Question : In a circle, a diameter AB and a chord PQ (which is not a diameter) intersect each other at X perpendicularly. If AX : BX = 3 : 2 and the radius of the circle is 5 cm, then the length of the chord PQ is:
Option 1: $2\sqrt{13}\;\mathrm{cm}$
Option 2: $5\sqrt{3}\;\mathrm{cm}$
Option 3: $4\sqrt{6}\;\mathrm{cm}$
Option 4: $6\sqrt{5}\;\mathrm{cm}$
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Correct Answer: $4\sqrt{6}\;\mathrm{cm}$
Solution :
We have, $\mathrm{AX:BX = 3:2}$ and the radius of the circle is $5\;\mathrm{cm}$. $\mathrm{AX=\frac{3}{5} \times 10 = 6}\;\mathrm{cm}$ $\mathrm{BX=\frac{2}{5} \times 10 = 4}\;\mathrm{cm}$ We know the theorem states that the two chords of a circle intersect within the circle, the product of the segments of one chord is equal to the product of the segments of the other chord. $\mathrm{PX\times QX = AX \times XB}$ $\mathrm{PX^2=6 \times 4}$ [Since $\mathrm{PX =QX}$] $\mathrm{PX=2\sqrt{6}}$ $\mathrm{PQ=2PX=4\sqrt{6}}$ Hence, the correct answer is $4\sqrt{6}\;\mathrm{cm}$.
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