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

$AL=LB=\frac{24}{2}cm=12cm$

and $CM=MD=\frac{10}{2}cm=5cm$.
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

x = 5

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

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

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

Hence,the radius of the circle is 13 cm.