\(\displaystyle \angle A \cong \angle D\); \(\displaystyle \angle B \cong \angle E\) since both are right angles.

Given that two pairs of corresponding angles are congruent and any one side of corresponding sides is congruent, it follows that the triangles are congruent. In the case of Statement I, the included sides are congruent, so by the Angle-Side-Angle Congruence Postulate, \(\displaystyle \bigtriangleup ABC \cong \bigtriangleup DEF\). In the case of the other two statements, a pair of nonincluded sides are congruent, so by the Angle-Angle-Side Congruence Theorem, \(\displaystyle \bigtriangleup ABC \cong \bigtriangleup DEF\). Therefore, the correct choice is I, II, or III.

\(\displaystyle \bigtriangleup ABC \cong \bigtriangleup DEF\), and corresponding parts of congruent triangles are congruent.

Since \(\displaystyle \angle B\) is a right angle, so is \(\displaystyle \angle E\). \(\displaystyle DE = AB\) and \(\displaystyle EF= BC\); since \(\displaystyle AB = BC\), it follows that \(\displaystyle DE = EF\). \(\displaystyle \bigtriangleup DEF\)  is an isosceles right triangle; consequently, \(\displaystyle m \angle A = m \angle C = 45^{\circ }\).

\(\displaystyle \bigtriangleup DEF\) is a 45-45-90 triangle with hypotenuse of length \(\displaystyle DF= AC = 20\). By the 45-45-90 Triangle Theorem, the length of each leg is equal to that of the hypotenuse divided by \(\displaystyle \sqrt{2}\); therefore, 

\(\displaystyle DE=EF = \frac{DF}{\sqrt{2}} = \frac{20}{\sqrt{2}} =\frac{20 \cdot \sqrt{2}}{\sqrt{2} \cdot \sqrt{2}} = \frac{20 \sqrt{2}}{2} = 10 \sqrt{2}\)

\(\displaystyle DE = 20\sqrt{2}\) is eliminated as the correct choice.

Also, the perimeter of \(\displaystyle \bigtriangleup DEF\) is

\(\displaystyle P = DE+EF+DF = 10 \sqrt{2} + 10 \sqrt{2} + 20 = 20+ 20 \sqrt{2} \ne 40\).

This eliminates the perimeter of \(\displaystyle \bigtriangleup DEF\) being 40 as the correct choice.

Also, \(\displaystyle m \angle F = 60^{\circ }\) is eliminated as the correct choice, since the triangle is 45-45-90.

The area of  \(\displaystyle \bigtriangleup DEF\) is half the product of the lengths of its legs:

\(\displaystyle A = \frac{1}{2} \cdot DE \cdot EF\)

\(\displaystyle = \frac{1}{2} \cdot 10 \sqrt{2}\cdot 10 \sqrt{2}\)

\(\displaystyle = \frac{1}{2} \cdot 10\cdot 10 \cdot \sqrt{2} \cdot \sqrt{2}\)

\(\displaystyle = \frac{1}{2} \cdot 10\cdot 10 \cdot 2\)

\(\displaystyle = 100\)

The correct choice is the statement that \(\displaystyle \bigtriangleup DEF\) has area 100.

A right triangle must have one right angle and two acute angles; this means that no angle of a right triangle can be obtuse. But since \(\displaystyle 120^{\circ } > 90^{\circ }\), it is obtuse. This makes it impossible for a right triangle to have a \(\displaystyle 120^{\circ }\) angle.

One of the angles of a right triangle is by definition a right, or \(\displaystyle 90^{\circ }\), angle, so this is the measure of one of the missing angles. Since the measures of the angles of a triangle total \(\displaystyle 180^{\circ }\), if we let the measure of the third angle be \(\displaystyle x\), then:

\(\displaystyle x + 68 + 90 = 180\)

\(\displaystyle x + 158 = 180\)

\(\displaystyle x + 158 - 158= 180 - 158\)

\(\displaystyle x = 22\)

The other two angles measure \(\displaystyle 22 ^{\circ }, 90^{\circ }\).

The total number of degrees in a triangle is \(\displaystyle 180\).

While \(\displaystyle 47^{\circ}\) is provided as the measure of one of the angles in the diagram, you are also told that the triangle is a right triangle, meaning that it must contain a \(\displaystyle 90^{\circ}\) angle as well. To find the value of \(\displaystyle x\), subtract the other two degree measures from \(\displaystyle 180\).

\(\displaystyle x=180-90-47=43^{\circ}\)

All the angles in a triangle must add up to 180 degrees.

\(\displaystyle 90^{\circ}+47^{\circ}+v=180^{\circ}\)

\(\displaystyle 137^{\circ}+v=180^{\circ}\)

\(\displaystyle 137^{\circ}+v-137^{\circ}=180^{\circ}-137^{\circ}\)

\(\displaystyle v=43^{\circ}\)

All the angles in a triangle adds up to \(\displaystyle 180^{\circ}\).

\(\displaystyle 90^{\circ}+32^{\circ}+w=180^{\circ}\)

\(\displaystyle 122^{\circ}+w=180^{\circ}\)

\(\displaystyle 122^{\circ}+w-122^{\circ}=180^{\circ}-122^{\circ}\)

\(\displaystyle w=58^{\circ}\)

All the angles in a triangle add up to \(\displaystyle 180\) degrees.

\(\displaystyle 90^{\circ}+44^{\circ}+y=180^{\circ}\)

\(\displaystyle 134^{\circ}+y=180^{\circ}\)

\(\displaystyle 134^{\circ}+y-134^{\circ}=180^{\circ}-134^{\circ}\)

\(\displaystyle y=46^{\circ}\)

All the angles in a triangle add up to \(\displaystyle 180^{\circ}\).

\(\displaystyle 90^{\circ}+59^{\circ}+z=180^{\circ}\)

\(\displaystyle 149^{\circ}+z=180^{\circ}\)

\(\displaystyle 149^{\circ}+z-149^{\circ}=180^{\circ}-149^{\circ}\)

\(\displaystyle z=31^{\circ}\)

Because an equilateral triangle has three sides that are the same length, divide the given perimeter by 3 to find the length of each side.

\(\displaystyle 2400\div3=800\)