Question

area of the sectors of 1540 square metre each subtends an angle of 50 degree at the centre of the circle find the radius.
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Answer 1

Given :

•Area of the sector =1540m²

• angle =50degree

To Find :

• The radius =?

Formula :

$\red \bigstar \boxed{ \tt{area \: of \: sector \: = \frac{ \theta}{360} \times \pi {r}^{2} }}$

Solution :

According to the given conditions :

$\\ \implies \sf \: area \: of \: sector\: = \frac{ \theta}{360} \times \pi {r}^{2} \\ \\ \implies \sf \: area \: of \: sector \: = 1540 \\ \\ \implies \sf \: \frac{ \theta}{360} \times \pi {r}^{2} = 1540 \\ \\ \implies \sf \: \frac{50}{360} \times \frac{22}{7} {r}^{2} = 1540 \\ \\ \implies \sf \: {r}^{2} = \frac{1540 \times 3 \times 7}{5 \times 22} \\ \\ \implies \sf \: {r}^{2} = 14 \times 36 \times 7 \\ \\ \implies \sf \: {r}^{2} = 14 \times 36 \times 7 \\ \\ \implies \sf \: r = \sqrt{14 \times 36 \times 7} \\ \\ \implies \sf \: r = \sqrt{2 \times 7 \times 6 \times 6 \times 7} \\ \\ \implies \sf \: r = 7 \times 6 \sqrt{2} \\ \\ \implies \sf \: r = 42 \sqrt{2} \\ \\ \implies \boxed{ \sf{r = 42 \sqrt{2} }} \\ \\ \\ \red \bigstar \sf \: thus \: the \: radius \: of \: the \: circle = 42 \sqrt{2}$