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 ratio of the area of the $\Delta ADE$ and $\Delta ABC$.

• A. 2 : 9
• B. 2 : 15
• C. 3 : 16
• D. 1 : 16

Answer 1

The correct option is D 1 : 16



We have,
$\frac{AD}{AB}=\frac{1}{4}\Rightarrow \frac{AB}{AD}=\frac{4}{1}\Rightarrow \frac{AD+DB}{AD}=\frac{4}{1}\frac{AD}{AD}+\frac{DB}{AD}=1+\frac{3}{1}\Rightarrow 1+\frac{DB}{AD}=1+\frac{3}{1}\frac{DB}{AD}=\frac{3}{1}\Rightarrow AD:DB=1:3$

Thus, the point D divides AB in the ratio 1 : 3
By section formula $(x,y)=(\frac{m{x}_2+n{x}_1}{m+n},\frac{m{y}_2+ny{+}_1}{m+n})\therefore D(x,y)=(\frac{(1\times 1)+(3\times 4)}{1+3},\frac{(1\times 5)+(3\times 6)}{1+3})=(\frac{1+12}{4},\frac{5+18}{4})=(\frac{13}{4},\frac{23}{4})$

Similarly, AE : EC=1 : 3 i.e., E divides AC in the ratio 1 : 3
$E(x,y)=(\frac{(1\times 7)+(3\times 4)}{1+3},\frac{1\times 2+3\times 6}{1+3})=(\frac{7+12}{4},\frac{2+18}{4})=(\frac{19}{4},5)$

Now, $ar\left(△ADE\right)$
$=\frac{1}{2}[4(\frac{23}{4}-5)+\frac{13}{4}(5-6)+\frac{19}{4}(6-\frac{23}{4})]=\frac{1}{2}\left[\right(23-20)+\frac{13}{4}(-1)+\frac{19}{4}\left(\frac{24-23}{4}\right)]=\frac{1}{2}(3-\frac{13}{4}+\frac{19}{16})=\frac{1}{2}\left[\frac{48-52+19}{16}\right]ar\left(△ADE\right)=\frac{15}{32}\text{sq. units}$

$ar\left(△ABC\right)=\frac{1}{2}\left[4\right(5-2)+1(2-6)+7(6-5\left)\right]=\frac{1}{2}\left[\right(4\times 3)+1(-4)+7\times 1]=\frac{1}{2}[12+(-4)+7]=\frac{1}{2}\left(15\right)ar\left(△ABC\right)=\frac{15}{2}\text{sq. units}$
Now, $\frac{ar\left(△ADE\right)}{ar\left(△ABE\right)}=\frac{\frac{15}{32}}{\frac{15}{2}}=\frac{15}{32}\times \frac{2}{15}=\frac{1}{16}$
$\therefore ar\left(\Delta ADE\right):ar\left(\Delta ABC\right)=1:16$