Question

In the given figure, diameter of the semi-circle is $6cm$ and $CD=1cm$.

Find $AB$

• A. $\sqrt{15}$
• B. $\sqrt{30}$
• C. $\sqrt{31}$
• D. None of these

Answer 1

The correct option is B

$\sqrt{30}$

Given: $AD=6cm$

$CD=1cm$

$\Rightarrow AC=AD-CD=6-1=5cm$

We have a property of chords which states that - If AB and CD are two chords of a circle intersecting at O, then AO x OB = CO x OD

Hence, we have $B{C}^2=AC\times CD$ (We have $B{C}^2$ as the perpendicular from the centre bisects the chord)

$B{C}^2=5\times 1$

$BC=\sqrt{5}$

In $△ACB$, applying Pythagoras Theorem, we get,

$A{B}^2=A{C}^2+B{C}^2$

$\Rightarrow A{B}^2=5^2+{\sqrt{5}}^2$

$=25+5=30$

$\Rightarrow AB=\sqrt{30}cm$