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Ex 12.3

Ex 12.3, 1
Deleted for CBSE Board 2023 Exams
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Ex 12.3, 2 Deleted for CBSE Board 2023 Exams

Ex 12.3, 3 Deleted for CBSE Board 2023 Exams

Ex 12.3, 4 Important Deleted for CBSE Board 2023 Exams

Ex 12.3, 5 Deleted for CBSE Board 2023 Exams

Ex 12.3, 6 Deleted for CBSE Board 2023 Exams

Ex 12.3, 7 Deleted for CBSE Board 2023 Exams

Ex 12.3, 8 Important Deleted for CBSE Board 2023 Exams

Ex 12.3, 9 Deleted for CBSE Board 2023 Exams

Ex 12.3, 10 Deleted for CBSE Board 2023 Exams

Ex 12.3, 11 Deleted for CBSE Board 2023 Exams

Ex 12.3, 12 Important Deleted for CBSE Board 2023 Exams

Ex 12.3, 13 Important Deleted for CBSE Board 2023 Exams

Ex 12.3, 14 Deleted for CBSE Board 2023 Exams

Ex 12.3, 15 Important Deleted for CBSE Board 2023 Exams

Ex 12.3, 16 Important Deleted for CBSE Board 2023 Exams

Chapter 12 Class 10 Areas related to Circles

Serial order wise

Last updated at May 29, 2018 by Teachoo

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