area of rectangle-see figure

Question

I am trying to find the area of the green rectangle. Triangle ABC has area $16$ and triangle CDE has area $4$. The two triangles have the same shape.

then $ab=32$, $cd=4$, and $\frac{a}{b} = \frac{c}{d}$ I need $b$ and $c$ to calculate the area of the rectangle. I need a fourth equation. I have tried $\frac{1}{2}(a+c)(b+d)=20 + bc$ (the area of the big triangle which includes the rectangle)

I have also tried some pythagaros, but I am stuck

Answer 1

use that $$\tan(\alpha)=\frac{b}{a},\tan(\alpha)=\frac{d}{c},\tan(\alpha)=\frac{d+b}{a+c}$$ and $$\alpha=\angle{CAB}=\angle {ECD}$$ and $$ab=32$$ and $$cd=8$$

Answer 2

$$\because \triangle ABC \sim \triangle CDE \\ \therefore \frac{a}{c}=\frac{b}{d}=k \\ a=kc, b = kd \\ \because S_{\triangle ABC}=16 \\ \therefore ab=32 \\ \because S_{\triangle CDE}=4 \\ \therefore cd=8 \\ ab=(kc)(kd)=k^2cd \\ 32=8k^2 \\ k=2 \quad(-2 \text{ rejected}) \\ S_\color{grey}{\text{gray rectangle}}=bc=(kd)c=k(cd)=2(8)=16$$

Answer 3

Hint: areas are proportional to the square of lengths in similar figures , so $\frac{AC}{CE}= \sqrt{\frac{16}{4}}=2\,$. It follows that $\frac{AC}{AE}=\frac{2}{3}\,$, so the area of the big right triangle (which includes both smaller triangles and the rectangle) is $S=S_{ABC} \cdot \left(\frac{AE}{AC}\right)^2 = 16 \cdot \frac{9}{4}=36\,$. Next, subtract the two known areas.

Answer 4

One more attempt:

1)$ab=32;$ 2) $cd=8; $

3) $(a+c)(d+b)=2 (16+4+ bc).$

$ad +ab +cd +cb = 40 +2bc;$

$ad + 32 +8 +cb = 40 +2bc;$

$ad=bc;$

1)×2):

$abcd=32×8 = 256;$

Since $ad = bc:$

$(bc)^2 = 256;$

$bc=16.$

Answer 5

Let the needed area be equal to $x$.

Thus, $$\frac{16+4+x}{16}=\left(\frac{AC}{AE}\right)^2=\left(1+\frac{CE}{AE}\right)^2=\left(1+\frac{1}{2}\right)^2=\frac{9}{4},$$ which gives $x=16.$