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Q23.

Expert-verified

Geometry

Found in: Page 486

Book edition Student Edition

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

Pages 227 pages

ISBN 9780395977279

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Short Answer

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.

$O M = 3, O A =6 and B C = 6 \sqrt{3}$
$Lateral Area = 45 \sqrt{3} and Volume = 36 \sqrt{3}$

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

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 $Δ V O M$ for the length of the sideOM .

$\begin{matrix} {(V M)}^{2} = {(V O)}^{2} + {(O M)}^{2} \\ l^{2} = h^{2} + {(O M)}^{2} \\ {(O M)}^{2} = 5^{2} - 4^{2} \\ {(O M)}^{2} = 25 - 16 \end{matrix}$

Further simplify,

$\begin{matrix} {(O M)}^{2} = 25 - 16 \\ O M = \sqrt{9} \\ O M = 3 \end{matrix}$

So, the length of the side $O M$ 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.

$\begin{matrix} A O = 2 (O M) \\ = 2 (3) \\ = 6 \end{matrix}$

So, the length of the side $A O$ 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:

$\begin{matrix} A M = \frac{\sqrt{3}}{2} (B C) \\ A O + O M = \frac{\sqrt{3}}{2} (B C) \\ 6 + 3 = \frac{\sqrt{3}}{2} (B C) \\ 9 = \frac{\sqrt{3}}{2} (B C) \end{matrix}$

Further simplify,

$\begin{matrix} 9 = \frac{\sqrt{3}}{2} (B C) \\ B C = 9 \times \frac{2}{\sqrt{3}} \\ B C = 6 \sqrt{3} \end{matrix}$

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.

$A = \frac{1}{2} p l$

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

$\begin{matrix} p = 3 (B C) \\ = 3 (6 \sqrt{3}) \\ = 18 \sqrt{3} \end{matrix}$

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.

$\begin{matrix} A = \frac{1}{2} p l \\ = \frac{1}{2} (18 \sqrt{3}) (5) \\ = 45 \sqrt{3} \end{matrix}$

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.

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

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

The area of the equilateral triangle is:

$\begin{matrix} B = \frac{\sqrt{3}}{4} {(B C)}^{2} \\ = \frac{\sqrt{3}}{4} {(6 \sqrt{3})}^{2} \\ = 27 \sqrt{3} \end{matrix}$

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.

$\begin{matrix} V = \frac{1}{3} B h \\ = \frac{1}{3} (27 \sqrt{3}) (4) \\ = 36 \sqrt{3} \end{matrix}$

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

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