Question

find the area of segment AYB if radius is 21 cm and angle AOB = 120​

Answer 1

Answer:

462 cm sqa.

Step-by-step explanation:

the formula of sector r of a circle I

angle upon 360× pie × radius sqa.

after putting the values we ge t 462

Answer 2

SOLUTION:-

Given:

•Radius of the circle is 21cm.

•Angle AOB=theta= 120°

Therefore,

Area of sector AOB;

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

&

Draw OD perpendicular to AB.

In ∆AOB,

OA= OB [radii of the circle]

•It's is a isosceles ∆

so, AD= DB

AB= 2AD

angle AOD= angle BOD

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

In ∆OAD,

$sin60 \degree = \frac{AD}{OA} \\ \\ = > \frac{ \sqrt{3} }{2} = \frac{AD}{21cm} \\ [cross \: multiplication] \\ = > 2AD = 21 \sqrt{3} \\ \\ = > AD = \frac{21 \sqrt{3} }{2}cm$

&

$cos60 \degree = \frac{</strong><strong>OD</strong><strong>}{</strong><strong>OA</strong><strong>} \\ \\ = > \frac{1}{2} = \frac{</strong><strong>OD</strong><strong>}{21} \\ \\ = > 2</strong><strong>O</strong><strong>D</strong><strong> = 21 \\ \\ = > </strong><strong>OD</strong><strong> = \frac{21}{2} cm$

We have, AB= 2AD

$= > 2 \times \frac{21 \sqrt{3} }{2} \\ \\ = > 21 \sqrt{3} cm$

Now,

Area of triangle OAB:

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

Area of segment AYB:

Area of sector AOB - Area of ∆AOB

$= > 462 {cm}^{2} - \frac{441 \sqrt{3} }{4} {cm}^{2} \\ \\ = > \frac{1848 - 441 \sqrt{3} }{4} \\ \\ = > \frac{1848 - 441 \times 1.732}{4} \\ \\ = > \frac{1848 - 76 3.812}{4} \\ \\ = > \frac{1084.18}{4} {cm}^{2} \\ \\ = > 271.05 {cm}^{2}$

Thank you.