Question

Area of a circle in which a chord of length root $2\sqrt 3 $ makes an angle 2$\dfrac{\pi }{{\text{3}}}$ at the centre is
A. ${\text{4}}\pi $
B. ${\pi ^{\text{2}}}$
C. $\dfrac{1}{2}$
D. $\sqrt 2 \pi $

Answer 1

Hint: To solve this question we use the basic theory of chapter circles as discussed below. As we know the perpendicular to the chord from the centre of the circle bisects the angle of a circle and also a perpendicular dropped from the centre of the circle to a chord bisects it. It means that both the halves of the chords are equal in length. By using these we will be able to solve this question.

Complete step-by-step answer:
Given a chord of length $2\sqrt 3 $ makes an angle 2$\dfrac{\pi }{{\text{3}}}$ at the centre.
we have to find the area of circle:

According to question-
$\angle {\text{AOC = }}$2$\dfrac{\pi }{{\text{3}}}$
And AC= $2\sqrt 3 $
Draw OB$ \bot $AC
Which results, AB= $\sqrt 3 $
And $\angle {\text{AOB = }}$$\dfrac{{\text{pi }}}{{\text{3}}}$
As we know, the sum of all interior angles of a triangle is ${180^0}$ always.
So, $\angle {\text{OAB = }}$$\dfrac{\pi }{6}$
Now, in $\Delta {\text{OBA}}$, OB$ \bot $AB
$ \Rightarrow $cos$\dfrac{\pi }{6}$= $\dfrac{{{\text{AB}}}}{{{\text{OA}}}}$
$ \Rightarrow $$\dfrac{{\sqrt 3 }}{2}$= \[\dfrac{{\sqrt 3 }}{{{\text{OA}}}}\]
$ \Rightarrow $OA= 2
So, the radius of the given circle is 2 units.
Its area= $\pi {{\text{r}}^{\text{2}}}$
= $\pi \times {{\text{2}}^{\text{2}}}$
= ${\text{4}}\pi $
Therefore, Area of a circle in which a chord of length root $2\sqrt 3 $ makes an angle 2$\dfrac{\pi }{{\text{3}}}$ at the centre is ${\text{4}}\pi $.


Note- The chord is a line segment that joins two points on the circumference of the circle. A chord only covers the part inside the circle. And diameter is also considered as a chord of the circle. The diameter is the longest chord possible in a circle and it divides the circle into two equal segments.