Question

Chord PQ of a circle of radius 6 cm subtends an angle of ${60}^o$ at the centre O. Find the area of the minor segment and major segment respectively.
(Use $\pi =3.14,\sqrt{3}=1.73$ )

• A. $3.27c{m}^2,113.04c{m}^2$
  
• B. $113.04c{m}^2,3.27c{m}^2$
• C. $3.27c{m}^2,109.77c{m}^2$
• D. $109.77c{m}^2,3.27c{m}^2$

Answer 1

The correct option is C $3.27c{m}^2,109.77c{m}^2$

Given:
Radius of the circle, r = 6 cm
$\angle POQ=\theta ={60}^o$

In $△$POQ, $\angle POQ={60}^o$ and OP = OQ, so $△$POQ is an equilateral triangle.

Area of the minor segment
= Area of sector POQ - Area of $△$POQ
$=\pi r^2\times \frac{\theta }{360}-\frac{\sqrt{3}}{4}\times r^2$

$=3.14\times 6^2\times \frac{60}{360}-\frac{1.73}{4}\times 6^2$

$=18.84-15.57$

$=3.27c{m}^2$

Now,
Area of the circle
$=\pi r^2$
$=3.14\times 6\times 6$
$=113.04c{m}^2$

$\therefore$ Area of the major segment
= Area of the circle − Area of the minor segment
$=113.04-3.27$
$=109.77c{m}^2$

Thus, the areas of the minor segment and major segment are $3.27c{m}^2$ and $109.77c{m}^2$ respectively.