Ex 12.3, 1
Find the area of the shaded region in figure, if PQ = 24 cm, PR = 7 cm and O is the centre of the circle.
Area of shaded region
= Area of semicircle – Area of ΔPQR
Since , QR is diameter,
It forms a semicircle.
We know that angle in a semicircle is a right angle.
Hence , ∠ RPQ = 90°
Hence, ΔRPQ is right triangle
Now , as per Pythagoras theorem
(Hypotenuse)2 = (Height)2 + (Base)2
(QR)2 = (PQ)2 + (PR)2
Putting values
(QR)2 = (24)2 + (7)2
(QR)2 = 576 + 49
(QR)2 = 625
QR = √625
QR = √(25×25)
QR = √((25)2)
QR = 25
Here, QR = diameter of the circle = 25 cm
So, radius = 𝑄𝑅/2 = 25/2 cm
Area of circle = 𝜋𝑟2
Area of semicircle = 1/2×area of circle
= 1/2×𝜋𝑟2
= 1/2×22/7×(25/2)^2
= 1/2×22/7×25/2×25/2
= (11 × 25 × 25)/28
= 6875/28 cm2
Area of Δ PQR
Δ PQR is a right angled triangle with Base PQ & Height PR
Area of Δ PQR = 1/2 × Base × Height
= 1/2×PQ×PR
= 1/2×24×7
= 12×7
= 84 cm2
Area of shaded region
= Area of semicircle – Area of ΔPQR
= 6875/28− 84
= (6875 − (84)(28))/28
= (6875 − 2352)/28
= 4523/28 cm2
Here, area of shaded region = 4523/28 cm2

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.