The area \(\displaystyle A\) of a right triangle with a base \(\displaystyle b\) and a height \(\displaystyle h\) can be found with the formula \(\displaystyle A=\frac{1}{2}bh\). Since the two legs of a right triangle are perpendicular to each other, we can use these as the base and height in the formula. Therefore:

\(\displaystyle A=\frac{1}{2}(7\:cm)(14\:cm)\)

\(\displaystyle A=\frac{1}{2}(98\:cm^2)\)

\(\displaystyle A=49\:cm\)

The hypotenuse of the smallest triangle can be calculated using the Pythagorean Theorem:

\(\displaystyle c = \sqrt{a^{2}+b^{2}} = \sqrt{3^{2}+4^{2}} = \sqrt{9+16} = \sqrt{25}=5\) inches.

Let \(\displaystyle c_{n}\) be the lengths of the hypotenuses of the triangles in inches. \(\displaystyle c_{1} = 5\) and \(\displaystyle c_{2} = 12\), so their common difference is

\(\displaystyle d = c_{2} - c_{1} = 12 - 5= 7\)

The arithmetic sequence formula is 

\(\displaystyle c_{n} = c_{1} + (n-1)d\)

The length of the hypotenuse of the largest triangle - the tenth triangle - can be found by substituting \(\displaystyle c_{1} = 5, n= 10, d = 7\):

\(\displaystyle c_{10} =5+ (10-1) 7 = 5 + 9 \cdot 7 = 5 + 63 = 68\) inches.

The largest triangle has hypotenuse of length 68 inches. Since the triangles are similar, corresponding sides are in proportion. If we let \(\displaystyle a\) and \(\displaystyle b\) be the lengths of the legs of the largest triangle, then

\(\displaystyle \frac{a}{3} = \frac{68}{5}\) 

\(\displaystyle \frac{a}{3} \cdot 3 = \frac{68}{5} \cdot 3\)

\(\displaystyle a = 40.8\)

Similarly,

\(\displaystyle \frac{b}{4} = \frac{68}{5}\)

\(\displaystyle \frac{b}{4} \cdot 4 = \frac{68}{5} \cdot 4\)

\(\displaystyle b = 54.4\)

The area of a right triangle is half the product of its legs:

\(\displaystyle A = \frac{ab}{2}= \frac{40.8 \cdot 54.4}{2} = 1,109.76\) square inches.

Divide this by 144 to convert to square feet:

\(\displaystyle 1,109.76 \div 144 \approx 7.7\)

Of the given responses, 8 square feet is the closest, and is the correct choice.

The length of the leg \(\displaystyle \overline{AB}\), which we will call \(\displaystyle a\), can be calculated by setting \(\displaystyle c = 15\), the length of hypotenuse \(\displaystyle \overline{AC}\), and \(\displaystyle b= 12\), the length of leg \(\displaystyle \overline{BC}\), and applying the Pythagorean Theorem:

\(\displaystyle a = \sqrt{c^{2} - b^{2}} = \sqrt{15^{2} - 12^{2}} =\sqrt{225- 144} = \sqrt{81} = 9\)

\(\displaystyle AD = AB = 9\). 

Construct the altitude of \(\displaystyle \bigtriangleup ABC\) - which is also that of \(\displaystyle \bigtriangleup ABD\), the shaded region - from \(\displaystyle B\) to \(\displaystyle \overline{AC}\). We call the length of this altitude \(\displaystyle h\) (height). The figure is seen below.

The area of \(\displaystyle \bigtriangleup ABC\) is one half this height multiplied by the corresponding base length \(\displaystyle AC\):

\(\displaystyle \frac{1}{2} \cdot AC \cdot h\)

The area of \(\displaystyle \bigtriangleup ABD\), the shaded region - is, similarly,

\(\displaystyle \frac{1}{2} \cdot AD \cdot h\)

Therefore, the fraction that \(\displaystyle \bigtriangleup ABD\) is of \(\displaystyle \bigtriangleup ABC\) is the fraction of their areas: 

\(\displaystyle \frac{\frac{1}{2} \cdot AD \cdot h}{\frac{1}{2} \cdot AC \cdot h} = \frac{AD }{AC}\)

Substituting the measures of the two segments:

\(\displaystyle \frac{AD}{AC} = \frac{9}{15} = \frac{3}{5}\)

We are looking for a Pythagorean triple - that is, three integers that satisfy the relationship \(\displaystyle a^{2}+b^{2} = c^{2}\)  . We know that \(\displaystyle c=13\), and the only Pythagorean triple with \(\displaystyle c=13\) is \(\displaystyle (5,12,13)\). The legs of the triangle are therefore 5 and 12, and the area of the right triangle is 

\(\displaystyle A = \frac{1}{2} bh= \frac{1}{2} \cdot 5\cdot 12 = 30\)

\(\displaystyle \bigtriangleup ABC\) has as its leg lengths 10 and 24, so the length of its hypotenuse, \(\displaystyle \overline{AC}\), is

\(\displaystyle AC = \sqrt{(AB)^{2}+(BC)^{2}} = \sqrt{10^{2}+24^{2}} = \sqrt{100+576} = \sqrt{676 } = 26\)

Its perimeter is the sum of its sidelengths:

\(\displaystyle P= 10+24+26 = 60\)

Its area is half the product of the lengths of its legs:

\(\displaystyle A = \frac{1}{2} \cdot 10 \cdot 24 = 120\)

\(\displaystyle \bigtriangleup GHI\) and \(\displaystyle \bigtriangleup JKL\) have the same perimeter and area, respectively, as \(\displaystyle \bigtriangleup ABC\); also, between \(\displaystyle \bigtriangleup MNO\) and \(\displaystyle \bigtriangleup ABC\), corresponding angles are congruent. In the absence of other information, none of these three triangles can be eliminated as being congruent to \(\displaystyle \bigtriangleup ABC\).

However, \(\displaystyle DF = 28\) and \(\displaystyle AC = 26\). Therefore, \(\displaystyle \overline{AC} \ncong \overline{DF}\). Since a pair of corresponding sides is noncongruent, it follows that \(\displaystyle \bigtriangleup DEF \ncong \bigtriangleup ABC\).

\(\displaystyle \angle A \cong \angle D\); \(\displaystyle \angle B \cong \angle E\) 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 \(\displaystyle 180\).

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