In order for two right triangles to be similar, their height-to-base ratios must be equal. Given a right triangle with a height $\displaystyle h=5$ and a base $\displaystyle b=20$, the ratio $\displaystyle \frac{h}{b}=\frac{5}{20}=\frac{1}{4}$. The only answer provided with a ratio of $\displaystyle \frac{1}{4}$ is $\displaystyle h=3, b=12$.

In order for two right triangles to be similar, their height-to-base ratios must be equal. Given a right triangle with a height $\displaystyle h$ of $\displaystyle 7$ and a base $\displaystyle b$ of $\displaystyle 42$,  the ratio $\displaystyle \frac{h}{b}=\frac{7}{42}=\frac{1}{6}$. The only answer provided with a ratio of $\displaystyle \frac{1}{6}$ is $\displaystyle h=5, b=30$.

Corresponding angles of similar triangles are congruent, so Statement 1 alone establishes that $\displaystyle \angle A \cong \angle D$ and $\displaystyle \angle C \cong \angle F$; similarly, Statement 2 alone establishes that $\displaystyle \angle A \cong \angle F$ and $\displaystyle \angle C \cong \angle D$. However, neither statement alone establishes the actual measure of any of the four acute angles. 

Assume both statements are true. From transitivity, it holds that $\displaystyle \angle A \cong \angle D \cong \angle C$. Specifically, the two acute angles of $\displaystyle \bigtriangleup ABC$, one of which is $\displaystyle \angle A$, are congruent. Since $\displaystyle \angle B$ is right, this makes $\displaystyle \bigtriangleup ABC$ and isosceles right triangle, and its acute angles, including $\displaystyle \angle A$, have measure $\displaystyle 45 ^{\circ }$. 

We need to use the Pythagorean theorem:

$\displaystyle a^2+b^2=c^2$

$\displaystyle a^2=c^2-b^2$

$\displaystyle a^2=10^2-8^2$

$\displaystyle a^2=100-64$

$\displaystyle a^2=36$

$\displaystyle a=6$

The area of the big square is 5, and area of the small square is 1. Therefore, the area of the four right triangles is $\displaystyle \dpi{100} \small 5-1=4$.

Since the four triangles are all exactly the same, the area of each of the right triangle is 1. We know that the longer side is 2 times the shorter side, so we can represent the shorter side as $\displaystyle \dpi{100} \small x$ and the longer side as $\displaystyle \dpi{100} \small 2x$. Then we are able to set up an equation:

$\displaystyle \dpi{100} \small \frac{1}{2}\times x\times 2x=1$. 

Therefore, x, the length of the shortest side of the right triangle, is 1.

This triangle can exist by the Triangle Inequality, since the sum of the lengths of its shortest two sides exceeds that of the longest side: 

$\displaystyle 9 + 12 = 21 > 16$

The sum of the squares of its shortest two sides is less than the square of that of its longest side:

$\displaystyle 9^{2}+12^{2} = 81+144= 225 < 256=16^{2}$

This makes the triangle obtuse.

Its sides are all of different measure, which makes the triangle scalene as well.

Obtuse and scalene is the correct choice.

We can determine the length of the side by using the Pythagorean Theorem:

$\displaystyle c^2=a^2+b^2$

where $\displaystyle c=5, a=4$

Our equation is then: 

$\displaystyle 5^2=x^2+4^2$

$\displaystyle x^2=25-16=9$

$\displaystyle x=3$

We can calculate the length of the side by using the pythagorean theorem: $\displaystyle c^2=a^2+b^2$

where our values are $\displaystyle c=12, a=8$

we can then solve for $\displaystyle x$:

$\displaystyle 12^2=8^2+x^2$

$\displaystyle x^2=12^2-8^2=80$

$\displaystyle x=8.94$

In order to find the length of the third side, we need to use the Pythagorean theorem. The hypotenuse, c, is 13, and the height, a, is 5, so we can simply plug in these values and solve for b, the length of the base of the right triangle:

$\displaystyle a^2+b^2=c^2$

$\displaystyle b^2=c^2-a^2=13^2-5^2=169-25$

$\displaystyle b^2=144\rightarrow b=12$

The shortest leg of a $\displaystyle \textup{30-60-90}$ triangle is one half the length of its hypotenuse. In this triangle, it is

$\displaystyle \frac{1}{2} \cdot 4^{t} = \frac{4^{t}}{2} = \frac{4^{t}}{ \sqrt{4}} = \frac{4^{t}}{4^{\frac{1}{2}} } = 4 ^{t-\frac{1}{2}}$