Question

SOLVE THIS ...................V ¥((¥¥¥£​

Answer 1

Given :-

•Let the vertices of the triangle be A(4,-6) , B(3,-2) and C(5,2).

•Let D be the midpoint of side BC of ΔABC.Therefore ,AD is the median in ΔABC.

SOLUTION :-

=> Coordinates of point D =(3+5/2 , -2+2/2)= (4,0)

=>Area of triangle = ½[x1(y2-y3)+x2(y3-y1)+x3(y1-y2)] -----------( equ.(1))

=>By substituting the values of vertices A, B, D in the equ. (1)

=>Area of ΔABD = ½[(4){(-2)-(0)} + (3){(0)-(-6)} + (4){(-6)} - {-2}]

=>Area of ΔABD = ½(-8+18-16)

=> .°. Area of ΔABD = 3 Square units.

=> By substituting the values of vertices A, D, C in the equ. (1)

=> Area of ΔABD = ½[(4){0-(-2)} + (4){(2)-(-6)} + (5){(-6)-0}]

=> Area of ΔABD = ½(-8+32-30)

=> .°. Area of ΔABD = 3 Square units.

=> However, area cannot be negative. Therefore, area of ΔADC is 3 square units.

=>Hence, clearly, median AD has divided MABC in two triangles of equal areas.