Question : Two circles each of radius 36 cm are intersecting each other such that each circle passes through the centre of the other circle. What is the length of the common chord to the two circles?
Option 1: $24 \sqrt{3}$ cm
Option 2: $12 \sqrt{3} $ cm
Option 3: $36 \sqrt{3}$ cm
Option 4: $16 \sqrt{3}$ cm
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Correct Answer: $36 \sqrt{3}$ cm
Solution : Given: Two circles each of radius 36 cm intersect each other such that each circle passes through the centre of the other circle. OO' = OC = 36 cm ⇒ OA = AO' = $\frac{36}{2}=18$ cm (radius) Using Pythagoras theorem in $\triangle AOC$, ${CA}^2+{OA}^2={OC}^2$. ⇒ ${CA}^2={OC}^2–{OA}^2$ ⇒ ${CA}^2={36}^2–{18}^2$ ⇒ ${CA}^2=1296–324=972$ ⇒ $CA=18\sqrt3$ cm The length of the common chord to the two circles = $2\times CA=2\times 18\sqrt3=36\sqrt3$ cm. Hence, the correct answer is $36\sqrt3$ cm.
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