Question

AB and CD are two parallel chords of a circle of length 24 cm and 10 cm respectively and lie on the same side of its centre O. If the distance between the chords is 7 cm, find the radius of the circle (in cm).

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Answer 1

Now, OL $\perp$ AB and OM $\perp$ CD

AL = LB = $\frac{24}{2}$ cm = 12 cm and CM = MD = $\frac{10}{2}$ cm = 5 cm. Also, distance between the chords = LM = 7 cm.
Join OA and OC.

Let OL = x cm. Then, OM = (x + 7) cm.
Let the radius of the circle be r cm, Then, OA = OC = r cm.

${OA}^2$ = ${OL}^2$ + ${AL}^2$ and ${OC}^2$ = ${OM}^2$ + ${CM}^2$

$r^2$ = $x^2$ + ${\left(12\right)}^2$….(i) and $r^2$ = ${(x+7)}^2$ + ${\left(5\right)}^2$ ….(ii)

$x^2$ + ${\left(12\right)}^2$ = ${(x+7)}^2$ + ${\left(5\right)}^2$ [From (i) and (ii)]

$x^2$ + 144 = $x^2$ + 14x + 74

$14x=70$

$\therefore x=5$

Substituting $x=5$ in (i),we get: $r^2$ =$x^2$ + ${\left(12\right)}^2$

$\Rightarrow r^2={\left(5\right)}^2$ + ${\left(12\right)}^2$ $=(25+144)=169$

$r={\left(169\right)}^{1/2}$ $=13cm$

Hence, the radius of the circle is $13cm$.