Question

A chord ab of a circle of radius 5.25 cm makes an angle of 60 degree at the center of the circle. Find the area of the major segment.

Answer 1

Answer:

72.12$cm^{2}$

Step-by-step explanation:

Area of the segment = $(\frac{\Theta }{360 } ) \times  \pi r^{2}$

where $\theta$  is in degrees.

(take π = 3.14)

Here, r = 5.25 cm

angle of the major segment = (360 - 60)°  = 300°

Now, area =  $(\frac{300 }{360 } ) \times  \pi (5.25)^{2}$

=   72.12$cm^{2}$

Answer 2

area of major segment is 84.125 cm ²

Step-by-step explanation:

The chord AB makes an angle of 60 degrees at the center

AB and the radii at the end points make an equilateral triangle

area of the minor sector

$\frac{\theta}{360}\pi r^2\\\\\frac{60}{360}\times\frac{22}{7}\times (5,25)^2= 14.4375$

area of minor sector = 14.4375 cm ²...1

length of the equilateral triangle  is a

then

area of equilateral triangle

$\frac{\sqrt{3}}{4}a^2$

area of triangle OPQ = $\frac{\sqrt{3}}{4}(5.25)^2= 11.93$

area of minor segment  = area of major sector - area of triangle OPQ

= 14.4375 - 11.93 = 2.5 cm ²

area of circle

$\pi r^2 \\\\\frac{22}{7}(5.25)^2= 86.62$

area of major segment = area of circle - area of minor segment

area of major segment = 86.62-2.5

area of major segment = 84.125 cm ²

hence , area of major segment is 84.125 cm ²

#Learn more:

Perimeter of a sector of a circle radius 5.25 cm is 16 cm find the arc of a circle​

https://brainly.in/question/14196782