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.


​

Answer 1

Answer:

• Shaded region area = 161.54cm².

Step-by-step explanation:

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,

➡QR² = PR²+PQ²

➡ Or, QR² = 7² +24²

➡ QR= 25 cm = Diameter

Hence, the radius of the circle = 25/2 cm

Now, the area of the semicircle = (πR²)/2

➡ (22/7)×(25/2)×(25/2)/2 cm²

➡ 13750/56 cm² = 245.54 cm²

Also, area of the ΔPQR = ½×PR×PQ

➡ (½)×7×24 cm²

➡ 84 cm²

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

➡ The area of the shaded region = 161.54 cm².

Answer 2

$\huge{\textbf{\textsf{{\color{navy}{Aɴ}}{\purple{sᴡ}}{\pink{ᴇʀ}}{\color{pink}{:}}}}}$

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,

QR2 = PR2+PQ2

Or, QR2 = 72+242

QR= 25 cm = Diameter

Hence, the radius of the circle = 25/2 cm

Now, the area of the semicircle = (πR2)/2

= (22/7)×(25/2)×(25/2)/2 cm2

= 13750/56 cm2 = 245.54 cm2

Also, area of the ΔPQR = ½×PR×PQ

=(½)×7×24 cm2

= 84 cm2

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

= 161.54 cm2