Practicing Success
Two circles of radius of 21 cm each intersect each other such that each passes through the centre of the other. What is the length of the common chord? |
$18\sqrt{3}$ cm 42 cm 27 cm $21\sqrt{3}$ cm |
$21\sqrt{3}$ cm |
OA = 21 cm, OE = \(\frac{1}{2}\)OC = \(\frac{R}{2}\) = \( {OA }^{2 } \) = \( {EA }^{2 } \) + \( {OE }^{2 } \) = \( {EA }^{2 } \) = \( {OA }^{2 } \) - \( {OE }^{2 } \) = \( {EA }^{2 } \) = \( {R }^{2 } \) - \( {R/2 }^{2 } \) = \( {EA }^{2 } \) = \( {R }^{2 } \) - \( {R/4 }^{2 } \) = EA = \(\sqrt {3/4 }\)\( {R }^{2 } \) = EA = \(\sqrt {3 }\)R/2 Now length of AB, = AB = 2 x E = 2 x \(\sqrt {3 }\)R/2 = AB = R\(\sqrt {3 }\) = 21 \(\sqrt {3 }\) Therefore, the length of the common chord is 21 \(\sqrt {3 }\). |