Question

A chord CD of a circle whose centre is O, is bisected at P by a diameter AB. Given, OA = OB = 15 cm and OP = 9 cm. Then the length of AD is: ​

Answer 1

Answer:

∴AD=  720 =26.83 cm

Step-by-step explanation:

Join BD,

∴BP=OB−OP=15−9=6 cm

In right △BPD,

BD  2  =BP  2  +PD  2

 

=(6)  2  +(12)  2  =180

In △ADB, ∠ADB=90  o  

(Angle in a semicircle is a right angle)

∴AB  2  =AD  2 +BD  2

 

⇒AD  2  =AB  2 −BD  2  =(30)  2  −180=720

∴AD= 720=26.83 cm  

Answer 2

Answer:

The length of the AD is = 26.83 cm

Step-by-step explanation:

In the given figure AB is a diameter

CD is a chord

OP bisected CD at p

Radius OA = radius OB = 15 cm

And OP = 9 cm

We know OP is a perpendicular bisector on CD

So, OD is a radius of the circle,

So, OD = 15 cm

Now OPD is a right angle triangle

So, OP^2 + PD^2 = OD^2

=> 9^2 +PD^2 = 15^2

=> PD^2 = 15^2 - 9^2

=> PD^2 = 225-81 = 144 = 12^2

=> PD = 12 cm

Now AP = AO+OP = 15 + 9 = 24 cm

Now APD is right angle triangle,

So, AP^2 +PD^2 = AD ^2

=> 24^2 + 12^2 = AD^2

=> AD^2 = 576+144 = 720

=> AD = √720 cm = 26.83 cm

So the length of AD = 26.83 cm