A chord 21 cm long is drawn in a circle of diameter 25 cm. The perpendicular distance of the chord from the centre is:
Answer & explanation
Correct answer: option 4
Let AB be the chord of length 21 cm.
= OD be the perpendicular distance
= AO be the radius of the circle
\( {OA }^{2 } \) = \( {OD }^{2 } \) + \( {AD }^{2 } \)
= \( {25/2 }^{2 } \) = \( {OD }^{2 } \) + \( {21/2 }^{2 } \)
= \(\frac{625}{4}\) = \( {OD }^{2 } \) + \(\frac{441}{4}\)
= \( {OD }^{2 } \) = \(\frac{625}{4}\) - \(\frac{441}{4}\) = \(\frac{184}{4}\) = 46
Therefore, OD is \(\sqrt {46 }\).