Question

The area of the triangle formed by joining the midpoints of the sides of the triangle ABC is. Provided, coordinates of A(-4, 3), B(2, 3), C(4, 5).

• A. 1.5 sq. units.
• B. 3 sq. units.
• C. 2.5 sq. units.

Answer 1

The correct option is A 1.5 sq. units.

The midpoints of AB, BC, and CA are

$(\frac{-4+2}{2},\frac{3+3}{2}),(\frac{4+2}{2},\frac{5+3}{2})and(\frac{-4+4}{2},\frac{3+5}{2})$
respectively and let’s denote them as D, E and F.

So D(-1, 3), E(3, 4) and F(0, 4) .
$x_1=-1,y_1=3,x_2=3,y_2=4,x_3=0,y_3=4.A\left(△\right)=\frac{1}{2}\left[x_1\right(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2\left)\right]=\frac{1}{2}[-1(4-4)+3(4-3)+0(3-4\left)\right]A\left(△\right)=1.5\text{sq.units}$

Area of a triangle is 1.5 sq. units.