Question

A chord which is the perpendicular bisector of a radius of length 12 cm in a circle, has length

• A. 27 cm
• B. $6\sqrt{3}$ cm
• C. $12\sqrt{3}$ cm
• D. $3\sqrt{3}$ cm
  

Answer 1

The correct option is C $12\sqrt{3}$ cm

Let $O$ denote the center of the circle, and let $OR$ and $AB$ be the radius and the chord which are perpendicular bisectors of each other at $M$.


Applying the Pythagorean theorem - to right triangle $OMA$ yields
$$\begin{array}{cc}(AM{)}^2& =(OA{)}^2-(OM{)}^2\\ & ={12}^2-6^2=108c{m}^2\\ AM& =6\sqrt{3}cm\end{array}$$ Thus the required chord has length $12\sqrt{3}c{m}^2$.
∴ (D) is the correct option.