Question

a chord of a circle of radius 14 cm substance 120 angle at the centre find the area of corresponding major segment of the circle​

Answer 1

a chord of a circle of radius 14 cm substance 120 angle at the centre find the area of corresponding major segment of the circle......

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In ∆ OVS

$\frac{ov}{os} = cos60 \\ \\ \frac{ov}{14} = \frac{1}{2} \\ \\ \boxed{ \underline \bold \red{ ov = > 7}} \\ \\ \frac{sv}{so} = sin60 \\ \\ \frac{sv}{14} = \frac{ \sqrt{3} }{2} \\ \\ \boxed{ \underline \bold \blue{sv = > 7 \sqrt{3} }} \\ \\ st = 2sv \\ \\ st = > 2 \times 7 \sqrt{3} \\ \\ \boxed { \underline \bold \pink{ st = > 14 \sqrt{3} }}$

$\boxed{ \underline \bold \green{area \: of \: triangle \: ost= \frac{1}{2} \times ov \times st}} \\ \\ = > \frac{1}{2} \times 7 \times 14 \sqrt{3} \\ \\ = > 49 \sqrt{3} \\ \\ = > 49 \times 1.73 \\ \\ = > 84.77 {cm}^{2}$

$\boxed{ \underline \bold \orange{area \: of \: triangle \: ost \: = > 84.77 {cm}^{2} }}$

$\boxed{ \underline \bold \red{area \: of \: sector \: osut = > \frac{theta}{360} \times \pi {r}^{2} }} \\ \\ = > \frac{120}{360} \times \frac{22}{7} \times 14 \times 14 \\ \\ = > 205.33 {cm}^{2}$

$\boxed{ \underline \bold \blue{area \: of \: sector \: osut = > 205.33 {cm}^{2} }}$

Area of sector SUT = Area of sector OSUT - ∆ OST

$= > 205.33 - 84.77 \\ \\ = > 120.56 {cm}^{2}$

$\boxed{ \underline \bold \green{area \: of \: segment \: sut = > 120.56 {cm}^{2} }}$

Hope it's helps you