Question : The length of the common chord of two circles of radii 15 cm and 13 cm, whose centres are 14 cm apart, is:
Option 1: 14 cm
Option 2: 12 cm
Option 3: 15 cm
Option 4: 24 cm
Correct Answer: 24 cm
Solution :
The radii of the circles are 15 cm and 13 cm.
Distance between the centres = 14 cm
AB = 15 cm and BC = 13 cm, and AC = 14 cm
Let OC = $x$ and AO = $(14 - x)$
Now in $\triangle$ COB,
$(OB)^{2} + x^{2} = (13)^{2}$
$⇒(OB)^{2} + x^{2} = 169$
$⇒(OB)^{2} = 169 - x^{2}$ ........(i)
Again in $\triangle$ AOB,
$(OB)^{2} + (14 - x)^{2} = (15)^{2}$
$⇒(OB)^{2} + (14 - x)^{2} = 225$
$⇒(OB)^{2} = 225 - (14 - x)^{2}$ .........(ii)
From equations (i) and (ii), we get,
$169 - x^{2} = 225 - (14 - x)^{2}$
$⇒(14 - x)^{2} - x^{2} = 225 – 169$
$⇒196 + x^{2} - 28x - x^{2} = 56$
$⇒196 - 28x = 56$
$⇒28x = 196 - 56$
$⇒28x = 140$
$⇒x = 5$
Putting the value of $x$ in equation (i),
$⇒(OB)^{2} = 169 - x^{2}$
$⇒(OB)^{2} = 169 - 5^{2}$
$⇒(OB)^{2} = 144$
$⇒(OB) = 12$
Since BD = OB × 2,
$\therefore$ BD = 12 × 2 = 24 cm
So, the length of the common chord is 24 cm.
Hence, the correct answer is 24 cm.
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