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Ex 12.2, 1 - A park, in shape of a quadrilateral ABCD - Ex 12.2

Ex 12.2, 1 - Chapter 12 Class 9 Herons Formula - Part 2
Ex 12.2, 1 - Chapter 12 Class 9 Herons Formula - Part 3
Ex 12.2, 1 - Chapter 12 Class 9 Herons Formula - Part 4

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Ex 12.2, 1 A park, in the shape of a quadrilateral ABCD, has ∠C = 90°, AB = 9 m, BC = 12 m, CD = 5 m and AD = 8 m. How much area does it occupy? Total Area of park = Area ∆ABD + Area ∆BCD Area of ∆BCD Since DC = 5 m and BC = 12 m, ∠C = 90°, Then ΔBCD is a right angled triangle Area of ∆BCD = 1/2 x base x height = 1/2 × 12 × 5 m2 = 30 m2 Area of ∆ABD Area of triangle = √(s(s−a)(s−b)(s −c)) Here, s is the semi-perimeter, and a, b, c are the sides of the triangle Here, a = 8m, b = 9m, c = BD Since, ∠C = 90° So,applying Pythagoras theorem, BD2 = BC2 + CD2 BD2 = (12)2 + (5)2 BD2 = 144 + 25 BD2 = 169 So, BD = √169 Hence, c = BD = 13 m s = (𝑎 + 𝑏 + 𝑐)/2 Area of Δ ABD = √(𝑠(𝑠 −𝑎)(𝑠 −𝑏)(𝑠 −𝑐)) Putting a = 8m, b = 9m, c = 13 m & s = 15 m Area of Δ ABD = √(15(15−8)(15−9)(15−13))m2 = √(15×7×6×2) m2 = √((3×5)×7×6×2) m2 = √((3×2)×6×(7×5)) m2 = √((6)×6×(35)) m2 = √((6)2× (35)) m2 = √62 × √35 = 6√35 m2 = (6 × 5.91) m2 = 35.46 m2 Area of the park = Area of ΔABD + Area of ΔBCD = 35.46 + 30 m2 = 65.46 m2 = 65.5 m2 (approx.)

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.