Question

Question 4
A chord of a circle of radius 10 cm subtends a right angle at the centre. Find the area of the corresponding: (i) minor segment (ii) major sector. $(Use\pi =3.14)$

Answer 1

Radius of the circle = 10 cm
Major sector is making ${360}^{\circ }-{90}^{\circ }={270}^{\circ }$
Area of the sector making angle ${270}^{\circ }$
$=\left(\frac{{270}^{\circ }}{{360}^{\circ }}\right)\times \pi r^{2}c{m}^2$
$=\left(\frac{3}{4}\right)\times {10}^2\pi =75\pi c{m}^2$
$=75\times 3.14c{m}^2=235.5c{m}^2$

$\therefore$ Area of the major sector $=235.5c{m}^2$
Height of $\Delta AOB=OA=10cm$
Base of $\Delta AOB=OB=10cm$
$Areaof\Delta AOB=\frac{1}{2}\times OA\times OB$
$=\frac{1}{2}\times 10\times 10=50c{m}^2$
Minor sector is making ${90}^{\circ }$
Area of the sector making angle ${90}^{\circ }$
$=\left(\frac{{90}^{\circ }}{{360}^{\circ }}\right)\times \pi r^{2}c{m}^2$
$=\left(\frac{1}{4}\right)\times {10}^2\pi =25\pi c{m}^2$
$=25\times 3.14c{m}^2$
$=78.5c{m}^2$
Area of the minor segment = Area of the sector making angle ${90}^{\circ }-Areaof\Delta AOB$
$=78.5c{m}^2-50c{m}^2=28.5c{m}^2$