Misc 13 - Class 12

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﷮𝑦 𝑑𝑥﷯ 𝑦→ equation of line AB Equation of line between A(2, 0) & B(4, 5) is 𝑦 − 0﷮𝑥 − 2﷯= 5 − 0﷮4 − 2﷯ 𝑦﷮𝑥 − 2﷯= 5﷮2﷯ y = 5﷮2﷯(x – 2) Area ABD = 2﷮4﷮𝑦 𝑑𝑥﷯ = 2﷮4﷮ 5﷮2﷯(x – 2)𝑑𝑥﷯ = 5﷮2﷯ 𝑥﷮2﷯﷮2﷯−2𝑥﷯﷮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﷮𝑦 𝑑𝑥﷯ 𝑦→ equation of line BC Equation of line between B(4, 5) & C(6, 3) is 𝑦 − 5﷮𝑥 − 4﷯= 3 − 5﷮6 − 4﷯ 𝑦 − 5﷮𝑥 − 4﷯= −2﷮2﷯ y – 5 = –1(x – 4) y – 5 = –x + 4 y = 9 – x Area BDEC = 4﷮6﷮𝑦 𝑑𝑥﷯ = 4﷮6﷮ 9−𝑥﷯﷯ 𝑑𝑥 =9 4﷮6﷮𝑑𝑥−﷯ 4﷮6﷮𝑥𝑑𝑥﷯ =9 𝑥﷯﷮4﷮6﷯− 𝑥﷮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﷮𝑦 𝑑𝑥﷯ 𝑦→ equation of line AC Equation of line between A(2, 0) & C(6, 3) is 𝑦 − 0﷮𝑥 − 2﷯= 3 − 0﷮6 − 2﷯ 𝑦﷮𝑥 − 2﷯= 3﷮4﷯ y = 3﷮4﷯ (x – 2) Area ACE = 2﷮6﷮𝑦 𝑑𝑥﷯ = 2﷮6﷮ 3﷮4﷯ 𝑥−2﷯ 𝑑𝑥﷯ = 3﷮4﷯ 𝑥﷮2﷯﷮2﷯−2𝑥﷯﷮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