Question

A chord of a circle of radius 21 cm subtends a right angle at the centre. Find the area of the

corresponding : minor segment. (Take

p = 3.14)​

Answer 1

$(\pi = 3.14) \\\\ Area of segment= {r}^{2} [ \frac{\piθ }{360 } - \frac{1}{2} \sinθ)] \\ \\ = ( {21})^{2} [ \frac{3.14 \times 90}{360} - \frac{1}{2} \times 1] \\\\ = 441[ \frac{3.14}{4} - \frac{1}{2} ] \\ \\ = 441( \frac{3.14 - 2}{4} ) \\ \\ = 441 \times [ \frac{1.14}{4} ] \\ \\ = \frac{502.72}{4} \\\\ = 125.685$

Answer 2

Step-by-step explanation:

(pi = 3.14)

Minor segment=r^ 2 [ pi theta 360 - 1 2 sin theta)]

=(21)^ 2 [ 3.14*90 360 - 1 2 *1]

=441[ 3.14 4 - 1 2 ]

=441( 3.14-2 4 )

=441*[ 1.14 4 ]

= 502.72 4

=125.685

Hope this helps you......