Question

A chord of a circle of radius 10cm subtends a right angle at the center. Find the area of the corresponding: (Use = 3.14) i. minor segment ii. major segment

Answer 1

radius r=10cm

angle =90

area of sector A=90/360pir^2

A=3.14x10x10/4

A=25x3.14

A=78.5sq.cm

Let the angle subtended and radius form an arc AOB,then

area of AOB=rxrsin90/2

AOB=10x10x1/2

ar. of AOB =50sq.cm

area of minor segment =78.5-50

=28.5aq.cm

Then area of circle =pir^2

=3.14x10x10

=314 Sq.cm

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

=314-28.5

=285.5sq.cm

Answer 2

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

Minor segment is shaded (Refer to the attachment)

Area of minor segment

= Area of the quadrant AOB− Area of the △AOB

$= \frac{1}{4} ×π {r}^{2} − \frac{1}{2} ×AO×OB </strong></p><p></p><p></p><p><strong>[tex]= \frac{1}{4} ×π {r}^{2} − \frac{1}{2} ×AO×OB$

$= \frac{1}{4} \times 3.14 \times {10}^{2} - \frac{1}{2} \times 10 \times 10$

$= 78.5 - 50$

$= 28.5 \: \: {cm}^{2}$

2.

Area of the major segment

= Area of the circle − Area of the minor segment

$= (\pi \: {r}^{2} - 28.5 \:) \: {cm}^{2}$

$= (\frac{22}{7} \times {10}^{2} - 28.5) \: \: {cm}^{2}$

$= (314.29 - 28.5) \: \: {cm}^{2}$

$\: = 285.79 \: \: {cm}^{2}$