Question

A chord of circle of radius 10 cm subtends a right angle at the centre.Find the area of the corresponding minor segment and hence find the area of the major sector?​

Answer 1

$\huge\mathfrak{Answer:}$

Given:

• We have been given that radius of circle is 10cm.
• Sector angle (θ) = 90°.

To Find:

• We need to find the area of minor segment and major segment.

Solution:

As it is given that length of radius (r) = 10cm and sector angle (θ) = 90°.

We know that the area of sector =

$\mapsto\sf{\dfrac{θ}{360} \times \pi {r}^{2} }$

$\sf{ = \dfrac{90}{360} \times 10 \times 10 \times \dfrac{314}{100} {cm}^{2} }$

$\sf{ = \dfrac{1}{4} \times 314 {cm}^{2} }$

$\sf{ = \dfrac{157}{2}{cm}^{2}}$

$\sf{= 78.5 {cm}^{2} }$

i) Area of the minor segment:

= [Area of minor sector] - [Area of right ΔAOB]

$\sf{ = (78.5 {cm}^{2}) - ( \dfrac{1}{2} \times 10 \times 10 {cm}^{2} }$

$\sf{ = 78.5 {cm}^{2} - 50 {cm}^{2} }$

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

ii) Area of major sector:

= [Area of the circle] - [ Area of the minor sector]

$\sf{ = \pi {r}^{2} - 78.5 {cm}^{2} }$

$\sf{ = \dfrac{314}{100} \times 10 \times 10 - 78.5}$

$\sf{ = (314 - 78.5) {cm}^{2} }$

$\sf{ = 235.5 {cm}^{2} }$

Answer 2

Answer:

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