Question

$O$ is the centre of the circle. $BC$ is a diameter of the circle. $OD\perp AB$. If $OD=4$ cm, $BD=5$ cm, then $CD$ is equal to

• A. $13$ cm
• B. $\sqrt{71}$ cm
• C. $\sqrt{89}$ cm
• D. None of these

Answer 1

The correct option is C $\sqrt{89}$ cm

Given- O is the centre of a circle whose diameter is BC. AB is a chord and OD$\perp$ AB. $BD=5cm$ and $OD=4cm$. CD has been joined.

To find out- $CD=?$

Solution- OD$\perp$AB.

$\therefore$ D is the mid point of AB since the perpendicular, dropped from the center of a circle to its any chord bisects the latter. So $AB=2BD=2\times 5cm=10cm$. And $BD=AD=5cm$. Now $\angle BAC={90}^o$ since angle in a semicircle$={90}^o$. $\therefore \Delta CAB\&\Delta CDB$ are right triangles with $BC\&DC$ as hypotenuses.

$\therefore$ By Pythagoras theorem, we have $OB=\sqrt{{BD}^2+O{D}^2}=\sqrt{5^2+4^2}cm=\sqrt{41}cm$.

But $BC=2OB(diameter=2radius)$.

$BC=2\times \sqrt{41}cm.\therefore AC=\sqrt{{BC}^2-A{B}^2}=\sqrt{{\left(\sqrt{41}\right)}^2-{10}^2}cm=8cm.$

So $CD=\sqrt{{AD}^2+A{C}^2}=\sqrt{5^2+8^2}cm=\sqrt{89}cm$.

Ans- Option C.