Question

without using cos and sin find the answer and tell me how to mark an answer as brainliest​

Answer 1

$\rule{260}{2}$

GIVEN :-

◙ Radius of the given circle = 12 cm

◙ Angle subtended by a chord of the circle at the centre = 120°

TO FIND :-

The area of the corresponding segment of the circle.

FORMULA TO BE USED :-

Area of segment =

$\displaystyle{\sf{\{(\frac{\pi}{360^\circ})\times \theta -sin\frac{\theta}{2}.cos\frac{\theta}{2}\}r^2 }}$

SOLUTION :-

Area of the corresponding segment is :-

$\displaystyle{\sf{\{(\frac{\pi}{360^\circ})\times 120^\circ -sin\frac{120^\circ}{2}.cos\frac{120^\circ}{2}\}r^2 }}\\\\\displaystyle{\sf{=\{(\frac{\pi}{3}) -sin60^\circ.cos60^\circ\}(12)^2 }}\\\\\displaystyle{\sf{=\{(\frac{\pi}{3}) -\frac{\sqrt{3} }{2} .\frac{1}{2} \}144 }}\\\\\displaystyle{\sf{=\{(\frac{\pi}{3}) -\frac{\sqrt{3} }{4} \}144 }}\\\\\displaystyle{\sf{=144\times (\frac{\pi}{3})-144\times \frac{\sqrt{3} }{4} }}\\\\\displaystyle{\sf{=48\pi-36\sqrt{3} }}$

[π = 3.14 , √3= 1.73]           ...(Given)

$\displaystyle{\sf{=48\times3.14-36\times1.73}}\\\\\displaystyle{\sf{=150.72-62.28}}\\\\\underbrace {\underline{\boxed{\boxed{\displaystyle{\sf{=88.44\ cm^2}}}}}}$

∴ Hence, the area of the corresponding segment of the circle is 88.44 cm².

$\rule{260}{2}$

Answer 2

Answer :

The area of the required segment is 88.44cm²

Given :

• The radius of the segment is 12cm
• The angle subtended is 120°

Formulae to be used :

$\sf \bullet \: \: Area \: \: of \: \: sector \: \: of \: angle\: \theta = \dfrac{\theta}{360\degree}\times \pi{r}^{2}$

$\sf \bullet \: \: Area \: \: of \: \: a \: \: right \: \: angled \: \: triangle = \dfrac{1}{2}\times bass \times height$

Solution :

Given,

the radius of the segment , r = 12cm

the angle subtended , θ= 120°

Area of the sector AOB is

$\sf ar(AOB) = \dfrac{120}{360}\times 3.14 \times (12)^{2} \\\\ \sf \implies ar(AOB) = \dfrac{1}{3}\times 3.14 \times 12 \times 12 \\\\ \sf ar(AOB) = 3.14 \times 12 \times 4 \\\\ \sf ar(AOB) = 150.72cm^{2}$

So C be the mid point of AB and

$\sf \angle AOC = \angle BOC = 60 \degree$

Let OC be x

So from ∆OCA we have :

$\sf \implies \dfrac{OC}{OA} = \cos60\degree \\\\ \sf \implies \dfrac{OC}{OA} = \dfrac{1}{2} \\\\ \sf \implies \dfrac{x}{12cm} = \dfrac{1}{2} \\\\ \sf \implies x = \dfrac{12}{2} \\\\ \sf \implies x = 6cm$

Now again ,

$\sf \implies \dfrac{AM}{OA} = \sin 60 \degree \\\\ \sf \implies \dfrac{AM}{12cm} = \dfrac{\sqrt{3}}{2} \\\\ \sf \implies AM = 6\sqrt{3}cm$

Therefore , we have

$\sf \implies AB = 2AM \\\\ \sf \implies AB = 2\times 6 \sqrt{3} \\\\ \sf \implies AB = 12\times 1.73\\\\ \sf \implies AB = 20.76cm$

Thus the area of ∆AOB is :

$\sf \implies ar.(\triangle AOB ) = \dfrac{1}{2}\times AB \times OC \\\\ \sf \implies ar.(\triangle AOB) = \dfrac{1}{2}\times 20.76\times6 \\\\ \sf \implies ar.(\triangle AOB ) = 62.28cm^{2}$

Therefore, area of the segment ADB is :

= ar.(segment AOB) - ar.(∆AOB)

= 150.72 - 62.28

= 88.44cm²