In $\triangle ABC$ where $AB = AC$, $D$ and $E$ are points on $AB$ and $AC$ respectively, such that $AB = 4BD$ and $AC = 4AE$.

Question

In $\triangle ABC$ where $AB = AC$, $D$ and $E$ are points on $AB$ and $AC$ respectively, such that $AB = 4BD$ and $AC = 4AE$. If area of quadrilateral $BCED$ is $52$ $cm^2$ and $[\triangle ADE] = x$ $cm^2$, find $x$.

What I Tried: Here is a picture :-

The only thing I am not able to figure out is that how to use the area of the quadrilateral using the available information in the picture. Suppose I am able to find the value of $y$, but how will I use that to find $[\Delta ADE]$ too? I cannot find a way to do that either.

Can anyone help me? Thank You.

Answer 1

You can use

$$ \dfrac{[ADE]}{[ABC]} = \dfrac{\tfrac{1}{2}\cdot AD \cdot AE \cdot \sin A}{\tfrac{1}{2}\cdot AB \cdot AC \cdot \sin A} $$

to know their ratios. And you know their difference.

Answer 2

You can connect $\overline{BE}$ and note that $$\frac{[\triangle ADE]}{[\triangle ABE]}=\frac{\overline{AD}}{\overline{AB}}=\frac{3y}{3y+y}=\frac{3}4$$ and also $$\frac{[\triangle ABE]}{[\triangle ABC]}=\frac{\overline{AE}}{\overline{AC}}=\frac{y}{y+3y}=\frac{1}4$$

Therefore you get $$\frac{[\triangle ADE]}{[\triangle ABC]}=\frac{3}{4}\cdot\frac{1}4=\frac{3}{16}=\frac{x}{x+52}\implies \color{red}{x=4}$$

Answer 3

Draw a parallel line to $BC$ from $D$ and it meets $AC$ at $F$.

Say $\triangle ABC = a$

$\triangle ADF = \frac{9}{16}a$ (both base and height ratio is $3:4$)

$\triangle EDF = \frac{2}{3} \triangle ADF = \frac{3a}{8}$ (same base but height is $2:3$, as $AE:EF = 2:3$)

So $52 = \triangle EDF + [BCDF] = \frac{3}{8}a + a - \frac{9}{16}a = \frac{13a}{16}$

So $a = 64$ $\triangle ADE = 64 - 52 = 12$

Answer 4

Join $DC$

let $ar(\Delta ADE)=p,ar(\Delta) DEC=q$

$$\frac{p}{q}=\frac{y}{3y}=\frac{1}{3}$$ $$\frac{ar(\Delta ADC)}{ar ( \Delta CDB)}=\frac{q+p}{{52-q}}=\frac{3y}{y}=3$$ $p=?,q=?$