Question

. Find the area of shaded region in the figurebelow if PQ = 16 cm, PR = 12 cm and O iscentre of the circle. [Take, it = 314) ]CBSE 2011OR​

Answer 1

EXPLANATION.

To find area of shaded region.

PQ = 16 cm. [Given].

PR = 12 cm. [Given].

O is Centre of the circle.

As we know that,

Formula of Pythagoras Theorem.

⇒ H² = P² + B².

Hypotenuse > Perpendicular > Base.

Using this formula in the equation, we get.

In RPQ.

⇒ (RQ)² = (PQ)² + (RP)².

⇒ (RQ)² = (16)² + (12)².

⇒ (RQ)² = 256 + 144.

⇒ (RQ)² = 400.

⇒ RQ = √400.

⇒ RQ = 20 cm.

RQ = Diameter of circle.

Radius of circle = 10 cm.

Area of shaded region = Area of semicircle - Area of triangle.

⇒ πr²/2 - 1/2 x b x h.

⇒ [3.14 x (10)²]/2 - 1/2 x 12 x 16.

⇒ [3.14 x 100]/2 - 6 x 16.

⇒ 157 - 96 = 61 cm².

Area of shaded region = 61 cm².

Answer 2

Given Data :

PQ = 16cm

PR = 12 cm

‘ O ’ is the centre of the circle

from the figure,

RQ is the Diameter of the circle as well as Hypotenuse of the Triangle

OR = OQ = Radius of the circle

To Find :

Area of The Shaded Region .

Solution :

It is a Right Angled Triangle .

.°. Hypotenuse > Perpendicular > Base .

H² = B² + P²

⇒ H² = (12)² + (16)²

⇒ H² = 400

⇒ H = 20cm

⇒ RQ = 20cm

⇒ Radius = 10cm

Area of the Circle = πr²

⇒ Area = 3.14 × (10)²

⇒ Area = 3.14 × 100

⇒ Area = 314 cm²

then,

Area of Semi circle = 314/2

⇒ Area of semicircle = 157 cm²

Area of Triangle = ½ × b × h

⇒ Area = ½ × 12 × 16

⇒ Area = 96cm²

Now Finding: Area of Shaded Region .

Area of Shaded Region = Area of semicircle - Area of of Triangle

⇒ 157 - 96

⇒ 61 cm²

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Area of Shaded Region = 61cm²