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?
Answer & explanation
Correct answer: option 4
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 \(\sqrt {3 }\).