Question

Col. Find the area of the minor segment of a
circle of radius 14cm, when the angle of
the corresponding sector is goº​

Answer 1

Answer:

Answer

Giventhat−

thecentralangle∠AOBofasector=60

o

andtheradius=14cm.

Tofindout−

theareaofthecorrespondingminorsegment.

Solution−

InΔAOBwehaveOA=OB.

∴∠OAB=∠OBA⟹∠OAB+∠OBA=2∠OABor2∠OBA.

Now∠AOB=60

o

.

∴2∠OAB=180

o

−60

o

⟹∠OAB=60

o

=∠OBA.

(theanglesumpropertyoftriangles).

SoΔAOBisequilateralonewithsidea=14cm.

∴ar.ΔAOB=

4

3

×a

2

=

4

3

×14

2

cm

2

=49

3

cm

2

.

Alsoar.sector=

360

o

θ

×π×r

2

whenθisthecentralangleand

r=radiusofthesector

=

360

o

60

o

×

7

22

×14

2

cm

2

=

3

308

cm

2

.

Nowar.minorsegment=ar.sector−ar.Δ

=

3

308

cm

2

−49

3

cm

2

=(

3

308

−49

3

)cm

2

.

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Answer 2

Given :

• Radius of the circle =14cm

• The angle of the corresponding sector =60 degree

To Find :

• The area of minor segment of the circle =?

Solution :

Let the centre of the circle be O.

Let the radii of the sector be OA and OB.

Given, $\tt{\angle AOB=60 \degree}$

In the ΔAOB ,OA = OB, so it is isosceles.

$\\ \implies\sf{\angle OAB= \angle OBA =\dfrac{ 180- \angle AOB}{2}=\dfrac{180-60}{2}=60 \degree}$

Hence , ΔAOB is Equilateral.

Hence OA=OB=14cm

$\\ \implies\sf{Area \: of \: ΔAOB=\dfrac{\sqrt{3}}{4}R^{2}}$

$\\ \\ \implies\sf{Area \: of \: sector \: ΔAOB=\pi R^{2} \times \dfrac{60 \degree}{360 \degree}}$

$\\ \\ \implies\sf{Area \: of \: sector \: ΔAOB=\dfrac{\pi R^{2}}{6}}$

$\\ \\ \implies\sf{Area \: of \: minor \: segment \: AB=\dfrac{\pi R^{2}}{6}-\dfrac{\sqrt{3}}{4}R^{2}}$

$\\ \\ \implies\sf{Area \: of \: minor \: segment \: AB=\dfrac{22}{7} \dfrac{(14)^{2}}{6} -\dfrac{\sqrt{3}}{4} (14)^{2} cm^{2}}$

$\\ \\ \implies\sf{ Area \: of \: minor \: segment \: AB=17.75cm^{2}}$

$\\ \red\bigstar\boxed{\bf{Area \: of \: minor \: segment \: AB=17.75cm^{2}}}$