Question : Two equal circles of radius 6 cm intersect each other such that each passes through the centre of the other. The length (in cm) of the common chord is:
Option 1: $5 \sqrt{3}$
Option 2: $6 \sqrt{3}$
Option 3: $4 \sqrt{3}$
Option 4: $3 \sqrt{3}$
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Correct Answer: $6 \sqrt{3}$
Solution :
Let the common chord AB bisect the distance between the centre XY at O.
⇒ XO = $\frac{\text{XY}}{2}$ = $\frac{6}{2}$ = 3 cm
⇒ AX = radius = 6 (The circles pass through each other’s centre)
Now, in $\triangle XAO$ right angled at O
⇒ AX
2
= AO
2
+ XO
2
⇒ 6
2
= AO
2
+ 3
2
⇒ AO
2
= 36 – 9
⇒ AO
2
= 27
⇒ AO = $3\sqrt{3}\ \text{cm}$
$\therefore$ AB = 2 × AO = $6\sqrt{3}\ \text{cm}$
Hence, the correct answer is $6\sqrt3\ \text{cm}$.
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