# Ex 8.2, 4 - Class 12

Last updated at May 29, 2018 by Teachoo

Last updated at May 29, 2018 by Teachoo

Transcript

Ex 8.2 , 4 Using integration find the area of region bounded by the triangle whose vertices are (– 1, 0), (1, 3) and (3, 2) Area of ∆ formed by points A(– 1, 0), B(1, 3) and C(3, 2) Step 1: Draw the figure Area ABD Area ABD= −11𝑦 𝑑𝑥 𝑦→ equation of line AB Equation of line between A(–1, 0) & B(1, 3) is 𝑦 − 0𝑥 − (−1)= 3 − 01 − (−1) 𝑦𝑥 + 1= 32 y = 32 (x + 1) Area ABD = −11𝑦 𝑑𝑥 = −11 32 𝑥+1 dx = 32 −11 𝑥+1 dx = 32 𝑥22+𝑥−11 = 32 122+1− −122+(−1) = 32 32− −12 = 32 × 2 = 3 Area BDEC Area BDEC = 13𝑦 𝑑𝑥 𝑦→ equation of line BC Equation of line between B(1, 3) & C(3, 2) is 𝑦 − 3𝑥 − 1= 2 − 33 − 1 𝑦 − 3𝑥 − 1= −12 2(y – 3) = –1(x – 1) 2y – 6 = –x + 1 2y = – x + 7 y = 12 (–x + 7) Area BDEC = 13𝑦 𝑑𝑥 = 13 12 −𝑥+7𝑑x = 12 −𝑥22+7𝑥13 = 12 −322+7(3)− −122+7(1) = 12 −92+21+ 12−7 = 12 −9 + 12+14 = 12 −82+14 = 12 −4+14 = 102 = 5 Area ACE Area ACE= −13𝑦 𝑑𝑥 𝑦→ equation of line AC Equation of line between A(–1, 0) & C(3, 2) is 𝑦 − 0𝑥 − (−1)= 2 − 03 − (−1) 𝑦𝑥 + 1= 24 y = 12 (x + 1) Area ACE = −13𝑦 𝑑𝑥 = −13 12 𝑥+1𝑑x = 12 𝑥22+𝑥−13 = 12 322+3− −122−1 = 12 92+3− 12−1 = 12 92+ 12+3+1 = 12 4+4 = 4 Hence, Required area = Area ABD + Area BDEC − Area ACE = 3 + 5 − 4 = 4 units

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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.