Question

The vertices of ∆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 $\Delta ADE$ and compare it with the area of ∆ABC.

Answer 1

We have,

$\frac{AD}{AB}=\frac{AE}{AC}=\frac{1}{4}\frac{AB}{AD}=\frac{AC}{AE}=4\frac{AD+DB}{AD}=\frac{AE+EC}{AE}=4\frac{1+DB}{AD}=\frac{1+EC}{AE}=4\frac{DB}{AD}=\frac{EC}{AE}=3\frac{AD}{DB}=\frac{AE}{EC}=\frac{1}{3}AD:DB=AE:EC=1:3$
$\Rightarrow$ D and E divide AB and AC respectively in the ratio 1:3

So, the co-ordinates of D and E are

$(\frac{1+12}{1+3},\frac{5+18}{1+3})=(\frac{13}{4},\frac{23}{4})and(\frac{7+12}{1+3},\frac{2+18}{1+3})=(\frac{19}{4},5)respectively$

We have ,

Area of $\Delta ABC=\frac{1}{2}|(4\times \frac{23}{4}+\frac{13}{4}\times 5+\frac{19}{4}\times 6)-(\frac{13}{4}\times 6+\frac{19}{4}\times \frac{23}{4}+4\times 5)|$

$=\frac{1}{2}|\frac{271}{4}-\frac{1069}{16}|=\frac{15}{32}sq.units$

Also , we have

Area of $\Delta ABC$ $=\frac{1}{2}|(4\times 5+1\times 2+7\times 6)-(1\times 6+7\times 5+4\times 2)|\frac{1}{2}|(20+2+42)-(6+35+8)|=\frac{15}{2}sq.units$

$\frac{Areaof\Delta ADE}{Areaof\Delta ABC}=\frac{\frac{15}{32}}{\frac{15}{2}}=\frac{1}{6}$

Hence, Area of $\Delta$ ADE : Area of $\Delta$ABC = 1:16