Chapter 9.4, Problem 25WE

To determine

To Find : The perimeter of ΔABC.

Answer to Problem 25WE

The perimeter of ΔABC is $18\sqrt{3}$ .

Explanation of Solution

Given information:

Three points A,B, C formed an equilateral triangle ΔABC.

  

O is the center point.

Radius of the circle is 6.

Consider the figure ;

  

In the above figure circle contains three points A, B and C these points formed an equilateral triangle ΔABC which is inscribed in circle.

Radius of the circle is $OB=OC=OA=6$

According to equilateral triangle, all three sides of triangle are equal in length and angle form by its all three vertex is of 60°.

This means $\angle A=\angle B=\angle C={60}^{\circ }\text{and }AB=AC=BC$

In triangle $\text{Δ}BOC$ angle $\angle OBD=\angle OCD={30}^{\circ }$

Because radius OB and OC bisects $\angle ABC={60}^{\circ }$ and $\angle ACB={60}^{\circ }$ into equal parts.

It is known that sum of angles of triangle is equal to 180° so in ΔBOC

  $\begin{array}{r}\hfill \angle BOC+\angle OBD+\angle OCD={180}^{\circ }\\ \hfill \angle BOC+{30}^{\circ }+{30}^{\circ }={180}^{\circ }\\ \hfill \angle BOC+{60}^{\circ }={180}^{\circ }\end{array}$

Subtract 60° from both sides of equation $\angle BOC+60°=180°$

  $\begin{array}{cc}\angle BOC+\cancel{{60}^{\circ }}-\cancel{{60}^{\circ }}& ={180}^{\circ }-{60}^{\circ }\\ \angle BOC& ={180}^{\circ }-{60}^{\circ }\\ \angle BOC& ={120}^{\circ }\end{array}$

And line segment OD bisects angle $\angle BOC=120°$ into equal parts.

That means $\angle BOD=\angle COD=60°$

Use trigonometric identity:

NA in form right triangle ΔODC

Here perpendicular $CD=x,hypotenuseOC=6$ and $\theta =60°$

  $sin60°=\frac{x}{6}$

Here measure of $sin60°=\frac{\sqrt{3}}{2}$

  $\begin{array}{l}sin60°=\frac{x}{6}\\ \frac{\sqrt{3}}{2}=\frac{x}{6}\end{array}$

Multiply both sides of equation $\frac{\sqrt{3}}{2}=\frac{x}{6}$ by 2 and 6;

  $\begin{array}{l}\frac{\sqrt{3}}{\cancel{2}}\times \cancel{2}\times 6=\frac{x}{\cancel{6}}\times 2\times \cancel{6}\\ 6\sqrt{3}=2x\end{array}$

Divide both sides of equation $6\sqrt{3}=2x$ by 2

  $\begin{array}{l}\frac{\cancel{6}\sqrt{3}}{\cancel{2}}=\frac{\cancel{2}x}{\cancel{2}}\\ 3\sqrt{3}=x\\ x=3\sqrt{3}\end{array}$

If $CD=3\sqrt{3}$ then also side $BD=3\sqrt{3}$ because $BD=CD$ as D is the midpoint of side BC.

Also

  $\begin{array}{l}\begin{array}{l}BC=CD+BD\hfill \\ BC=3\sqrt{3}+3\sqrt{3}\hfill \end{array}\\ BC=6\sqrt{3}\end{array}$

Also if $BC=6\sqrt{3}$ then $AB=AC=NA$ because all three sides of equilateral triangle are equal in length.

To find perimeter of ΔABC

Perimeter=3a where a=NA

Perimeter $\begin{array}{l}\Delta ABC=3\times 6\sqrt{3}\hfill \\ =18\sqrt{3}\hfill \end{array}$

Therefore, perimeter of ΔABC is $18\sqrt{3}$ .