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 $\Delta ABC$.

Answer 1

In$△ABC$ $A(4,6),B(1,5)$ and $C(7,2)$

Area of $△ABC=\frac{1}{2}\left[x_1\right(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2\left)\right]$

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

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

A line is drawn to intersect sides $AB$ and $AC$ at $D$ and $E$

Then $\frac{AD}{AB}=\frac{1}{4}$

The point $D$ divide the line $AB$ in $AD$ and $DB$.The ratio of $AD$ and $DB=m_1:m_2=1:3$

$\therefore$Co-ordinates of $D$=$(\frac{m_1{x}_2+m_2{x}_1}{m_1+m_2},\frac{m_1{y}_2+m_2{y}_1}{m_1+m_2})$

Here, $(x_1,y_1)=(4:6)$ and $(x_2,y_2)=(1,5)$

$=(\frac{1\times 1+4\times 3}{1+3},\frac{1\times 5+3\times 6}{3+1})$

$=(\frac{13}{4},\frac{23}{4})$

The point $E$ divide the line $A$C in $AE$ and $EC$. The ratio of $AE$ and $EC$=$m_1:m_2=1:3$

$\therefore$ Co-ordinates of $E=(\frac{m_1{x}_2+m_2{x}_1}{m_1+m_2},\frac{m_1{y}_2+m_2{y}_1}{m_1+m_2})$

Here, $(x_1,y_1)=(4:6)$ and $(x_2,y_2)=(7,2)$

$=(\frac{1\times 7+3\times 4}{1+3},\frac{1\times 2+3\times 6}{3+1})$

$=(\frac{19}{4},5)$

Now the area of $\Delta ADE=\frac{1}{2}\left[4\right(\frac{23}{4}-5)+\frac{13}{4}(5-6)+\frac{19}{4}(6-\frac{23}{4}\left)\right]$

$=\frac{1}{2}[\frac{12}{4}-\frac{13}{4}+\frac{19}{16}]$

$=\frac{1}{2}\left[\frac{48-52+19}{16}\right]$

$=\frac{1}{2}\times \frac{15}{16}=\frac{15}{32}$ sq.unit

Hence, Area $\left(△ADE\right):$Area ($△ABC$)=$\frac{15}{32}:\frac{15}{2}=\frac{1}{32}:\frac{1}{2}=1:16$