equal to12 .

The length of the sideBM is the half of the length of the sideBC asM is the midpoint of the side. So, the length of the sideBM is equal to6 .

The length of each lateral side of pyramid is also equal. So, the length of the sideVB is equal to .

Use the Pythagoras theorem$\Delta VBM$ in for the length of side AO .

$\begin{array}{c}{\left(VB\right)}^2={\left(BM\right)}^2+{\left(VM\right)}^2\\ {10}^2={\left(6\right)}^2+l^2\\ l=\sqrt{100-36}\\ l=8\end{array}$

So, the slant height of the pyramid is equal to8 .

As AM is the altitude of the equilateral triangle and the relationship between the side of equilateral triangle and altitude is given below simplify it for the length of side of equilateral triangle.

$\begin{array}{c}AM=\frac{\sqrt{3}}{2}\left(BC\right)\\ =\frac{\sqrt{3}}{2}\left(12\right)\\ =6\sqrt{3}\end{array}$

So, the length of side AM is equal to$6\sqrt{3}$ .

The length of the side OM is one third of the length of side AM. The length of the sideOM is equal to $2\sqrt{3}$.

Use the Pythagoras theorem in$\Delta VOM$ for the height of pyramid.

$\begin{array}{c}{\left(VM\right)}^2={\left(VO\right)}^2+{\left(OM\right)}^2\\ 8^2=h^2+{\left(2\sqrt{3}\right)}^2\\ h=\sqrt{64-12}\\ h=\sqrt{52}\end{array}$

So, the height h of the pyramid is equal to$\sqrt{52}$ .

The formula for the lateral area of the pyramid is half the product of perimeter of the base and slant height.

$\text{A}=\frac{1}{2}pl$

The length of each side of equilateral triangle is equal to12 . So, the perimeter of the equilateral triangle is equal to$3\left(12\right)=36$ .

Substitute36 for p and 8 for l in the formula for lateral area.

$\begin{array}{c}\text{A}=\frac{1}{2}pl\\ =\frac{1}{2}\left(36\right)\left(8\right)\\ =144\end{array}$

The formula for the volume of pyramid is the one third of the product of base area and height.

$\text{V}=\frac{1}{3}\text{B}h$

Simplify the area of equilateral triangle with side12 .

$\begin{array}{c}\text{B}=\frac{\sqrt{3}}{4}{a}^2\\ =\frac{\sqrt{3}}{4}{\left(12\right)}^2\\ =\frac{\sqrt{3}}{4}\left(144\right)\\ =36\sqrt{3}\end{array}$

Substitute$36\sqrt{3}$for B and$\sqrt{52}$for h in the formula for volume of pyramid.

$\begin{array}{c}\text{V}=\frac{1}{3}\text{B}h\\ =\frac{1}{3}\left(36\sqrt{3}\right)\left(\sqrt{52}\right)\\ =12\sqrt{3}\times 2\sqrt{13}\\ =24\sqrt{39}\end{array}$