Question : Two chords of a circle, $\overline{\mathrm{AB}}$ and $\overline{\mathrm{CD}}$, meet outside the circle at the point $\mathrm{P}$. If $\mathrm{m}(\overline{\mathrm{AP}})=200 \mathrm{~mm}, \mathrm{~m}(\overline{\mathrm{AB}})$ $=120 \mathrm{~mm}$, and $\mathrm{m}(\mathrm{CP}) = 160 \mathrm{~mm}$, what is the length of $\mathrm{CD}$?
Option 1: 100 mm
Option 2: 75 mm
Option 3: 60 mm
Option 4: 150 mm
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Correct Answer: 60 mm
Solution :
Let the chords $AB$ and $CD$ meet at $P$.
We know that if two chords of a circle intersect externally, then the product of the lengths of the segments are equal.
So, $AP\times BP = CP\times DP$
⇒ $AP\times (AP – AB) = CP\times (CP – CD)$
⇒ $200\times (200 – 120) = 160\times (160 – CD)$
⇒ $CD = 60$ mm
Hence, the correct answer is 60 mm.
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