Question : PQ and RS are two parallel chords of a circle such that PQ is 48 cm and RS is 40 cm. If the chords are on the opposite sides of the centre and the distance between them is 22 cm, what is the radius (in cm) of the circle?
Option 1: 25
Option 2: 24
Option 3: 35
Option 4: 22
Correct Answer: 25
Solution :
Given: PQ = 48 cm, RS = 40 cm and MN = 22 cm
Now, QM = $\frac{48}{2}$ = 24 cm and RN = $\frac{40}{2}$ = 20 cm
Let OM = $x$ cm then ON = (22 – $x$) cm
Also, let the radius of the circle be $r$ cm.
From $\triangle$QOM, we get,
$r^2=24^2+x^2$ -----------------------(1)
From $\triangle$RON, we get,
$r^2=20^2+(22-x)^2$ ---------------------(2)
From equation (1) and (2) we get,
$24^2+x^2=20^2+(22-x)^2$
⇒ $(22-x)^2-x^2=24^2-20^2$
⇒ $(22-x+x)(22-x-x)=(24+20)(24-20)$
⇒ $(22-2x)×22=44×4$
⇒ $x=7$
So, the radius of the circle, $r=\sqrt{24^2+7^2}=25$ cm
Hence, the correct answer is 25.
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