Question

The vertices of $\Delta ABC=areA(4,6),B(1,5)andC(7,2)$. 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 $\Delta ABC$

Answer 1

Concept:
Application:

We have,

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

$\Rightarrow \frac{AB}{AD}=\frac{AC}{AE}=4$

$\Rightarrow \frac{AD+DB}{AD}=\frac{AE+EC}{AE}=4$

$\Rightarrow 1+\frac{DB}{AD}=1+\frac{EC}{AE}=4$

$\Rightarrow \frac{DB}{AD}=\frac{EC}{AE}=3$

$\Rightarrow \frac{AD}{DB}=\frac{AE}{EC}=\frac{1}{3}$

$\Rightarrow AD:DB=AE:EC=1:3$

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,

$\therefore$ Area of $\Delta ADE$

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

$\Rightarrow Areaof\Delta ADE=\frac{1}{2}|(\frac{92}{4}+\frac{65}{4}+\frac{114}{4})-(\frac{78}{4}+\frac{437}{16}+20)|$

$\Rightarrow Areaof\Delta ADE=\frac{1}{2}|\frac{271}{4}-\frac{1069}{16}|$

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

Also, we have

$\therefore Areaof\Delta ABC$

$=\frac{1}{2}|(4\times 5+1\times 2+7\times 6)-(1\times 6+7\times 5+4\times 2)|$

$\Rightarrow Areaof\Delta ABC=\frac{1}{2}|(20+2+42)-(6+35+8)|$

$\Rightarrow Areaof\Delta ABC=\frac{1}{2}|64-49|=\frac{15}{2}$sq. units

$\therefore \frac{Areaof\Delta ADE}{Areaof\Delta ABC}=\frac{\frac{15}{32}}{\frac{15}{2}}=\frac{1}{16}$

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