Question

Find the area of the triangle formed by joining the mid points of the sides of the triangle ABC. Here the coordinates of the points are A (-4, 3), B (2, 3) and C (4, 5).

• A. 2
• B. 0.5
• C. 1
• D. 1.5

Answer 1

The correct option is D

1.5

The mid points of AB, BC and CA are ( $\frac{-4+2}{2}$, $\frac{3+3}{2}$), ( $\frac{4+2}{2}$, $\frac{3+5}{2}$) and ( $\frac{-4+4}{2}$, $\frac{5+3}{2}$) respectively and lets denote them as D, E and F. So D(-1, 3), E(3, 4) and F(0, 4).

Suppose $x_1$ = -1, $y_1$ = 3

$x_2$ = 3, $y_2$ = 4

$x_3$ = 0, $y_3$ = 4

Area of a triangle = $\frac{1}{2}$[ $x_1$($y_2$ - $y_3$) + $x_2$( $y_3$ - $y_1$) + $x_3$($y_1$ - $y_2$)]

= $\frac{1}{2}$[ -1(4 - 4) + 3(4 - 3) + 0(3 - 4)]

= 1.5 sq.units

Area of a triangle = 1.5 sq.units