Chapter 10.7, Problem 20PSB

To determine

Find the perimeter of ABCD.

Answer to Problem 20PSB

  $6+6\sqrt{3}$

Explanation of Solution

Given information:

ABCD is a kite, with $\overline{AB}\cong \overline{BC}$ , $\overline{AD}\cong \overline{CD}$ , and $m\angle B=120$ . The radius of the circle is 3.

  

Calculation:

Consider the given figure,

  

  $\overline{AB}\cong \overline{BC}$ , $\overline{AD}\cong \overline{CD}$ , $m\angle B=120$ (given)

Two congruent chords correspond congruent arc,

  $mA\stackrel{⌢}{B}=mB\stackrel{⌢}{C},mA\stackrel{⌢}{D}=mC\stackrel{⌢}{D}$

The measure of the angle with vertex on the circle is half the measure of the intercepted arc,

  $\begin{array}{l}m\angle ABC=120°=\frac{mA\stackrel{⌢}{D}+mC\stackrel{⌢}{D}}{2}=\frac{2mA\stackrel{⌢}{D}}{2}\\ mA\stackrel{⌢}{D}=mC\stackrel{⌢}{D}=120°\\ mA\stackrel{⌢}{B}+mB\stackrel{⌢}{C}+mA\stackrel{⌢}{D}+mC\stackrel{⌢}{D}=360°\\ putvalues\\ \left(2\right)120°+2mA\stackrel{⌢}{B}=360°\\ 2mA\stackrel{⌢}{B}=360°-240°\\ 2mA\stackrel{⌢}{B}=120°\\ mA\stackrel{⌢}{B}=60°\end{array}$

Now again the measure of the angle with vertex on the circle is half the measure of the intercepted arc,

  $\begin{array}{l}m\angle ADB=\frac{mA\stackrel{⌢}{B}}{2}=\frac{60°}{2}=30°\\ m\angle BAD=\frac{mB\stackrel{⌢}{C}+mC\stackrel{⌢}{D}}{2}=\frac{60°+120°}{2}=90°\\ BD=6\\ AB=\frac{BC}{2}=\frac{6}{2}=3\\ A{B}^2+A{D}^2=B{D}^2\\ 3^2+A{D}^2=6^2\\ A{D}^2=36-9=27\\ AD=3\sqrt{3}\\ 2AB+2AD=6+2.3\sqrt{3}=6+6\sqrt{3}\\ Perimeter=6+6\sqrt{3}\end{array}$

Hence, the perimeter of ABCD is $6+6\sqrt{3}$