Find the area of the major segment APB of a circle of radius 35 cm and ∠AOB = 90°

Solution:

Area of minor segment = Area of sector AOBC − Area of right triangle AOB

$=\frac{90^{\circ}}{360^{\circ}} \times \pi(\mathrm{OA})^{2}-\frac{1}{2} \times \mathrm{OA} \times \mathrm{OB}$

$=\frac{1}{4} \times \frac{22}{7} \times(35)^{2}-\frac{1}{2} \times 35 \times 35$

$=\frac{1}{4} \times \frac{22}{7} \times(35)^{2}-\frac{1}{2} \times 35 \times 35$

$=962.5-612.5$

 

$=350 \mathrm{~cm}^{2}$

Area of major segment APB = Area of circle − Area of minor segment

$=\pi(\mathrm{OA})^{2}-350$

$=\frac{22}{7} \times(35)^{2}-350$

$=3850-350$

 

$=3500 \mathrm{~cm}^{2}$

Hence, the area of major segment is 3500 cm2