**Misc 13**

Last updated at Dec. 8, 2016 by Teachoo

Last updated at Dec. 8, 2016 by Teachoo

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= 24𝑦 𝑑𝑥 𝑦→ equation of line AB Equation of line between A(2, 0) & B(4, 5) is 𝑦 − 0𝑥 − 2= 5 − 04 − 2 𝑦𝑥 − 2= 52 y = 52(x – 2) Area ABD = 24𝑦 𝑑𝑥 = 24 52(x – 2)𝑑𝑥 = 52 𝑥22−2𝑥24 = 52 422−2 ×4− 222−2 ×2 = 52 8−8−2+4 = 52 ×2 =5 Area BDEC Area BDEC = 46𝑦 𝑑𝑥 𝑦→ equation of line BC Equation of line between B(4, 5) & C(6, 3) is 𝑦 − 5𝑥 − 4= 3 − 56 − 4 𝑦 − 5𝑥 − 4= −22 y – 5 = –1(x – 4) y – 5 = –x + 4 y = 9 – x Area BDEC = 46𝑦 𝑑𝑥 = 46 9−𝑥 𝑑𝑥 =9 46𝑑𝑥− 46𝑥𝑑𝑥 =9 𝑥46− 𝑥2246 =9 6−4− 12 62− 42 =9 ×2− 12 36−16 =18−10 = 8 Area ACE Area ACE= 26𝑦 𝑑𝑥 𝑦→ equation of line AC Equation of line between A(2, 0) & C(6, 3) is 𝑦 − 0𝑥 − 2= 3 − 06 − 2 𝑦𝑥 − 2= 34 y = 34 (x – 2) Area ACE = 26𝑦 𝑑𝑥 = 26 34 𝑥−2 𝑑𝑥 = 34 𝑥22−2𝑥26 = 34 622−2 ×6− 222−2 ×2 = 34 362−12−2+4 = 34[8] =6 Hence Area Required = Area ABD + Area BDEC – Area ACE = 5 + 8 – 6 = 7 square units

About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He provides courses for Mathematics from Class 9 to 12. You can contact him here.