Question

find the area of the shadded region , given that radius of the circle is 10cm and triangle OAB is equilateral triangle , (shadded region is circle other than the triangle OAB)​

Answer 1

Answer:

A

8

11

B

8

3

Given,

A(6,3),B(−3,5),C(4,−2)andD(x,3x)

By using area of triangle =

2

1

∣x

1

(y

2

−y

3

)+x

2

(y

3

−y

2

)+x

3

(y

1

−y

2

)∣

Area of ΔDBC =

∣

∣

∣

∣

∣

2

(x)(5+2)+(−3)(−2−3x)+4(3x−5)

∣

∣

∣

∣

∣

=

∣

∣

∣

∣

∣

2

7x+6+9x+12x−20

∣

∣

∣

∣

∣

=

∣

∣

∣

∣

∣

2

28x−14

∣

∣

∣

∣

∣

Area of ΔABC =

∣

∣

∣

∣

∣

2

(6)(5+2)+(−3)(−2−3)+4(3−5)

∣

∣

∣

∣

∣

=

∣

∣

∣

∣

∣

2

42+15−8

∣

∣

∣

∣

∣

=

2

49

Given:

AreaofABC

AreaofDBC

=

2

1

2

49

∣

∣

∣

∣

∣

2

28x−14

∣

∣

∣

∣

∣

=

2

1

∣28x14∣=

2

49

28x14=±

2

49

Taking a positive sign

28x14=

2

49

56x−28=49

56x=77

x=

56

77

=

8

11

Taking a negative sign

28x14=−

2

49

56x−28=−49

56x=−21

x=

56

21

=

8

3

So, x=

8

11

or x=

8

3

Answer 2

Given:

The radius of the circle = $10$ cm

$\triangle OAB$ is an equilateral triangle

To find:

The area of the shaded region

Step-by-step explanation:

Area of the shaded region = Area of the major sector - Area of an equilateral triangle

Area of an equilateral triangle = $\frac{\sqrt{3} }{4} \;a^{2} \;cm^{2}$

⇒ $\frac{\sqrt3}4\;\times\;10\;\times\;10\;=\;43.3\;cm^2$

Area of major sector = $\frac\theta{360^\circ}\;\times\;\mathrm\pi\;\times\;\mathrm r^2$

⇒ $\frac{300^\circ}{360^\circ}\;\times\;\frac{22}7\;\times\;100\;=\;261.90\;\mathrm{cm}^2$

Area of the shaded region = $261.90\;\mathrm{cm}^2\;-\;43.3\;\mathrm{cm}^2$

⇒ Area of the shaded region = $218.60\;cm^{2}$

Answer:

Therefore, the area of the shaded region is $218.60\;cm^{2}$