Question

A chord of a circle of radius 15 cm subtends an angle of 60° at the centre. Find the areas of the corresponding minor and major segments of the circle.
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Answer 1

.Area of sector ABCD =πr^2¢/360°

=)22/7×15×15×60°/360°

=)117.85.

Area of ∆ABC.

=1/2r^2sin¢

=)1/2×15×15×√3/2

=)225√3/4

=)97.3125.

minor segment BDC=Area of sector ABDC-Area of triangle ABC.

=117.85-97.3125

=)20.53 ___1)

and again ,Area of circle =πr^2¢

=)22/7×15×15

=)707.14.cm^2.

.

so, major segment BDC

=)Area of circle - minor segment.

=)707.14-20.5

=)686.642 Ans

$\: \:$

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Answer 2

Answer~

Given:-

• Radius = 15cm
• Angle A = Angle B

a) Area of minor segment

• Area of minor sector

$\longrightarrow { \frac{ \theta}{360 \degree} \times \pi \: r ^2}$

$\longrightarrow \: \frac{60 \degree}{360 \degree} \times (15)^2 \times 3.14cm^2$

$\longrightarrow \bold \red{117.75cm^2}$

• Area of triangle, here we can see the triangle is equilateral.

$\longrightarrow \: \frac{ \sqrt{3} }{4} \times (oa)^2$

$\longrightarrow \: \frac{ \sqrt{3} }{4} \times (15)^2$

$\longrightarrow \bold\red {97.3cm^2}$

$\bold \blue{Area \: of \: minor \: segment \: = \: Area \: of \: sector \: - \: Area \: of \: triangle.}$

$\blue \: \bold {Area \: of \: minor \: segment =}117.75 - 97.3cm \: sq.$

$\longrightarrow \: 20.4cm \: sq.$

So we got area of minor segment as 20.4 cm sq.

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b) Area of major segment

• Area of major sector
• Angle 360-60=300

$\longrightarrow { \frac{ \theta}{360\degree} \times \pi \: r ^2}$

$\longrightarrow { \frac{ 300 \degree}{360\degree} \times 3.14 \times \: (15)^2}$

$\longrightarrow \bold \red {588.75cm \: sq}.$

Area of major segment = Area of major sector+ Area of equilateral∆AOB

$\longrightarrow \: 588.75cm \: sq + 97.3cm \: sq.$

$\longrightarrow \bold \red{686.05cm \: sq.}$