# Misc 14

Last updated at March 11, 2017 by Teachoo

Last updated at March 11, 2017 by Teachoo

Transcript

Misc 14 Using the method of integration find the area of the region bounded by lines: 2 + = 4, 3 2 =6 and 3 +5=0 Step 1: Make Figure 2 + = 4 3 2 = 6 3 + 5 = 0 Step 2: Find intersecting Points A & B Point A Point A is intersection of lines x 3y + 5 = 0 & 2x + y = 4 Now, x 3y + 5 = 0 x = 3y 5 Putting x = 3y 5 in 2x + y = 4 2(3y 5) + y = 4 6y 10 + y = 4 7y = 14 y = 2 Putting y = 2 in x 3y + 5 = 0 x 3(2) + 5 = 0 x 6 + 5 = 0 x = 1 So, point A (1, 2) Point B Point B is intersection of lines x 3y + 5 = 0 & 3x 2y = 6 Now, x 3y + 5 = 0 x = 3y 5 Putting x = 3y 5 in 3x 2y = 6 3(3y 5) 2y = 6 9y 15 2y = 6 7y = 21 y = 3 Putting y = 3 in x 3y + 5 = 0 x 3(3) + 5 = 0 x 9 + 5 = 0 x = 4 So, point A (4, 3) Finding area Area ABED Area ABED = 1 4 Equation of AB 3 +5=0 +5=3 + 5 3 = = + 5 3 Therefore, Area ABED = 1 4 +5 3 = 1 3 1 4 +5 = 1 3 2 2 +5 1 4 = 1 3 4 2 2 +5.4 1 2 2 +5.1 = 1 3 8+20 1 2 5 = 1 3 45 2 = 15 2 Area ACD Area ACD = 1 2 Equation of line AC 2 + =4 =4 2 Area ACD = 1 2 4 2 = 4 2 2 2 1 2 = 4 2 1 2 = 4.2 2 2 4.1 1 2 = 8 4 4+1 = 1 Area CBE Area CBE = 2 4 Equation of line BC 3 +2 =6 3 6=2 3 6 2 = = 3 6 2 Therefore, Area CBE = 2 4 3 6 2 = 1 2 2 4 3 6 = 1 2 3 2 2 6 2 4 = 1 2 3.4 2 2 6.4 3.2 2 2 6.2 = 1 2 24 24 6+12 =3 Hence Area Required = Area ABED Area ACD Area CBE = 15 2 1 3 = 15 2 4 = 15 8 2 = 7 2 square units

Class 12

Important Question for exams Class 12

- Chapter 1 Class 12 Relation and Functions
- Chapter 2 Class 12 Inverse Trigonometric Functions
- Chapter 3 Class 12 Matrices
- Chapter 4 Class 12 Determinants
- Chapter 5 Class 12 Continuity and Differentiability
- Chapter 6 Class 12 Application of Derivatives
- Chapter 7 Class 12 Integrals
- Chapter 8 Class 12 Application of Integrals
- Chapter 9 Class 12 Differential Equations
- Chapter 10 Class 12 Vector Algebra
- Chapter 11 Class 12 Three Dimensional Geometry
- Chapter 12 Class 12 Linear Programming
- Chapter 13 Class 12 Probability

About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.