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Question : In the given figure, BD passes through centre O, AB = 12 and AC = 8. What is the radius of the circle?

Option 1: $3\sqrt{2}$
Option 2: $4\sqrt{3}$
Option 3: $3\sqrt{5}$
Option 4: $3\sqrt{3}$

Team Careers360 8th Jan, 2024

Answer (1)

Team Careers360 15th Jan, 2024

Correct Answer: $3\sqrt{5}$

Solution :
Given: AB = 12 and AC = 8
Let CD = $a$.
We know that,
$AC×AD = AB^2$
⇒ $8×(8+a) = 12^2$
⇒ $64+8a = 144$
⇒ $8a = 80$
$\therefore a = 10$
So, CD = 10
Now, in $\triangle ABD, \angle B = 90^\circ$
AD = 18; AB = 12.
Let BD = $x$
Using Pythagoras theorem, we get,
$18^2 = 12^2+x^2$
⇒ $324 = 144+x^2$
⇒ $180 = x^2$
$\therefore x = 6\sqrt5$
Thus BD = $6\sqrt5$ is the diameter.
$\therefore$ Radius = $\frac{1}{2}×6\sqrt5 = 3\sqrt5$
Hence, the correct answer is $3\sqrt5$.

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