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
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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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