Question

(2)
In figure 7.29, P is the centre of the
circle of radius 6 cm. Seg QR
is a tangent at Q. If PR = 12,
find the area of the shaded region.
(V3 = 1.73)
Fig. 7.29​

Answer 1

Explanation:

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Answer 2

Explanation:

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Since line drawn from the centre of a circle to the tangent is perpendicular to the tangent.

∴OQ perpendicular to QR

△OQR is a right angled triangle.

∴OR

2

=QR

2

+OQ

2

⇒QR=

OR

2

−OQ

2

=

20

2

−10

2

=

400−100

=

300

=10

3

∴ Area of △ OQR=

2

1

×QR×OQ

=

2

1

×10

3

×10

=86.5cm

2

Also,let ∠QOR=θ

sinθ=

OR

QR

=

20

10

3

θ=60

o

∴ Area of sector OQT=π(10)

2

×

360

60

=

7

22

×100×

360

60

=52.33cm

2

Area of the shaded region = Area of △ OQR - Area of sector OQT

= (86.5−52.33)cm

2

= 34.17cm

2