Find the area $OFE$ in the figure below

Question

In the figure $ABCD$ is a square with side $4$ and $CE=\frac{4}{5}$. Calculate the area of ​​the $\triangle OFE$

I try:

$S\triangle DCE = \dfrac{0.8.4}{2} = 1,6$

$DE^2 = 4^2+(\frac{4}{5})^2 \implies DE = \dfrac{4\sqrt{26}}{5} $

$S\triangle COD = \dfrac{2\sqrt2.2\sqrt2}{2}=4 \implies S\triangle COH = 2 $

$S\triangle OHE=\dfrac{2.1,2}{2}=1,2 \implies S\triangle OCE=2-1,2= 0,8$

Answer 1

From $F$, draw a line to the right parallel to $DC$, with it meeting $CH$ at $G$, as shown in the diagram below.

Since $\triangle FGC$ is an isosceles right-angled triangle,

$$\lvert FG\rvert = \lvert GC\rvert \tag{1}\label{eq1A}$$

Also, $\triangle FGE \sim \triangle DCE$, so then

$$\frac{\lvert FG\rvert}{\lvert EG\rvert} = \frac{\lvert DC\rvert}{\lvert CE\rvert} \;\;\to\;\; \lvert FG\rvert = 5\lvert EG\rvert \tag{2}\label{eq2A}$$

Using \eqref{eq1A}, we then get

$$\lvert GC\rvert + \lvert EG\rvert = 0.8 \;\;\to\;\; 6\lvert EG\rvert = 0.8 \;\;\to\;\; \lvert EG\rvert = \frac{2}{15} \tag{3}\label{eq3A}$$

With this in \eqref{eq2A}, we have

$$\lvert FG\rvert = \frac{2}{3} \tag{4}\label{eq4A}$$

Using the one-half base times height triangle area formula, with $A$ meaning area, gives

$$A(\triangle FCE) = \frac{1}{2}\left(\frac{4}{5}\right)\left(\frac{2}{3}\right) = \frac{4}{15} \tag{5}\label{eq5A}$$

Then using this and your result, we get that

$$\begin{equation}\begin{aligned} A(\triangle OFE) & = A(\triangle OCE) - A(\triangle FCE) \\ & = \frac{4}{5} - \frac{4}{15} \\ & = \frac{8}{15} \end{aligned}\end{equation}\tag{6}\label{eq6A}$$

Answer 2

$CF$ is angle bisector of $\triangle CDE$. Therefore,$\frac{CD}{DF}=\frac{CE}{CF}\implies DF=\frac{5DE}6$ and $FE=\frac{DE}6$.

We can find area of $\triangle CDF$ and $\triangle CDE$ by using $sin\angle CDE=\frac{CE}{DE}$.

Now ar$(\triangle CEF)$=ar$(\triangle CDE)-$ar$(\triangle CDF)$.

Area of $\triangle OHE$ and $\triangle OHC$ can be easily calculated and from this we can obtain area of $\triangle OFE$ .

Answer 3

There are many ways to approach this, but here's one that aims for efficiency:

$\triangle CFE$ has base $0.8$ and height $x$. $\triangle CEO$ has base $0.8$ and height $2$. That's good. Very good.

$\triangle FGD$ is similar to $\triangle ECD$ because straight line $DE$. Therefore:

$\dfrac{0.8}4 = \dfrac{x}{4-x}$