Question 10 - Area between curve and curve - Chapter 8 Class 12 Application of Integrals
Last updated at April 16, 2024 by Teachoo
Area between curve and curve
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Area between curve and curve
Last updated at April 16, 2024 by Teachoo
Question 10 Using the method of integration find the area of the triangle ABC, coordinates of whose vertices are A(2, 0), B (4, 5)& C (6, 3) Area of formed by points A(2, 0), B (4, 5)& C (6, 3) Step 1: Draw the figure Area ABD Area ABD= 2 4 equation of line AB Equation of line between A(2, 0) & B(4, 5) is 0 2 = 5 0 4 2 2 = 5 2 y = 5 2 (x 2) Area ABD = 2 4 = 2 4 5 2 (x 2) = 5 2 2 2 2 2 4 = 5 2 4 2 2 2 4 2 2 2 2 2 = 5 2 8 8 2+4 = 5 2 2 =5 Area BDEC Area BDEC = 4 6 equation of line BC Equation of line between B(4, 5) & C(6, 3) is 5 4 = 3 5 6 4 5 4 = 2 2 y 5 = 1(x 4) y 5 = x + 4 y = 9 x Area BDEC = 4 6 = 4 6 9 =9 4 6 4 6 =9 4 6 2 2 4 6 =9 6 4 1 2 6 2 4 2 =9 2 1 2 36 16 =18 10 = 8 Area ACE Area ACE= 2 6 equation of line AC Equation of line between A(2, 0) & C(6, 3) is 0 2 = 3 0 6 2 2 = 3 4 y = 3 4 (x 2) Area ACE = 2 6 = 2 6 3 4 2 = 3 4 2 2 2 2 6 = 3 4 6 2 2 2 6 2 2 2 2 2 = 3 4 36 2 12 2+4 = 3 4 [8] =6 Hence Area Required = Area ABD + Area BDEC Area ACE = 5 + 8 6 = 7 square units