The vertices of a $\triangle ABC$ are $A(4, 6), B(1, 5)$ and $C(7, 2)$. A line is drawn to intersect sides $AB$ and $AC$ at $D$ and $E$ respectively. such that $\frac{AD}{AB}=\frac{AE}{AC}=\frac{1}{4}$. Calculate the area of the $∆ADE$ and compare it with the area of $∆ABC$.

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$\frac{AD}{AB} =\frac{AE}{AC} =\frac{1}{4}$.

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 $\triangle A B C$,

$\frac{A D}{A B}=\frac{A E}{E C}=\frac{1}{4}$

$\angle D A E=\angle B A C$              (Common)

Therefore, by AAA property, we get,

$\triangle A D E \sim \Delta A B C$

Therefore,

$\frac{\text { Area } \triangle A D E}{\text { Area } \Delta A B C}=(\frac{A D}{A B})^{2}=(\frac{1}{4})^{2}$

$=\frac{1}{16}$

$\frac{\operatorname{Ar} \triangle A D E}{(\frac{15}{2})}=\frac{1}{16}$

Area $\triangle A D E=\frac{1\times15}{16 \times 2}$

$=\frac{15}{32}$ sq. units

Area $A D E:$ Area $A B C=\frac{15}{32}: \frac{15}{2}$

$=1: 16$

The area of triangle $ADE$ is $\frac{15}{32}$ sq. units and the area of triangle $ADE$ is $\frac{1}{16}$ times the area of triangle $ABC$. 

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Updated on: 10-Oct-2022

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