If $D( −51,\ 25 )$, $E( 7,\ 3)$ and $F( 27,\ 27)$ are the mid-points of sides of $\vartriangle ABC$, find the area of $\vartriangle ABC$.
Given: Mid-points of the sides of a triangle are: $( -\frac{1}{2},\ \frac{5}{2}),\ ( 7,\ 3)$ and $( \frac{7}{2},\ \frac{7}{2})$
To do: To find the area of the triangle.
Solution:
If midpoint of any triangle are given then area of triangle is $4$ times area of triangle form by joining midpoints of triangle
Now using the above concept
First we will find area $( \vartriangle DEF)=\frac{1}{2}[(-\frac{1}{2})( 3-\frac{7}{2})+7(\frac{7}{2}-\frac{5}{2})+\frac{7}{2}( \frac{5}{2}-3)]$
$=\frac{1}{2}( \frac{1}{4}+7-\frac{7}{4})$
$=\frac{11}{4}$ square unit
Now area of $ABC=4\times\frac{11}{4}=11$ square unit
 
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