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^{2 }= PR^{2}+PQ^{2}

Or, QR^{2 }= 7^{2}+24^{2}

QR= 25 cm = Diameter

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

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

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

= \( \dfrac{13750}{56}
\) cm^{2 }= 245.54 cm^{2}

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

=(\( \dfrac{1}{2}\))×7×24 cm^{2}

= 84 cm^{2}

Hence, the area of the shaded region = 245.54 cm^{2}-84 cm^{2}

= 161.54 cm^{2}

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

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

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

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)

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