Question

The vertices of a $\Delta ABC$ 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 $\Delta ADE$ and compare it with the area of $\Delta ABC$. (Recall Theorem 6.2 and Theorem 6.6).

Answer 1

Given that:
$\frac{AD}{AB}=\frac{AE}{AC}=\frac{1}{4}$
$\frac{AD}{AD+DB}=\frac{AE}{AE+EC}=\frac{1}{4}$
$\frac{AD}{DB}=\frac{AE}{EC}=\frac{1}{3}$
Therefore, D and E are two points on side AB and AC respectively
such that they divide side AB and AC in a ratio of 1:3.
Coordinates of Point $D=(\frac{1\times 1+3\times 4}{1+3},\frac{1\times 5+3\times 6}{1+3})$
=$(\frac{13}{4},\frac{23}{4})$
Coordinates of Point $E=(\frac{1\times 7+3\times 4}{1+3},\frac{1\times 2+3\times 6}{1+3})$
=$(\frac{19}{4},\frac{20}{4})$
Area of a triangle =$\frac{1}{2}\left[x_1\right(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2\left)\right]$
Area of $\Delta ADE=\frac{1}{2}[4(\frac{23}{4}-\frac{20}{4})+\frac{13}{4}(\frac{20}{4}-6)+\frac{19}{4}(6-\frac{23}{4})]$
=$\frac{1}{2}[3-\frac{13}{4}+\frac{19}{16}]=\frac{1}{2}\left(\frac{48-52+19}{16}\right)=\frac{15}{32}sq.units$
Area of $\Delta ABC=\frac{1}{2}\left[4\right(5-2)+1(2-6)+7(6-5\left)\right]$
=$\frac{1}{2}[12-4+7]=\frac{15}{2}sq.units$
Clearly, the ratio between the areas of $\Delta ADE$ and $\Delta ABC$ is 1:16.