Question

Q.10 In the adjoining figure O is the centre of a circle if OA = 10 cm, OB = 15 cm and
angle BOD = 720 then the area of shaded region is( cm)
(A) 5 pai
(B)10 pai
(C) 25 pai
(D) 35 pai​

Answer 1

Step-by-step explanation:

Area of the shaded region = area of the circle −area of the triangle ABC−area of the quadrant COD

Here, in triangle ABC ∠CAB=90

o

(angle in a semicircle)

Hence triangle ABC is right-angled at A

Then applying Pythagoras theorem,

BC

2

=AC

2

+AB

2

BC

2

=24

2

+7

2

BC

2

=625

BC=25 cm

Therefore diameter of circle =25cm

i.e., Radius=

2

25

=12.5 cm

Then area of circle =πr

2

=490.625 cm

2

Area of triangleABC=

2

1

×base×height

=84 cm

2

Area of quadrant =

4

1

×πr

2

=122.65625 cm

2

Therefore area of shaded region =490.625−84−122.65625=283.96875 cm

2

Answer 2

Given :- In the adjoining figure O is the centre of a circle if OA = 10 cm, OB = 15 cm and angle BOD = 72°

To Find :- Area of shaded region ?

Solution :-

We know that,

• Area of sector = (Angle at centre/360°)πr²

So,

→ Radius of sector OAC = 10 cm

→ Angle at centre = 72°

then,

→ Area of sector OAC = (72°/360°) * π * (10)²

→ Area of sector OAC = (1/5) * π * 100

→ Area of sector OAC = 20π cm²

similarly,

→ Radius of sector OBD = 15 cm

→ Angle at centre = 72°

then,

→ Area of sector OBD = (72°/360°) * π * (15)²

→ Area of sector OBD = (1/5) * π * 225

→ Area of sector OBD = 45π cm²

therefore,

→ Shaded Area = Area of sector OBD - Area of sector OAC

→ Shaded Area = 45π - 20π

→ Shaded Area = 25π cm² (C) (Ans.)

Hence, area of shaded region is equal to 25π cm² .

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