Ex 7.3, 5 - Chapter 7 Class 10 Coordinate Geometry (Term 1)

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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)]