Question

In the given figure, lengths of the chords $AB$ and $CD$ are $12cm$ and $18cm$ respectively and distance between them is $15cm$. Find the radius of the circle.

Answer 1

Let the radius of the circle be $r$.

Given, AB = 12 cm, CD = 18 cm and FE = 15 cm

We know that perpendicular to the chord from the centre bisects the chord.

$\Rightarrow AF=FB=\frac{AB}{2}=6cm$

and $CE=ED=\frac{CD}{2}=9cm$

Let $OE$ be $x$ cm. Then $OF=15-x$ cm.

Applying Pythagoras theorem to $△AFO$,

$A{O}^2=A{F}^2+O{F}^2$

$\Rightarrow r^2=6^2+(15-x{)}^2$ ... (i)

Applying Pythagoras theorem to $△CEO$,

$O{C}^2=C{E}^2+O{E}^2$

$\Rightarrow r^2=9^2+x^2$ ... (ii)

Equating equations (i) and (ii), we get

$6^2+(15-x{)}^2=9^2+x^2$

$\Rightarrow 36+(15-x{)}^2=81+x^2$

$\Rightarrow 36+225+x^2-30x=81+x^2$

$\Rightarrow 30x=180$

$\Rightarrow x=\frac{180}{30}=6$

$\Rightarrow r^2=9^2+6^2=117$

$\Rightarrow$ Radius, $r$ $=\sqrt{117}cm.$