Question

Find the area of the shaded region in Fig. 12.19, if PQ = 24 cm, PR = 7 cm and O is the centre of the circle.

[ NCERT ]​

Answer 1

Answer:

area of shaded region = area of semicircle - area of ∆ pQR

since QR is diameter

it from semicircle

We know that angle in a semicircle is a right angle

hence ,<PRQ = 90°

hence <RPQ= 90 °

now as per phytogoras theoram

( QR) square = ( PQ) square + ( pR) square

QR = 25

radius = 25/2

area of semicircle = 1/3 π r square

= 6875/28 cm square

area of PQR = 1/2 base x height

= 1/2 x PQ x PR

= 84 cm square

area of shaded region = area of semicircle - area of ∆ PQR

= 6875/28 - 84

= 4523/28 cm square

area of shaded region = 4523/28 cm square

Answer 2

S O L U T I O N :

Given,

• PQ = 24 cm, PR = 7 cm.
• Point O is the centre of the circle.

To Find,

• The area of the shaded region.

Explanation,

Point P is in the semi - circle,

=> P = 90°

Here, Hypotenuse of the circle = Diameter of the circle.

=> QR = Diameter

In ΔQPR,

• Angle P is 90°.

According to the Pythagoras theorem,

=> QR² = PR² + PQ²

=> QR² = 7² + 24²

=> QR² = 625

=> QR = 25 cm

We know that,

Radius = Diameter/2

=> 25/2

=> 12.5 cm

As we know that,

Area of the semicircle = πR²/2

=> 3.14 × (12.5)²/2

=> 3.14 × 156.25/2

=> 490.625/2

=> 245.3125 cm²

Now,

Area of ΔPQR = ¹/2 × Base × Height

=> ¹/2 × 7 × 24

=> 7 × 12

=> 84 cm²

So,

Area of shaded region = Area of the semicircle - Area of ΔPQR

=> 245.3125 - 84

=> 161.3125 cm²

Therefore,

The area of shaded region is 161.3125 cm².