Imagine a right triangle as a square cut in half at a diagonal angle.

When figuring out the area, you figure it out the same way as finding the area of a square, but after multiplying length x width, divide the answer by 2.

\(\displaystyle 14 \cdot 13 = 182inches^2\)

\(\displaystyle 182/2=91 inches^2\)

The formula for the area of a right triangle is 

\(\displaystyle A=\frac{1}{2}XY\).

Plugging in the values given, 

\(\displaystyle A=\frac{1}{2}*5*12=30\).

The area \(\displaystyle A\) of a right triangle is given by \(\displaystyle A=\frac{bh}{2}\), where \(\displaystyle b\) represents the length of the triangle's base and \(\displaystyle h\) represents the length of the triangle's height. The base \(\displaystyle b\) and the height \(\displaystyle h\) of the triangle given in the problem are \(\displaystyle 12\) and \(\displaystyle 5\) units long, respectively. Hence, the area \(\displaystyle A\) of the triangle can be calculated as follows:

\(\displaystyle A=\frac{bh}{2}= \frac{(12)(5)}{2} = 30\).

Hence, the area of a right triangle with base length \(\displaystyle 12\) units and height \(\displaystyle 5\) units is \(\displaystyle 30\) square units.

Right triangles are congruent if both the hypotenuse and one leg are the same length. These triangles are congruent by HL, or hypotenuse-leg.

Two right triangles can have all the same angles and not be congruent, merely scaled larger or smaller. If all the side lengths are multiplied by the same number, the angles will remain unchanged, but the triangles will not be congruent.

To determine the answer choice that does not lead to congruence, we should simply use process of elimination.

If \(\displaystyle AC=16\), then subtracting tells us that \(\displaystyle DC=8\).; therefore \(\displaystyle \overline{AD}\cong \overline{DC}\). Given the fact that reflexively \(\displaystyle \overline{BD}\cong\overline{BD}\) and that both \(\displaystyle \angle BDA\) and \(\displaystyle \angle BDC\) are both right angles and thus congruent, we can establish congruence by way of Side-Angle-Side.

Similarly, if \(\displaystyle AB=10\), then \(\displaystyle \overline{AB}\cong \overline{BC}\), and given the other information we determined with our last choice, we can establish conguence by way of Hypotenuse-Leg.

If \(\displaystyle \angle BAC\cong \angle BCA\), given what we already know we can establish congruence by Angle-Angle-Side

Finally, if \(\displaystyle \overrightarrow{BD}\) is an angle bisector, then our two halves are congruent. \(\displaystyle \angle ABD\cong \angle CBD\). Given what we know, we can establish congruence by Angle-Side-Angle

The only remaining choice is the case where \(\displaystyle m\angle ABC=100^o\). This does not tell us how the two parts of this angle are related, we lack enough information for congruence.

Since we know that \(\displaystyle \overleftarrow{A}\overrightarrow{B}\parallel \overleftarrow{D}\overrightarrow{E}\), we know that \(\displaystyle \angle B\) is also a right angle and is thus congruent to \(\displaystyle \angle D\).

We are given that \(\displaystyle \overline{BC}\cong \overline{CE}\).  Furthermore, since \(\displaystyle \angle BCA\) and \(\displaystyle \angle DCE\) are vertical angles, they are also congruent.  

Therefore, we have enough evidence to conclude congruence by Angle-Side-Angle.  Vertex \(\displaystyle A\) matches up with \(\displaystyle D\), vertex \(\displaystyle B\) matches up with \(\displaystyle E\), and \(\displaystyle C\) matches up to \(\displaystyle C\). Thus, our congruence statement should look the following

\(\displaystyle \bigtriangleup ABC\cong \bigtriangleup DEC\)

We know that congruent triangles have equal corresponding angles and equal corresponding sides.  We are given that the corresponding sides are equal and are in the ratio of \(\displaystyle 1:2:\sqrt{3}\).  A triangle whose sides are in this ratio is a \(\displaystyle 30^{\circ}-60^{\circ}-90^{\circ}\), where the shortest side lies opposite the \(\displaystyle 30^{\circ}\) angle, the longest side is the hypotenuse and lies opposite the right angle, and the third side lies opposite the \(\displaystyle 60^{\circ}\) angle. (Remember \(\displaystyle \sqrt{3}< 2\).) So we know the corresponding angles are equal. Therefore, the triangles are congruent.

We know that congruent triangles have equal corresponding angles and equal corresponding sides.  We are given that the corresponding sides are equal and are in the ratio of \(\displaystyle \sqrt{3}:2\sqrt{3}:3\).

Simplify the ratio by dividing by \(\displaystyle \sqrt{3}\)

\(\displaystyle \frac{\sqrt{3}}{\sqrt{3}}:\frac{2\sqrt{3}}{\sqrt{3}}:\frac{3}{\sqrt{3}}\)

\(\displaystyle \frac{\sqrt{3}}{\sqrt{3}}=1\)

\(\displaystyle \frac{2\sqrt{3}}{\sqrt{3}}=2\)

\(\displaystyle \frac{3}{\sqrt{3}}\cdot\frac{\sqrt{3}}{\sqrt{3}}=\frac{3\sqrt{3}}{3}=\sqrt{3}\)

Thus, the corresponding sides are in the ratio \(\displaystyle 1:2:\sqrt{3}\) and we know both triangles are \(\displaystyle 30^{\circ}-60^{\circ}-90^{\circ}\) triangles. Since the corresponding angles and the corresponding sides are equal, the triangles are congruent.

We know that congruent triangles have equal corresponding angles and equal corresponding sides.  We are given that the corresponding sides are equal, and the measures of two angles.  We know that \(\displaystyle \angle C=60^{\circ}\) and \(\displaystyle \angle A'=30^{\circ}\) because the sum of the angles of a triangle must equal \(\displaystyle 180^{\circ}\). So the corresponding angles are also equal.  Therefore, the triangles are congruent.