Question

A circle has two equal chords AB and AC, chord AD bisects BC in E. If $AC=12$ and $AE=8cm$, then the measure of AD is ?

• A. $24$ cm
• B. $18.5$ cm
• C. $18$ cm
• D. $19$ cm

Answer 1

Answer: (C)

$18$ cm

Given, $CE=BE$ and $AC=AB=12cm$ and $AE=8$

We know, $\angle AEC=\angle AEB={90}^o$

$\angle CAE=\angle BAE=\theta \cos \theta =\frac{8}{12}=\frac{2}{3},\sin \theta =\frac{\sqrt{5}}{3}\sin \theta =\frac{\sqrt{5}}{3}=\frac{CE}{12}\Rightarrow CE=4\sqrt{5}CB=2CE=8\sqrt{5}\sqrt{{\left(OB\right)}^2-{\left(BE\right)}^2}=\left(OA\right)-\left(EA\right)r^2-{\left(4\sqrt{5}\right)}^2={(r-8)}^2{r}^2-80=r^2+64-16rr=9cmOA=2r=18cm$