Question

In the given figure, O is the centre of the circle with radius 7 cm. If OC = CD, then the length of chord AB is

• A.
  $3.5\sqrt{3}cm$
• B.
  $7\sqrt{5}cm$
• C.
  $7\sqrt{3}cm$
• D.
  $14cm$

Answer 1

The correct option is C

$7\sqrt{3}cm$
Given that $\Delta OCA$ is a right-angled triangle.

In $\Delta OCA,$ by Pythagoras theorem,


$A{C}^2=O{A}^2–O{C}^2\Rightarrow A{C}^2=(7{)}^2-{\left(\frac{7}{2}\right)}^2(\because OC=\frac{OD}{2}=dfrac72cm)\Rightarrow A{C}^2=49-\frac{49}{4}A{C}^2=\frac{(196-49)}{4}\Rightarrow A{C}^2=\frac{147}{4}\Rightarrow AC=\frac{\sqrt{147}}{2}\Rightarrow AC=\frac{7\sqrt{3}}{2}$

Since the perpendicular from the centre of a circle to a chord bisects the chord.

$\therefore AB=2AC=2\times \frac{7\sqrt{3}}{2}cm=7\sqrt{3}cm$

Hence, the correct answer is option (c).