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Q23.

Expert-verifiedFound in: Page 486

Book edition
Student Edition

Author(s)
Ray C. Jurgensen, Richard G. Brown, John W. Jurgensen

Pages
227 pages

ISBN
9780395977279

**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.**

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

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

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 .

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.

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$6\sqrt{3}$ .

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

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.

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$45\sqrt{3}$ .

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$36\sqrt{3}$ .

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