Question

find the area of segment if minor segment with radius 6cm chord subtends an angle of 60degree​

Answer 1

24.01cm²

Given:-

• Radius of circle = 6cm

• Angle subtended by minor segment is of 60°

To find:-

• Area of minor segment

Solution:-

★ $\sf Area_{(minor\;segment)} = Area_{(major\; segment)} - Area_{(Triangle)}$

$\sf Area_{(minor\;segment)} = \pi r^2 \dfrac{ \theta}{ 360^\circ} - \dfrac{1}{2} r^2 \sin{ \theta}$

$:\implies\sf \dfrac{22}{7} \times 6 \times 6 \times \dfrac{60^\circ}{ 360^\circ} - \dfrac{1}{2} \times 6 \times 6 \times \sin{60}$

$:\implies\sf \dfrac{132}{7} - 9 \sqrt{3}$

Put value of $\sqrt{3}$ = 1.73

$:\implies\sf \dfrac{132}{7} - 9 \times 1.73$

$:\implies\sf {\underline{24.01\;cm^2}}$

$\rule{200}3$

Answer 2

Answer:-

${ \underline {\sf{ \: Area \: of \: minor \: segment }}}= \\ \\ \longmapsto \sf \: \frac{πr^2θ}{360} - \frac{1}{2} r^2 sinθ \\ \\ \sf \longmapsto\frac{22}{7} \frac{6 \times 6 \times 60}{360} - \frac{1}{2} \times 12 \sin60^{\circ}$

$\\ \\ \sf \longmapsto\frac{22}{7} \frac{6 \times 6 \times 60}{360} - \frac{1}{2} \times 12 \sin60^{\circ}$

$\sf \longmapsto \dfrac{132}{7} -9 \sqrt{3}$

${ \red{ \sf \: Putting \: the \: value \: of \sqrt{3} =1.73}}$

$\longmapsto \sf24.01 {cm}^{2}$