Question : Two circles having equal radii intersect each other such that each passes through the centre of the other. The sum of the diameter of these two circles is 84 cm. What is the length of the common chord?
Option 1: $21 \sqrt{3}\ \text{cm}$
Option 2: $14 \sqrt{3}\ \text{cm}$
Option 3: $28 \sqrt{3}\ \text{cm}$
Option 4: $24 \sqrt{3} \ \text{cm}$
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Correct Answer: $21 \sqrt{3}\ \text{cm}$
Solution :
Given: Sum of the diameter of circles = 84 cm
Diameter of one circle = $\frac{84}{2}=42$
Radius = $\frac{42}{2}=21$
According to the question,
AD = DB
OP = 21 cm
Again OA = PA = 21 cm
Here, $\angle ADO=90^\circ$
OD = PD = $\frac{21}{2}$
In $\triangle$OAD, using Pythagorean theorem we get,
$AD=\sqrt{OA^2-OD^2}$
$=\sqrt{21^2-(\frac{21}{2})^2}$
$=\sqrt{441-(\frac{441}{4})}$
$=\sqrt{\frac{1323}{4}}$
$={\frac{21\sqrt3}{2}}$
So, the length of the common chord AB = $2\times{\frac{21\sqrt3}{2}}=21 \sqrt{3}\ \text{cm}$
Hence, the correct answer is $21 \sqrt{3} \ \text{cm}$.
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