Question

hello

give an answer fast​

Answer 1

$\huge \mathbb \fcolorbox{black}{aqua}{ANSWER}$

↝ Area of shaded region = Area of semi circle - Area of ∆PQR

↝ As QR is diameter , it forms semicircle .

↝ As we know angle in the semicircle is right angle . Hence , ∠RPQ is 90°

↝ Therefore , ∆PQR is right angled triangle

By Pythagoras Theorem

↝ ${(Hypotenuse)}^{2} = {(Base)}^{2} + {(Height)}^{2}$

↝${(QR)}^{2} = {(PR)}^{2} + {(PQ)}^{2}$

↝ ${(QR)}^{2} = {(24)}^{2} + {(7)}^{2}$

↝ ${(QR)}^{2} = 576 + 49$

↝ ${(QR)}^{2} = 625$

↝ QR = √625

↝ QR = 25 cm

↝ Diameter of the circle = QR = 25 cm

↝ Radius = $\frac{25}{2}$

↝ Area of semicircle = $\frac{1}{2} × π {r}^{2}$

↝ Area of semicircle = $\frac{1}{2}×\frac{22}{7} × \frac{25}{2} × \frac{25}{2}$

↝ = $\frac{11 × 25 × 25 }{28}$

↝ = $\frac{6875}{28}$

↝ Area of semicircle = $\frac{6875}{28}$

_____________________________________

In ∆PQR

↝ Area of ∆PQR = $\frac{1}{2} × base × height$

↝ = $\frac{1}{2} × PQ × PR$

↝ = $\frac{1}{2} × 24 × 7$

↝ Area of ∆ PQR = 84 ${cm}^{2}$

_______________________________________

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

↝ = $\frac{6875}{28} - 84$

↝ = $\frac{6875 - 84 × 28}{28}$

↝ = $\frac{4523}{28}$

↝ Area of shaded region = $\frac{4523}{28} {cm}^{2}$

_______________________________________

$<marquee>$✯✯ Hope It Helps✯✯

Answer 2

HAPPY BIRTHDAY BHAII

HOPE UR UPCOMING DAYS BRINGS U LOADS OF HAPPINESS...

ALWAYS BE HAPPY BHAI...

AND MAY GOD GIFT U WHAT U WANT ..

A VERY HAPPY BIRTHDAY BHAI ⭐⭐