What is the area of the minor segment of a circle of radius 14 cm when its central angle is 60?

Find the area of the minor segment of a circle of radius 14 cm, when its central angle is 60˚. Also find the area of the corresponding major segment.[use π=22/7]

Radius of the circle = 14 cmCentral Angle, 𝜽 = 60°,

Area of the minor segment

`=theta/360^@xxpir^2-1/2r^2sintheta`

`=60^@/360^@xxpixx14^2-1/2xx14^2xxsin60^@`

`=1/6xx22/7xx14xx14-1/2xx14xx14xxsqrt3/2`

`=(22xx14)/3-49sqrt3`

`=(22xx14)/3-(147sqrt3)/3`

`=(308-147sqrt3)/3 cm^2`

Area of the minor segment= `(308-147sqrt3)/3 cm^2`

Concept: Areas of Sector and Segment of a Circle

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Answer

Hint: Analyze the situation with a diagram. Calculate the area of the minor sector of the circle first and then subtract the area of the triangle formed by the chord joining with the center from it.According to the information given in the question, consider the above figure. We have to calculate the area of the minor segment of the circle divided by chord AB. Clearly, the minor segment of the circle is lying at the bottom of chord AB. AB is subtending angle ${60^ \circ }$ at the center and the radius of the circle is 14 cm.Area of circle $ = \pi \times {\left( {14} \right)^2} = 196\pi $Area of sector OAB of circle $ = \dfrac{{{{60}^ \circ }}}{{{{360}^ \circ }}} \times 196\pi = \dfrac{1}{6} \times 196\pi $Now, $\Delta OAB$ is an isosceles triangle because OA and OB are the radius of the circle. We know that:Area of triangle $ = \dfrac{1}{2}ab\sin \theta $, applying this formula $\Delta OAB$, we’ll get:$ \Rightarrow $Area ($\Delta OAB$) $ = \dfrac{1}{2} \times OA \times OB \times \sin {60^ \circ } = \dfrac{1}{2} \times 14 \times 14 \times \dfrac{{\sqrt 3 }}{2}$,$ \Rightarrow $Area ($\Delta OAB$) $ = 49\sqrt 3 $Now the area of minor segment is equal to the difference between the area of sector OAB and area of triangle AOB:$ \Rightarrow $Area of minor segment $ = \dfrac{1}{6} \times 196\pi - 49\sqrt 3 = \dfrac{{196 \times 22}}{{6 \times 7}} - 84.87$,$ \Rightarrow $Area of minor segment $ = 102.67 - 84.87 = 17.8$Thus the area of the minor segment is 17.8 square cm.Note: If $\theta $is the angle subtended by a sector of a circle of radius r at the center, then the area of the sector is given as:$ \Rightarrow $Area of sector $ = \dfrac{\theta }{{{{360}^ \circ }}} \times \pi {r^2}$

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60 Questions 60 Marks 60 Mins

Given:

Radius of circle = 14 cm

Central angle = 60°

Formula used:

Area of minor segment = \(πr^2 × \frac{θ}{360°} - \frac{1}{2}r^2sinθ\)

where r is the radius

and θ is the angle of the segment

Sin 60° = \(\frac{\sqrt3}{2}\)

Calculation:

According to the question,

Area of segment = \(\frac{22}{7} × 14^2 × \frac{60°}{360°} - \frac{1}{2} × 14^2sin60°\)

⇒ \(22× 28× \frac{1}{6} - 98× \frac{\sqrt3}{2}\)

⇒ 102.67 - 49 × 1.73

⇒ 102.67 - 84.87 = 17.79 ≈ 17.8 cm2

∴ The area of minor segment is 17.8 cm2

Alternate Method

Given:

Radius of circle = 14 cm

Central angle = 60°

Formula used:

Area of the circle = πr2

Area of sector = \(πr^2 × \frac{θ}{360°}\) 

Calculation:

According to the diagram,

Area of the circle = πr2

⇒ 22/7 × (14)2 = 616 cm2

Area of the sector AOB = \(πr^2 × \frac{θ}{360°}\)

⇒ \(\frac{22}{7} × 14^2 × \frac{60°}{360°}\)

⇒ 616 × 1/6 = 102.67 cm2

Area of Δ AOB = \(\frac{\sqrt3}{4} \times (AO)^2\)

⇒ Area of the minor segment = Area of the sector AOB - Area of Δ AOB

⇒ 102.67 - 84.87 = 17.79 ≈ 17.8 cm2

∴ The area of minor segment is 17.8 cm2

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