Question

If D $(\frac{-1}{5},\frac{5}{2})$, E (7, 3) and F $\frac{7}{2},\frac{7}{2}$ are the mid points of sides of $△ABC$, find the area of $△ABC$.

Answer 1

The area of the triangle formed by joining midpoints of a triangle is $\frac{1}{4}$(Area of triangle)
Area of DEF = $\frac{1}{4}$ Area of ABC
Area of ABC = 4 Area of DEF

Area of a triangle whose sides are $(x_1,y_1),(x_2,y_2),(x_3,y_3)$
Area = $\frac{1}{2}[(x_1{y}_2-x_2{y}_1)+(x_2{y}_3-x_3{y}_2)+(x_3{y}_1-x_1{y}_3)]$
Area of DEF = $\frac{1}{2}[\frac{-3}{5}-\frac{35}{2}+\frac{49}{2}-\frac{21}{2}+\frac{35}{4}+\frac{7}{10}]$
Area of DEF = $\frac{1}{2}\left[\frac{107}{20}\right]$

=$\frac{107}{40}$ sq units.

Area of ABC = 4 (Area of DEF) = 4 x $\frac{107}{40}$ = 10.7 sq units.