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.

Asked by Pragya Singh | 1 year ago |  117

1 Answer

Solution :-

Here, P is in the semi-circle and so,

P = 90°

So, it can be concluded that QR is hypotenuse of the circle and is equal to the diameter of the circle.

QR = D

Using Pythagorean theorem,

QR= PR2+PQ2

Or, QR= 72+242

QR= 25 cm = Diameter

Hence, the radius of the circle = \( \dfrac{25}{2}\) cm

Now, the area of the semicircle = \( \dfrac{πR^2}{2}\)

= (\( \dfrac{22}{7}\))×(\( \dfrac{25}{2}\))×(\( \dfrac{25}{2}\dfrac{1}{2}\)) cm2

\( \dfrac{13750}{56} \) cm= 245.54 cm2

Also, area of the ΔPQR = \(\dfrac{1}{2}\)×PR×PQ

=(\( \dfrac{1}{2}\))×7×24 cm2

= 84 cm2

Hence, the area of the shaded region = 245.54 cm2-84 cm2

= 161.54 cm2

Answered by Abhisek | 1 year ago

Related Questions

Prove that the area of a circular path of uniform width hsurrounding a circular region of radius r is πh(2r + h).
 

Class 10 Maths Areas Related to Circles View Answer

A road which is 7 m wide surrounds a circular park whose circumference is 352 m. Find the area of the road.
 

Class 10 Maths Areas Related to Circles View Answer

The outer circumference of a circular race-track is 528 m. The track is every­where 14 m wide. Calculate the cost of levelling the track at the rate of 50 paise per square metre 
 

Class 10 Maths Areas Related to Circles View Answer

A circular pond is of diameter 17.5 m. It is surrounded by a 2 m wide path. Find the cost of constructing the path at the rate of ₹25 per square metre (Use π = 3.14)

Class 10 Maths Areas Related to Circles View Answer

A path of width 3.5 m runs around a semi­circular grassy plot whose perimeter is 72 m. Find the area of the path.

Class 10 Maths Areas Related to Circles View Answer