Question

The semicircle is divided into two sectors whose angles are in the ratio 4:5. find the ratio of theirs areas .

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Answer 1

Given :-

The semicircle is divided into two sectors whose angles are in the ratio 4:5

To Find :-

The ratio of theirs areas

$\large\underline{\bold{Solution-}}$

$\begin{gathered}\begin{gathered}\bf \:Let \: sector \: central \: angle \: be - \begin{cases} &\sf{4x} \\ &\sf{5x} \end{cases}\end{gathered}\end{gathered}$

and

$\bf \: Let \sf \: radius \: of \: semicircle \: be \: r \: units.$

We know,

$\bf \: Area_{(sector)} = \dfrac{ \theta}{360 \degree} \times \pi \: {r}^{2}$

So,

$\bf \: \dfrac{Area_{(sector \: 1)}}{Area_{(sector \: 2)}} = \sf \: \dfrac{\dfrac{ 4 \: \cancel{x}}{ \cancel{360 \degree}} \times \cancel{\pi \: {r}^{2}}}{\dfrac{ 5 \: \cancel{x}}{\cancel{360 \degree}} \times\cancel{ \pi \: {r}^{2}}}$

$\bf \: \dfrac{Area_{(sector \: 1)}}{Area_{(sector \: 2)}} = \sf \: \dfrac{4}{5}$

Additional Information :-

$\boxed{ \bf{ \: Area_{(sector)} = \dfrac{ \theta}{360 \degree} \times \pi \: {r}^{2}}}$

$\boxed{ \bf{Area_{(sector)} = \dfrac{1}{2} lr}}$

$\boxed{ \bf{ \: Perimeter_{(sector)} = 2r + l}}$

$\boxed{ \bf{ \: Length \: of \: arc_{(sector)} = \dfrac{ \theta}{360 \degree} \times 2\pi \: {r}}}$

$\boxed{ \bf{ \: Area_{(major - sector)} = \dfrac{ (360 \degree - \theta)}{360 \degree} \times \pi \: {r}^{2}}}$