AB is the diameter of a circle and AD is a chord. If AB = 26 cm, AD = 24 cm, the distance of AD from the centre of the centre of the circle is

Topic Pythagoras property

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8 cm

5 cm

12 cm

10 cm

answer is B.

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Detailed Solution

Here AB = 26 cm and AD = 24 cm.
Now, joining DB and drawing a perpendicular from O to AD, where perpendicular meeting AD at C.
Distance of chord AD from the centre of the circle = OC.
seo

∠ADB = 90° (Due to Angle in semicircle)
Also, ∠ACO = 90° (Due to CO is perpendicular to AD)
In triangle ADB, ∠B = 90°
Then, AD2+DB2=AB2 Using Pythagoras Theorem
Putting values of AD and AB,
(24)2+DB2=(26)2
Rearranging the terms
⇒DB2=(26)2−(24)2
⇒DB2=676−576=100
⇒DB=√100=10
Thus, DB = 10 cm.
Now, in triangles ACO and ADB :
∠A = ∠A (Due to Common in both triangles)
∠ACO = ∠ADB (Each equal to 90°)
By AA similarity criteria :