Question

A chord of a circle of radius 14cm subtends an angle of 60. At the centre ofvthe circle find the area ofvthe corresponding minor segment

Answer 1

Answer:

Given The radius of the circle is = 14cm✓

And the chord substends 60° At the centre ✓

So we can find out the area of the minor segment by using the formula

$\frac{theta}{360} \times \pi \: r ^{2} - r ^{2} \times sin \: \frac{theta}{2} \times cos \: \frac{theta}{2}$

So now let us find the area of the minor segment which is as follows:

$= > \frac{60}{360} \times \frac{22}{7} \times 196 - 196 \times sin \frac{60}{2} \times cos \frac{60}{2}$

$= > \frac{1}{6} \times \frac{22}{7} \times 196 - 196 \times \frac{1}{2} \times \frac{ \sqrt{3} }{2} .. \\ \\ (because \: sin \: 30 = \frac{1}{2} and \: cos30 = \frac{ \sqrt{3} }{2})$

$= > \frac{4312}{42} - 196 \times \frac{ \sqrt{3} }{4}$

$= > \frac{4312}{42} - \frac{333.2}{4}$

LCM of 42 and 4 is = 84

$= > \frac{4312 \times 2 - 333.2 \times 21}{84}$

$= > \frac{8624 - 6997.2}{84}$

$= > \frac{1626.8}{84}$

$= > 19.36cm ^{2}$

Therefore the area of the minor segment is 19.36 cm² ✓✓✓

Answer 2

Answer:

correct answer is

Step-by-step explanation:

your answer is 79.36