Question : Two circles having equal radii intersect each other such that each passes through the centre of the other. The length of the common chord is 24 cm, so what will be the diameter of each circle?
Option 1: $16 \sqrt{3}$ cm
Option 2: $8 \sqrt{3}$ cm
Option 3: $12 \sqrt{3}$ cm
Option 4: $10 \sqrt{3}$ cm
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Correct Answer: $16 \sqrt{3}$ cm
Solution :
AD = DB
Let, $O{_1}O{_2} = 2x$
Again $O{_1}A = O{_2}A = 2x$ [Radius of the circle]
$\angle ADO{_1} = 90°$
$O{_1}D = O{_2}D = \frac{2x}{2} = x$
$AD = \frac{1}{2}AB = 12$
Using the Pythagorean theorem,
We get,
$AD^2 = AO{_1}^2 +O{_1}D^2 $
⇒ $(12)^2 = (4x^2 - x^2)$
⇒ $12 = \sqrt{3x^2}$
⇒ $x = 4\sqrt3$
⇒ Radius = $2x$ = 8$\sqrt3$
⇒ Diameter = 2$\times$radius = $8\sqrt{3} × 2 = 16\sqrt3$
Hence, the correct answer is $16 \sqrt{3}$ cm.
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