Question

Find the area of the major segment APB, in Fig 5, of a circle of radius 35 cm and AOB = ${90}^{\circ }$
$[Use\pi =\frac{22}{7}]$

Answer 1

Radius of the given circle = 35 cm

Area of the minor segment = Area of sector OAB - area of $\Delta$AOB
Area of section $OAB=\frac{\theta }{{360}^{\circ }}\pi r^2$
$=\frac{{90}^{\circ }}{{360}^{\circ }}\times \frac{22}{7}\times (35cm{)}^2$
$=\frac{1}{4}\times \frac{22}{7}\times 1225c{m}^2$
$=962.5c{m}^2$
$Areaof\Delta AOB=\frac{1}{2}\times OA\times OB$
$=\frac{1}{2}\times 35cm\times 35c{n}^2$
$=612.5c{m}^2$
$\therefore$ Area of the minor segment = 962.5 $c{m}^2$ - 612.5 $c{m}^2$ = 350 $c{m}^2$
Area of the major segment = Area of the circle - area of the minor segment
Area of the circle $=\pi r^2=\frac{22}{7}\times (35cm{)}^2=3850c{m}^2$
Thus, the area of the major segment APB = 3850 $c{m}^2$ - 350 $c{m}^2$ = 3500 $c{m}^2$