In the given figure, AB is the diameter of the circle with centre O. A chord CD is bisected by the diameter at P. If OA = OB = 15 cm and OP = 9 cm. Find the length of chord AD.

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Solution

Join OC and AD
OA = OB = OC = 15cm (radii)

In right angled $△$ OCP
$O C^{2} = O P^{2} + C P^{2}$
$C P^{2} = O C^{2} - O P^{2} = (15)^{2} - (9)^{2}$
$C P^{2} = 225 - 81 = 144$
$C P = \sqrt{144} c m .$
CP= 12 cm
CP = PD = 12 cm

Now in right angled $△$ APD
PD = 12 cm
AP = AO + OP = 15 + 9 = 24 cm
$A D^{2} = A P^{2} + P D^{2} = (24)^{2} + (12)^{2} = 576 + 144 = 720$

AD = $\sqrt{720}$ cm.
$= 12 \sqrt{5}$ cm.

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In the given figure, AB is the diameter of the circle with centre O. A chord CD is bisected by the diameter at P. If OA = OB = 15 cm and OP = 9 cm. Find the length of chord AD.

Join OC and AD OA = OB = OC = 15cm (radii) In right angled △ OCP OC2=OP2+CP2 CP2=OC2–OP2=(15)2–(9)2 CP2=225–81=144 CP=√144cm. CP= 12 cm CP = PD = 12 cm Now in right angled △ APD PD = 12 cm AP = AO + OP = 15 + 9 = 24 cm AD2=AP2+PD2=(24)2+(12)2=576+144=720 AD =√720 cm. =12√5 cm.