Question

find the area of the segment of a circle given that the angle of the sector is 120 and radius of circle is 21cm. ​

Answer 1

Answer:

462 cm²

Step-by-step explanation:

Area of sector = Angle/360 × πr²

= 120/360 × 22/7 × 21 × 21

= 1/3 × 22 × 3 × 21

= 22 × 21

= 462 cm²

Answer 2

ANSWER:-

Given:

The angle of the sector is 120° & radius of circle is 21cm.

To find:

The area of the segment of a circle.

Solution:

We have,

•Radius of the circle= 21cm.

•Let angle AOB= 120°

•Let area of sector AOB:

We know that, area of the sector of a circle is;

$= > \frac{ \theta}{360 \degree} \pi {r}^{2}$

So,

$= > \frac{120 \degree}{360 \degree} \times \frac{22}{7} \times 21cm \times 21cm \\ \\ = > \frac{1}{3} \times 22 \times 3cm \times 21cm \\ \\ = > 462 {cm}^{2}$

⚫Draw OD perpendicular to AB

In ∆OAB,

•OA= OB [Radii of the circle]

Therefore,

∆OAB is an isosceles ∆.

=) AD= DB

[Isosceles ∆, altitude from the vertex bisect the base]

So,

=) AB= 2AD

angle AOD = angle BOD

$= > \frac{1}{2} \times 120 \degree \\ \\ = > 60 \degree$

In ∆OAD,

$sin60 \degree = \frac{AD}{OA} \\ \\ = > \frac{ \sqrt{3} }{2} = \frac{AD}{21} \\ \\ = > 2AD = 21 \sqrt{3} \\ \\ = > AD = \frac{21 \sqrt{3} }{2} cm \\ and \\ cos60 \degree = \frac{OD}{OA} \\ \\ = > \frac{1}{2} = \frac{OD}{21} \\ \\ = > 2OD= 21 \\ \\ = > OD = \frac{21}{2} cm$

So,

AB= 2AD

AB= 2× 21√3/2

AB= 21√3cm

Now,

Area of ∆OAB,

$= > \frac{1}{2} \times AB \times OD \\ \\ = > \frac{1}{2} \times 21 \sqrt{3} \times \frac{21}{2} \\ \\ = > \frac{441 \sqrt{3} }{4} {cm}^{2}$

⚫Area of segment AYB:

=)Area of sector - Area of ∆OAB

$= > 462 {cm}^{2} - \frac{441 \sqrt{3} }{4} {cm}^{2} \\ \\ = > \frac{1848 - 441 \times 1.732}{4} {cm}^{2} \ \\ \\ \ = > \frac{1848 - 763.812}{4} {cm}^{2} \\ \\ = > \frac{1084.18}{4} {cm}^{2} \\ \\ = > 271.04 {cm}^{2}$

Thus,

The area of the segment of a circle is 271.04cm².

Hope it helps ☺️