Area of combination of figures : circle based

Chapter 12 Class 10 Areas related to Circles
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### Transcript

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 