Question

AB and CD are two parallel chords of a circle such that AB = 10 cm and CD = 24 cm. If the chords are on the opposite sides of the centre and the distance between them is 17 cm, find the radius of the circle.

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Answer 1

Let O be the centre of the given circle and let its radius be r cm. Draw OP$\perp$AB and OQ$\perp$CD. Therefore, points P, O and Q are collinear. So, PQ = 17 cm
Let OP = x cm. Then, OQ = (17 – x) cm
Join OA and OC. Then, OA = OC = r

Since the perpendicular from the centre to a chord of the circle bisects the chord.
$AP=PB=5cm$ and $CQ=QD=12cm$
In right triangles OAP and OCQ, we have
$O{A}^2=O{P}^2+A{P}^2$ and $O{C}^2=O{Q}^2+C{Q}^2$
$r^2=x^2+5^2$ ....(i)
and, $r^2=(17-x{)}^2+{12}^2$ ....(ii)
$x^2+5^2=(17-x{)}^2+{12}^2$[On equating the values of $r^2$]
$x^2+25=289-34x+x^2+144$
$34x=408$
$x=12cm$
Putting x = 12 cm in equation (i), we get
$r^2=122+52=169$
$r=13cm$
Hence, the radius of the circle is 13 cm.