Find the area of the major segment APB of a circle of radius 35 cm and ∠AOB = 90°, as shown in the given figure.
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
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