Question

In the figure, the vertices of $ABC$ are $A(4,6),B(1,5)$ and $C(7,2).$ A line-segment $DE$ is drawn to intersect the sides $AB$ and $AC$ at $D$ and $E$ respectively such that $\frac{AD}{AB}=\frac{AE}{AC}=\frac{1}{3}$. Calculate the area of $\Delta ADE$.

• A. $\frac{15}{32}$ sq. units
• B. $\frac{65}{23}$ sq. units
  
• C. $\frac{5}{6}$ sq. units
• D. $6$ sq. units

Answer 1

The correct option is C $\frac{5}{6}$ sq. units

Area of a triangle having vertices $(x_1,y_1),(x_2,y_2)$ and $(x_3,y_3)$ is given by

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

$\therefore$Area of a triangle $ABC$ is given by

$A=\frac{1}{2}[4(5-2)+1(2-6)+7(6-5)]$

$A=\frac{1}{2}[12-4+7]=\frac{15}{2}$sq.units

In $△ADE$ and $△ABC,$

$\frac{AD}{AB}=\frac{AE}{EC}=\frac{1}{3}$

and $\angle DAE=\angle BAC$(Common)

Hence,$△ADE\sim △ABC$ by $AA$

or $\frac{Areaof△ADE}{Areaof△ABC}={\left(\frac{AD}{AB}\right)}^2={\left(\frac{1}{3}\right)}^2=\frac{1}{9}$

$\frac{Areaof△ADE}{\frac{15}{2}}=\frac{1}{9}$

$Areaof△ADE=\frac{15}{2}\times \frac{1}{9}=\frac{5}{6}$sq.units