Question : The diameter $PQ$ of a circle with centre $O$ is perpendicular to the chord $RS$. $PQ$ intersects $RS$ at $T$. If $RS=16$ cm and $QT=4$ cm, what is the length (in cm) of the diameter of the circle?
Option 1: 20
Option 2: 48
Option 3: 24
Option 4: 10
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Correct Answer: 20
Solution : Given: RS = $16$ cm QT = $4$ cm Join OR, which is a radius. Let the radius be $r$ cm. So, OT = $r-4$ cm RT = $\frac{1}{2}×RS =\frac{1}{2}×16=8$ cm From $\triangle$ ORT, $OR^2=OT^2+RT^2$ ⇒ $r^2=(r-4)^2+8^2$ ⇒ $8r=16+64$ ⇒ $8r=80$ ⇒ $r=10$ Diameter = $2r=2×10=20$ cm So, the diameter of the circle is 20 cm. Hence, the correct answer is 20.
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