Question

If D $(-1/2 , 5/2)$ , E (7 , 3) and F $(7/2 , 7/2)$ are the mid-points of the sides of triangle ABC , then find the area of the triangle ABC.

Answer 1

First to find the area we need to find the vertices of the triangle.

Midpoint D = x1+x2 = -1........(1)

y1+y2 = 5.............................(4)

Midpoint E = x1+x3 = 14.........(2)

y1+y3 = 6...............................(5)

Midpoint F = x2+x3 = 7............(3)

y2+y3 = 7...............................(6)

Now let us find the x terms first.

(1) ........ x1 = -1-x2
Sub in (2) ........ -1-x2 + x3 = 14........(7)
Now subtracting (7)  and (3)
we get, 2x3 = 22
x3 = 11
Sub x3 in (2) ........x1+11 = 14
x1 = 3
and so x2 = -4

now let us find y terms.

(3) ............y1 = 5 - y2
sub in (5).........5-y2+y3 = 6.........(8)
Now subtracting (8) and (6) we get,
y3 = 4
now sub y3 in (5)...... y1 = 2
and so
y2 = 3

now the vertices are (3,2) (-4,3) (11,4)

Area of the triangle = (3(3-4) + (-4) (4-2) + 11(2-3))/2
= (-3 -8 -11)2
= - 11

Remove the negative sign (-) from the number -11. 

The area of the triangle is 11 sq.units.

Because the formula calls for absolute value, you simply remove the negative sign.

Answer 2

If DEF is the triangle formed by joining the mid-points of sides of the triangle ABC, then the area of triangle ABC is 4 times the area of triangle DEF.
So find the area of triangle DEF first and multiply with 4 to get the area of ABC.

Area of a triangle with vertices (x₁,y₁); (x₂,y₂) and (x₃,y₃) is given by
Area = [ x₁(y₂ - y₃) + x₂(y₃ - y₁) + x₃(y₁ - y₂) ] / 2

Vertices of DEF are $D( -\frac{1}{2}, \frac{5}{2} ); E(7,3)\ and\ F( \frac{7}{2},\frac{7}{2} )$

So Area of triangle DEF

$[-\frac{1}{2}(3- \frac{7}{2})+7( \frac{7}{2} - \frac{5}{2})+\frac{7}{2}(\frac{5}{2}-3)]/2$

$[-\frac{1}{2}(\frac{6-7}{2})+7( \frac{7-5}{2})+\frac{7}{2}(\frac{5-6}{2})]/2$

$=[-\frac{1}{2}(-\frac{1}{2})+7(1)+\frac{7}{2}(-\frac{1}{2})]/2\\ \\ =(\frac{1}{4} +7- \frac{7}{4})/2 \\ \\ =\frac{1+28-7}{4*2} \\ \\= \frac{22}{8} \\ \\= \frac{11}{4}$

So area of DEF is $\frac{11}{4}$
So area of ABC = $4*\frac{11}{4}=\boxed{11\ sq.\ units}$