Question 10 - Miscellaneous - Chapter 8 Class 12 Application of Integrals

Last updated at April 16, 2024 by

Misc 13 - Using integration, find area of triangle ABC - Miscellaneous

Misc 13 - Chapter 8 Class 12 Application of Integrals - Part 2
Misc 13 - Chapter 8 Class 12 Application of Integrals - Part 3
Misc 13 - Chapter 8 Class 12 Application of Integrals - Part 4
Misc 13 - Chapter 8 Class 12 Application of Integrals - Part 5 Misc 13 - Chapter 8 Class 12 Application of Integrals - Part 6 Misc 13 - Chapter 8 Class 12 Application of Integrals - Part 7

Transcript

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

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.