Which triangles in the figure above are congruent and/or similar? Find the value of x, angle /_ACE and the area hat(AEC), and BFDC.

Aug 11, 2016

Assuming:

$A E \bot C E$ and

$B F \bot A C$

There are three similar (not congruent) triangles: $\Delta A C E$, $\Delta A B F$ and $\Delta F D E$

$x = 13$

$\angle A C E = {45}^{o}$

${S}_{A E C} \approx 187.88$

${S}_{B F D C} \approx 102.88$

Explanation:

There are three triangles:
$\Delta A C E$, $\Delta A B F$ and $\Delta F D E$.

Given that all of them are right triangles and all of them are isosceles. Therefore, all of them have acute angles of ${45}^{o}$ and, therefore, all are similar to each other.

Since all three hypotenuses are different, there are no congruent pairs of them.

Length of $x$.
Since $\Delta A B F$ is a right triangle with catheti $x$ and $x$ and hypotenuse $13 \sqrt{2}$, Pythagorean Theorem gives:
${x}^{2} + {x}^{2} = {\left(13 \sqrt{2}\right)}^{2}$
$\implies 2 {x}^{2} = 338$
$\implies {x}^{2} = 169$
$\implies x = 13$

Angle $\angle A C E = {45}^{o}$ because, as we stated above, triangle $\Delta A C E$ is right isosceles.

Area of $\Delta A E C$.
${S}_{A E C} = \frac{1}{2} \cdot A E \cdot C E =$
$= \frac{1}{2} \cdot {\left(13 \sqrt{2} + 1\right)}^{2} = \frac{1}{2} \left(338 + 26 \sqrt{2} + 1\right) \approx 187.88$

Area of $B F D C$.
We have to subtract areas of $\Delta A B F$ and $\Delta F D E$ from the area of $\Delta A C E$ to get the area of $B F D C$.
${S}_{B F D C} = {S}_{A E C} - {S}_{A B F} - {S}_{F D E} =$
$= \frac{1}{2} \left(338 + 26 \sqrt{2} + 1\right) - \frac{1}{2} \cdot {x}^{2} - \frac{1}{2} \cdot {1}^{2} =$
$= \frac{1}{2} \left(339 + 26 \sqrt{2} - 169 - 1\right) = \frac{1}{2} \left(169 + 26 \sqrt{2}\right) \approx 102.88$

Aug 11, 2016

We have a triangle $\Delta F E D$ such that $\overline{F E} = \overline{E D} = 1$ and
$\overline{F D} = 2$ defining a null area triangle!

Explanation:

We have a triangle $\Delta F E D$ such that $\overline{F E} = \overline{E D} = 1$ and
$\overline{F D} = 2$ defining a null area triangle!