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In figure, the vertices of $\vartriangle ABC$ 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 $\vartriangle ADE$ and compare it with area of $\vartriangle ABC$.
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Given:
The vertices of $\vartriangle ABC$ 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$\frac{AD}{AB} =\frac{AE}{AC} =\frac{1}{3}$.
To do:
We have to calculate the area of $\vartriangle ADE$ and compare it with area of $\vartriangle ABC$.
Solution:
We know that,
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 triangle $A B C =\frac{1}{2}[4(5-2)+1(2-6)+7(6-5)]$
$=\frac{1}{2}[12+(-4)+7]$
$=\frac{15}{2} \text { sq. units. }$ In $\triangle A D E$ and $\triangle A B C$,
$\frac{A D}{A B}=\frac{A E}{E C}=\frac{1}{3}$
$\angle D A E=\angle B A C$ (Common) Therefore, by AAA property, we get,
$\triangle A D E \sim \Delta A B C$
Therefore,
$\frac{\text { Area } \triangle A D E}{\text { Area } \Delta A B C}=(\frac{A D}{A B})^{2}=(\frac{1}{3})^{2}=\frac{1}{9}$
$\frac{\operatorname{Ar} \triangle A D E}{(\frac{15}{2})}=\frac{1}{9}$
Area $\triangle A D E=\frac{1.5}{2 \times 9}$
$=\frac{5}{6}$ sq. units
Area $A D E:$ Area $A B C=\frac{5}{6}: \frac{15}{2}$
$=1: 9$
The area of triangle $ADE$ is $\frac{5}{9}$ sq. units and the area of triangle $ADE$ is $\frac{1}{9}$ times the area of triangle $ABC$.
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