Last updated at May 29, 2018 by Teachoo

Transcript

Ex 12.3, 15 In figure, ABC is a quadrant of a circle of radius 14 cm and a semicircle is drawn with BC as diameter. Find the area of the shaded region. Area of shaded region = Area of semicircle BEC – (Area of quadrant ABDC – Area of Δ ABC) Area quadrant ABDC Radius = 14 cm Area of quadrant ABDC = 1/4 (area of circle) = 1/4×(πr2) = 1/4×22/7×(14)^2 = 1/4×22/7×14×14 = 154 cm2 Area triangle Δ ABC Since ABDC is a quadrant ∠ BAC = 90° Hence Δ ABC is a right triangle with Base AC & Height AB So, Area Δ ABC = 1/2 × Base × Height = 1/2 × AC × AB = 1/2 × 14 × 14 = 7 × 14 = 98 cm2 Area semicircle BEC Here, Diameter = BC Finding BC first Since Δ ABC is a right triangle Applying Pythagoras theorem in right triangle ABC (Hypotenuse)2 = (Height)2 + (Base)2 (BC)2 = (AB)2 + (AC)2 (BC)2 = (14)2 + (14)2 (BC)2 = (14)2 × 2 BC = √("(14)2 × 2" ) BC = √("(14)2" ) × √("2" ) BC = 14 × √("2" ) BC = 14 √2 cm So, Diameter = BC = 14√2 cm Radius = r = 𝐷𝑖𝑎𝑚𝑒𝑡𝑒𝑟/2 = 𝐵𝐶/2 = (14√2)/2 = 7 √2 cm Area of semicircle BEC = 1/2× Area of circle = 1/2 (𝜋𝑟2) = 1/2×22/7×(7√2)^2 = 1/2×22/7×7×7×√2×√2 = 1/2×22×7×2 = 154 cm2 ∴ Area of semi circle BEC = 154 cm2 Area of shaded region = Area of semicircle BEC – (Area of quadrant ABDC – Area of Δ ABC) = 154 – (154 – 98) = 154 – 154 + 98 = 98 cm2 Hence, area of shaded region = 98 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 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.