# Ex 7.3, 5

Last updated at May 29, 2018 by Teachoo

Last updated at May 29, 2018 by Teachoo

Transcript

Ex 7.3 , 5 You have studied in Class IX, (Chapter 9, Example 3), that a median of a triangle divides it into two triangles of equal areas. Verify this result for ABC whose vertices are A(4, 6), B(3, 2) and C(5, 2) Let ABC be as shown in the figure Let AD be the median which divides BC into two equal parts, BD & CD Hence, Coordinates of D = (( 1 + 2)/2, ( 1 + 2)/2) = ((3+5)/2, ( 2 + 2)/2) = (8/2, 0/2) = (4, 0) Now, we need to prove that Area ABD = Area ACD Solving L.H.S Area of triangle ABD = 1/2 [ x1(y2 y3) + x2(y3 y1) + x3(y1 y2) ] Here x1 = 4 , y1 = 6 x2 = 3 , y2 = 2 x3 = 4 , y3 = 0 Putting values Area of triangle ABD = 1/2 [ 4( 2 0) + (3)(0 ( 6)) + 4( 6 ( 2)) ] = 1/2 [ 4( 2) + 3(0 + 6) + 4( 6 + 2)] = 1/2 [ 4( 2) + 3(6) + 4( 4)] = 1/2 [ 8 + 18 16] = 1/2 [ 6] = 3 Now, since area cannot be negative Area of triangle ABC = 3 square units Solving R.H.S Area of triangle ACD = 1/2 [ x1(y2 y3) + x2(y3 y1) + x3(y1 y2) ] Here x1 = 4 , y1 = 6 x2 = 5 , y2 = 2 x3 = 4 , y3 = 0 Area of triangle ACD = 1/2 [ 4(2 0) + (5)(0 ( 6)) + 4( 6 2) ] = 1/2 [ 4(2) + 5(0 + 6) + 4( 8)] = 1/2 [ 4(2) + 5(6) + 4( 8)] = 1/2 [ 8 + 30 32] = 1/2 [6] = 3 square units = R.H.S Hence, Area of ABD = Area of ACD Hence proved Area of triangle ABC = 1/2 [ 4(2) + 5(0 + 6) + 4( 8)]

Class 10

Important Questions for Exam - Class 10

- Chapter 1 Class 10 Real Numbers
- Chapter 2 Class 10 Polynomials
- Chapter 3 Class 10 Pair of Linear Equations in Two Variables
- Chapter 4 Class 10 Quadratic Equations
- Chapter 5 Class 10 Arithmetic Progressions
- Chapter 6 Class 10 Triangles
- Chapter 7 Class 10 Coordinate Geometry
- Chapter 8 Class 10 Introduction to Trignometry
- Chapter 9 Class 10 Some Applications of Trignometry
- Chapter 10 Class 10 Circles
- Chapter 11 Class 10 Constructions
- Chapter 12 Class 10 Areas related to Circles
- Chapter 13 Class 10 Surface Areas and Volumes
- Chapter 14 Class 10 Statistics
- Chapter 15 Class 10 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.