A chord of a circle of radius 10 cm subtends a right angle at the center. Find the area of the corresponding : 

(i) minor segment 

(ii) major sector. (Use π = 3.14)

detailed solution

1

We know that, the formula for the area of the sector of a circle

Area of the sector = $\frac{\theta }{{360}^0}$ $\times$ ${\mathrm{\pi r}}^2$

Area of the segment  = Area of the sector - Area of the triangle

Radius = r = 10 cm, Angle =  θ  = 90°

In the given figure,

AB is the chord that forms a right angle at the center.

Thus we get the formulas as follows,

 Area of minor segment APB = Area of sector OAPB - Area of right triangle AOB

 Area of major segment AQB = ${\mathrm{\pi r}}^2$ - Area of minor segment APB

Area of the right triangle ΔAOB = 1/2 $\times$ OA $\times$ OB

(i) Area of minor segment APB = Area of sector OAPB - Area of right ΔAOB

= θ/360° × $\mathrm{\pi }$r² - 1/2 × OA × OB

= 90°/360° × $\mathrm{\pi }$r² - 1/2 $\times$ r $\times$ r

= 1/4 $\mathrm{\pi }$r² -1/2 r²

= r² (1/4 $\mathrm{\pi }$ - 1/2)

= r² (3.14 - 2)/4

= (r² $\times$ 1.14)/4

= (10 $\times$ 10 $\times$ 1.14)/4 

= 28.5 cm²

(ii) Area of major sector AOBQ = $\mathrm{\pi }$r² - Area of minor sector OAPB

= $\mathrm{\pi }$r² - θ/360° $\times$ $\mathrm{\pi }$r²

= $\mathrm{\pi }$r² (1 - 90°/360°)

= 3.14 $\times$10² $\times$ 3/4

= 235.5 cm²