Question

In fig. 8, the vertices of $\Delta ABC$ 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 $\Delta ADE$ and compare it with area of $\Delta ABC$.

Answer 1

We have, the vertices of $\Delta ABC$ as $A(4,6),B(1,5)$ and $C(7,2)$



And $\frac{AD}{AB}=\frac{AE}{AC}=\frac{1}{3}$

Then, coordinates of D are

$(\frac{1\left(1\right)+2\left(4\right)}{1+2},\frac{1\left(5\right)+2\left(6\right)}{1+2})$

$(\frac{1+8}{3},\frac{5+12}{3})i.e.,D\left(3\frac{17}{3}\right)$

and coordinates of E are

$(\frac{1\left(7\right)+2\left(4\right)}{1+2},\frac{1\left(2\right)+2\left(6\right)}{1+2})$

$(\frac{7+8}{3},\frac{2+12}{3})i.e.,E\left(5\frac{14}{3}\right)$

Now, Area of $\Delta ADE$

= $\frac{1}{2}[4(\frac{17}{3}-\frac{14}{3})+3(\frac{14}{3}-6)+5(6-\frac{17}{3})]$

= $\frac{1}{2}\left[4\right(1)+3(-\frac{4}{3})+5\left(\frac{1}{3}\right)]$

= $\frac{5}{6}$ units

and Area of $\Delta ABC$

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

= $\frac{1}{2}\left[4\right(3)+1(-4)+7(1\left)\right]=\frac{15}{2}$ units.

$\therefore \frac{\text{ar}\left(\Delta ADE\right)}{\text{ar}\left(\Delta ABC\right)}=\frac{5/6}{15/2}=\frac{1}{9}$

i.e., ar $\left(\Delta ADE\right):ar\left(\Delta ABC\right)=1:9$