A chord AB of a circle whose centre is O, is bisected at E by a diameter CD. OC= OD = 15cm and OE = 9 cm. Find the length of AB

Rectangle

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Circle

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Solution

The correct option is D
Circle

Given OE = 9 cm, OA = 15cm

Consider $△ O A E$ , $∠ O E A = 90^{o}$

Applying Pythagoras theorem

$A E^{2} = O A^{2} - O E^{2}$

$A E^{2} = 15^{2} - 9^{2}$

$A E^{2} = 225 - 81 = 144$

$\Rightarrow A E = 12 c m$

$∴ A B = 2 \times 12 = 24 c m$ [the perpendicular from the centre bisects the chord]

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A chord AB of a circle whose centre is O, is bisected at E by a diameter CD. OC= OD = 15cm and OE = 9 cm. Find the length of AB

The correct option is D Circle Given OE = 9 cm, OA = 15cm Consider △OAE, ∠OEA=90o Applying Pythagoras theorem AE2 = OA2 − OE2 AE2 = 152 − 92 AE2 = 225 − 81 =144 ⇒AE=12cm ∴AB=2 × 12 = 24cm [the perpendicular from the centre bisects the chord]