Question

In the figure triangle ABC is inscribed in a circle with cen-tre O. If CA$=$CB$=15$cm and AB$=6$cm, find the radius of the circle.

Answer 1

Let D be the midpoint of AB, and let CD produced meet the circle in E.

Join AE.

Let $CD=h,DE=x,AE=l$.

We have $AC=BC=15,AD=3$.

$\because CD$ is the perpendicular bisector of $AB$, $CD$ passes through the centre $O$ of the circle.

$\therefore CE$ is a diameter of the circle

$\therefore \angle CAE={90}^o$ ....... (angle in a semi circle is a right angle)

From the right-angled $\Delta ADC$,

$h^2=A{C}^2-A{D}^2={15}^2-3^2=225-9=216$

$\therefore h=\sqrt{216}=6\sqrt{6}$.

From the right-angled $\Delta CAE$, we have

$(h+x{)}^2=C{A}^2+A{E}^2$

$={15}^2+l^2$

$={15}^2+3^2+x^2$.

$\therefore h^2+x^2+2hx=225+9+x^2$.

$⟹216+2hx=234$;

$⟹2hx=18$

$⟹x=\frac{18}{2h}=\frac{9}{6\sqrt{6}}=\frac{3}{2\sqrt{6}}$.

Therefore, diameter CE$=h+x=6\sqrt{6}+\frac{3}{2\sqrt{6}}=\frac{72+3}{2\sqrt{6}}=\frac{75}{2\sqrt{6}}$.

So, radius of the circle$=\frac{diameter}{2}=\frac{75}{4\sqrt{6}}$cm.