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Ex 12.2, 2 - Find area of a quadrilateral ABCD in which - Ex 12.2

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

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Ex 12.2, 2 Find the area of a quadrilateral ABCD in which AB = 3 cm, BC = 4 cm, CD = 4 cm, DA = 5 cm and AC = 5 cm. Area of Quadrilateral = Area of ΔABC + Area of ΔADC Area of ΔABC 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 = 3 , b = 4, c = 5 S = (𝑎 + 𝑏 + 𝑐)/2 Area of Δ ABC = √(𝑠(𝑠 −𝑎)(𝑠 −𝑏)(𝑠 −𝑐)) Putting a = 3 , b = 4, c = 5 & s = 6 = √(6(6 −3)(6 −4)(6 −5)) cm2 = √(6×3×2×1) cm2 = √(6×6) cm2 = √((6)2) = 6 cm2 ∴ Area of ΔABC = 6 cm2 Area of ΔADC 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 = 5 , b = 4, c = 5 S = (𝑎 + 𝑏 + 𝑐)/2 Area of Δ ADC = √(𝑠(𝑠 −𝑎)(𝑠 −𝑏)(𝑠 −𝑐)) Putting a = 5 , b = 4, c = 5 & s = 7 = √(7(7 −5)(7 −4)(7 −5)) cm2 = √(7×2×3×2) = √((2×2)×7×3) = √((22)×7×3) = √((2)2) × √(7×3) = 2 × √21 = 2√21 cm2 = (2 × 4.58) = 9.16 cm2 Thus, Area Δ ADC = 9.16 cm2 Area of Quadrilateral = Area of ΔABC + Area of ΔADC = (6 + 9.16) cm2 = (6 + 9.16) cm2 = 15.16 cm2 -= 15.2 cm2 (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.