Q23. - Geometry

Question

Find the following for the regular triangular pyramid.

a.If h = 4 and l = 5, find OM, OA and BC.

b.Find the lateral area and the volume.

Step-by-Step Solution

Answer

1. $OM=3\text{,\hspace{0.17em}}OA\text{=6\hspace{0.17em} and }BC=6\sqrt{3}$
   
2. $\text{Lateral\hspace{0.17em}\hspace{0.17em}Area}=45\sqrt{3}\text{ and Volume}=36\sqrt{3}$
   

Step 1. Given information.

The length of h is equal to 4 and the length of l is equal to5 in the given pyramid:

Step 2. Determine OM .

Use the Pythagoras theorem in$\Delta VOM$ for the length of the sideOM .

 

 $\begin{array}{c}{\left(VM\right)}^2={\left(VO\right)}^2+{\left(OM\right)}^2\\ l^2=h^2+{\left(OM\right)}^2\\ {\left(OM\right)}^2=5^2-4^2\\ {\left(OM\right)}^2=25-16\end{array}$

Further simplify,

$\begin{array}{c}{\left(OM\right)}^2=25-16\\ OM=\sqrt{9}\\ OM=3\end{array}$

 

So, the length of the side$OM$ is equal to3 .

Step 3. Determine AO .

The length of the sideAO is always the double the length of the sideOM in an equilateral triangle.

 $\begin{array}{c}AO=2\left(OM\right)\\ =2\left(3\right)\\ =6\end{array}$

 

So, the length of the side$AO$ is equal to 6.

Step 4. Determine BC .

The sideAM is the altitude of the equilateral triangleABC with side BC. The relation between both is given below:

$\begin{array}{c}AM=\frac{\sqrt{3}}{2}\left(BC\right)\\ AO+OM=\frac{\sqrt{3}}{2}\left(BC\right)\\ 6+3=\frac{\sqrt{3}}{2}\left(BC\right)\\ 9=\frac{\sqrt{3}}{2}\left(BC\right)\end{array}$

 

Further simplify,

$\begin{array}{c}9=\frac{\sqrt{3}}{2}\left(BC\right)\\ BC=9\times \frac{2}{\sqrt{3}}\\ BC=6\sqrt{3}\end{array}$

 

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

Step 1. Given information.

The length of h is equal to4 and the length of l is equal to 5in the given pyramid:

Step 2. Determine l .

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

 $\text{A}=\frac{1}{2}pl$

 

The perimeter of the base is the three times the sum of the length of sideBC of equilateral triangleABC .

 $\begin{array}{c}p=3\left(BC\right)\\ =3\left(6\sqrt{3}\right)\\ =18\sqrt{3}\end{array}$

 

As found in part(a), the length l is equal to 5.

Step 3. Determine lateral area.

Substitute5 for l and$18\sqrt{3}$ for in the formula for lateral area.

 $\begin{array}{c}\text{A}=\frac{1}{2}pl\\ =\frac{1}{2}\left(18\sqrt{3}\right)\left(5\right)\\ =45\sqrt{3}\end{array}$

So, the lateral area of the pyramid is equal to$45\sqrt{3}$ .

Step 4. Determine the volume.

The formula for the volume of the pyramid is equal to the one third of the product of area of base and the length h.

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

 

The base of the pyramid is equilateral triangle with side equal to$6\sqrt{3}$ .

 

The area of the equilateral triangle is:

 $\begin{array}{c}\text{B}=\frac{\sqrt{3}}{4}{\left(BC\right)}^2\\ =\frac{\sqrt{3}}{4}{\left(6\sqrt{3}\right)}^2\\ =27\sqrt{3}\end{array}$

 

As found in part(a), the length h is equal to 4.

 

Substitute4 for hand $27\sqrt{3}$for B in the formula for volume.

$\begin{array}{c}\text{V}=\frac{1}{3}\text{B}h\\ =\frac{1}{3}\left(27\sqrt{3}\right)\left(4\right)\\ =36\sqrt{3}\end{array}$

 

So, the volume of the pyramid is equal to$36\sqrt{3}$ .