Misc 13 - Chapter 8 Class 12 Application of Integrals (Term 2)
Last updated at May 29, 2018 by
Last updated at May 29, 2018 by
Transcript
Misc 13 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
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