Question : ABC is an isosceles triangle inscribed in a circle. If AB = AC = $12\sqrt{5}$ cm and BC = 24 cm, then the radius of circle is:
Option 1: 10 cm
Option 2: 15 cm
Option 3: 12 cm
Option 4: 14 cm
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Correct Answer: 15 cm
Solution :
Given, AB = AC = 1$2\sqrt{5}$cm and BC = 24 cm
Join OB, OC, and OA.
Draw AD$\perp$ BC which will pass through centre O
OD bisects BC in D as perpendicular from the centre to a chord bisects the chord.
So, BD = CD = 12 cm
Using Pythagoras theorem in $\triangle$ ABD,
AB
2
= AD
2
+ BD
2
Or, $(12\sqrt{5})^{2}$ = AD
2
+ 12
2
Or, AD
2
= 576
Or, AD = 24 cm
Let the radius of the circle be OA = OB = OC = r
So, OD = AD – AO = 24 – r
Using Pythagoras theorem in $\triangle$ OBD,
r
2
=12
2
+ (24 – r)
2
Or, r
2
= 144 + 576 + r
2
– 48r
Or, r = 15 cm
Hence, the correct answer is 15 cm.
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