Example 15 - If A(-5, 7), B(-4, -5), C(-1, -6), D(4, 5) - Examples

  1. Chapter 7 Class 10 Coordinate Geometry
  2. Serial order wise

Transcript

Example 15 If A( 5, 7), B( 4, 5), C( 1, 6) and D(4, 5) are the vertices of a quadrilateral, find the area of the quadrilateral ABCD. Let the vertices of quadrilateral be A( 5, 7) , B( 4, 5) C( 1, 6) , D(4, 5) Joining AC, There are 2 triangles formed ABC & ACD Hence, Area of quadrilateral ABCD = Area of ABC + Area of ADC Finding area ABC Area of triangle ABC = 1/2 [ x1(y2 y3) + x2(y3 y1) + x3(y1 y2) ] Here x1 = 5 , y1 = 7 x2 = 4 , y2 = 5 x3 = 1 , y3 = 6 Putting values Area of triangle ABC = 1/2 [ 5( 5 ( 6)) + ( 4)( 6 7) + ( 1)(7 ( 5)) ] = 1/2 [ 5( 5 + 6) 4( 13) + ( 1)(7 + 5)] = 1/2 [ 5(1) 4( 13) + ( 1)(12)] = 1/2 [ 5 + 52 12] = 1/2 [35] square units Similarly, Finding area ADC Area of triangle ADC = 1/2 [ x1(y2 y3) + x2(y3 y1) + x3(y1 y2) ] Here x1 = 5 , y1 = 7 x2 = 4 , y2 = 5 x3 = 1 , y3 = 6 Area of triangle ADC = 1/2 [ 5(5 ( 6)) + 4( 6 7) + ( 1)(7 5) ] = 1/2 [ 5(5 + 6) + 4( 13) + ( 1)(2)] = 1/2 [ 5(11) + 4( 13) + ( 1)(2)] = 1/2 [ 55 52 2] = 1/2 [ 109] But area cannot be negative, So, Area of triangle ADC = 1/2 [ 109] square units Area of quadrilateral ABCD = Area of triangle ABC + Area of triangle ADC = 1/2 [ 35 + 109] = 1/2 [ 144] = 72 square units

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