Question

Given $BD=12$ and $AC=3$ in the circle with center $A$. Find the radius.

• A. $3$
• B. $3\sqrt{5}$
• C. $4$
• D. $4\sqrt{5}$

Answer 1

Answer: (B)

$3\sqrt{5}$

Perpendicular from the centre to chord bisects the chord

Thus $BC=CD$

Given, $BD=12$

Therefore, $BC+CD=12$

$\Rightarrow BC+BC=12$

$\Rightarrow 2BC=12$

$\Rightarrow BC=\frac{12}{2}=6$

Now applying pythagoras theorem in $△ABC$

Thus ${AB}^2={AC}^2+{BC}^2$

$\Rightarrow {AB}^2=(3{)}^2+{\left(6\right)}^2\Rightarrow {AB}^2=45\Rightarrow AB=3\sqrt{5}$