Question

There are two concentric circles, each with center O and of radii 10 cm and 26 cm respectively. Find the length of the chord AB of the outer circle which touches the inner circle at P.

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Answer 1

Join OP and OB

OP is $\perp$ to AB and it bisects AB

$\therefore$ AP = PB

Applying Pythagoras theorem to ΔOPB

${PB}^2$ = ${OB}^2$ – ${OP}^2$

${PB}^2$ = ${26}^2$ – ${10}^2$

${PB}^2$ = 676 – 100

${PB}^2$ = 567

PB = 24 cm

Therefore the length of the chord is 2 x 24 = 48 cm.