Question

O is the center of the circle. AB is a chord of the circle. $$OM\perp AB$$, if AB = 20 cm, OM = $$2\sqrt{11}$$ cm, then the radius of the circle is

A. 15 cm
B. 12 cm
C. 10 cm
D. 11 cm

Answer 1

Hint: In this question the length of AB and OM is given and we have to find the radius of the given circle. So for this let us draw the diagram,

So here we draw a line OA, which is the radius of the circle and we have to find the length of OA in orders to find the radius.
So for this we need to know the Pythagorean theorem, which states that “the squire of the hypotenuse is equal to the sum of the squares of the base and height”.

Complete step by step answer:
Given,
OM = $$2\sqrt{11}$$ cm, AB = 20 cm and OM is perpendicular to AB.
As we know that when we draw a perpendicular line from the centre of a circle to a chord, then the line bisects the chord.
i.e, AM = MB = $$\dfrac{20}{2} =10$$ cm
Now the $$\triangle OMA$$ is forming a right angle triangle where OA is the hypotenuse, AM is the base and OM is height.
Therefore by Pythagorean theorem,
$$\left( \text{Hypotenuse} \right)^{2} =\left( \text{Base} \right)^{2} +\left( \text{Height} \right)^{2} $$

$$\Rightarrow \left( OA\right)^{2} =\left( AM\right)^{2} +\left( OM\right)^{2} $$
$$\Rightarrow \left( OA\right)^{2} =\left( 10\right)^{2} +\left( 2\sqrt{11} \right)^{2} $$
$$\Rightarrow \left( OA\right)^{2} =100+2^{2}\left( \sqrt{11} \right)^{2} $$ [$$\because \left( ab\right)^{n} =a^{n}\cdot b^{n}$$]
$$\Rightarrow \left( OA\right)^{2} =100+4\times 11$$ [since $$\left( \sqrt{a} \right)^{2} =a$$]
$$\Rightarrow \left( OA\right)^{2} =100+44$$
$$\Rightarrow \left( OA\right)^{2} =144$$
$$\Rightarrow OA=\sqrt{144}$$
$$\Rightarrow OA=\sqrt{12\times 12}$$
$$\Rightarrow OA=12$$ [$$\because \sqrt{a\times a} =a$$]
Therefore the radius of the circle is 12 cm.
So, the correct answer is “Option B”.


Note: To solve this type of question you need to know that the distance from the centre to any point on the circumference of a circle is called the radius of that circle, so that is why we considered OA as radius. Also you can solve the above problem by taking the triangle $$\triangle OBM$$ instead of $$\triangle OAM$$.