Find the following for the regular triangular pyramid.

If VA = 5 and h = 3, find the slant height, the lateral area and the volume.

The length ofVA is equal to5 and length of h is equal to3 in the given pyramid:

Use the Pythagoras theorem in$\Delta AOV$ for the length of sideAO .

$\begin{array}{c}{\left(AV\right)}^2={\left(AO\right)}^2+{\left(VO\right)}^2\\ 5^2={\left(AO\right)}^2+3^2\\ AO=\sqrt{25-9}\\ AO=4\end{array}$

So, the length of the sideAO is equal to4 .

The length of the sideOM is half the length of the sideAO . So, the length of the sideOM is equal to2 .

The length of the total sideAM is the sum of the lengths of sidesAO and OM. So, the length of theAM side is equal to$4+2=6$ .

AsAM 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.

$\begin{array}{c}BC=\frac{2}{\sqrt{3}}\left(AM\right)\\ =\frac{2}{\sqrt{3}}\left(6\right)\\ =4\sqrt{3}\end{array}$

So, the length of each side of equilateral triangle is equal to$4\sqrt{3}$ .

The length of the side MC is the half of the length of the side BC as M is the midpoint of the side. So, the length of the sideMC is equal to$2\sqrt{3}$ .

The length of each lateral edge of pyramid is equal. So, the length of side VC is also equal to5 .

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

$\begin{array}{c}{\left(VC\right)}^2={\left(VM\right)}^2+{\left(MC\right)}^2\\ 5^2=l^2+{\left(2\sqrt{3}\right)}^2\\ l=\sqrt{25-12}\\ l=\sqrt{13}\end{array}$

So, the slant height l of the pyramid is equal to$\sqrt{13}$ .

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 to $2\sqrt{3}$. So, the perimeter of the equilateral triangle is equal to$3\left(4\sqrt{3}\right)=12\sqrt{3}$ .

Substitute$6\sqrt{3}$ for P and$\sqrt{13}$ for l in the formula for lateral area.

$\begin{array}{c}\text{A}=\frac{1}{2}pl\\ =\frac{1}{2}\left(12\sqrt{3}\right)\left(\sqrt{13}\right)\\ =6\sqrt{39}\end{array}$

So, the lateral area of the pyramid is equal to$6\sqrt{39}$ .

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 side$4\sqrt{3}$ .

$\begin{array}{c}\text{B}=\frac{\sqrt{3}}{4}{a}^2\\ =\frac{\sqrt{3}}{4}{\left(4\sqrt{3}\right)}^2\\ =\frac{\sqrt{3}}{4}\left(48\right)\\ =12\sqrt{3}\end{array}$

Substitute$12\sqrt{3}$ for B and 3 for h in the formula for volume of pyramid.

$\begin{array}{c}\text{V}=\frac{1}{3}\text{B}h\\ =\frac{1}{3}\left(12\sqrt{3}\right)3\\ =12\sqrt{3}\end{array}$

So, the volume of the pyramid is equal to$12\sqrt{3}$ .