Corresponding angles of similar triangles are congruent, so, since \(\displaystyle \bigtriangleup ABC \sim \bigtriangleup DEF\), and \(\displaystyle \angle A\) is right, it follows that 

\(\displaystyle m \angle D = m \angle A = 90 ^{\circ }\)

\(\displaystyle \angle D\) is a right angle of a right triangle \(\displaystyle \bigtriangleup DEF\). The other two angles must be acute - that is, with measure less than \(\displaystyle 90 ^{\circ }\) -  so \(\displaystyle m \angle D > m \angle E\).

The figure referenced is below:

\(\displaystyle \overarc{ACB}\) has measure \(\displaystyle 270 ^{\circ }\), so its corresponding minor arc, \(\displaystyle \overarc{A B}\), has measure \(\displaystyle 360 ^{\circ } - 270 ^{\circ } = 90^{\circ }\). The inscribed angle that intercepts this arc, which is \(\displaystyle \angle C\), has measure half this, or \(\displaystyle 45 ^{\circ }\). Since \(\displaystyle \angle B\) is a right angle, the other acute angle, \(\displaystyle \angle A\), has measure 

\(\displaystyle m \angle A = 90 ^{\circ } - m \angle C = 90 ^{\circ } - 45^{\circ } = 45 ^{\circ }\)

Therefore, \(\displaystyle m \angle A = 45 ^{\circ } = m \angle C\).

By the Converse of the Pythagorean Theorem, a triangle is right if and only if the sum of the squares of the lengths of the smallest two sides is equal to the square of the longest side. Compare the quantities \(\displaystyle (AB)^{2}+ ( BC )^{2}\) and \(\displaystyle (AC)^{2}\)

\(\displaystyle (AB)^{2}+ ( BC )^{2} = 36 ^{2} + 48^{2} = 1,296 + 2,304 = 3,600\)

\(\displaystyle (AC)^{2} = 60^{2} = 3,600\)

\(\displaystyle (AB)^{2}+ ( BC )^{2} =(AC)^{2}\), so \(\displaystyle \bigtriangleup ABC\) is right, with the right angle opposite longest side \(\displaystyle \overline{AC}\). Thus, \(\displaystyle \angle B\) is right and has degree measure 90.

Let \(\displaystyle s\) be the length of a leg of the right triangle. Then the sidelength of the square is \(\displaystyle \frac{2}{3}s\) .

(a) The square has area \(\displaystyle A = \left ( \frac{2}{3} s \right ) ^{2} = \frac{4}{9} s ^{2}\)

(b) The isosceles right triangle has base and height \(\displaystyle s\) area \(\displaystyle A = \frac{1}{2}s^{2}\)

\(\displaystyle \frac{1}{2} > \frac{4}{9}\), so (b) is greater.

Each triangle is a right triangle with legs along the \(\displaystyle x\)- and \(\displaystyle y\)-axes, so the area of each can be calculated by taking one-half the product of the two legs. 

(a) The horizontal and vertical legs have measures 18 and \(\displaystyle b\), respectively, so the triangle has area \(\displaystyle A = \frac{1}{2} \cdot 18 \cdot b = 9b\).

(b) The horizontal and vertical legs have measures \(\displaystyle 2b\) and 9, respectively, so the triangle has area \(\displaystyle A = \frac{1}{2} \cdot 2b \cdot 9= 9b\).

The areas are equal.

\(\displaystyle \Delta MEC\) is a right triangle with right angle \(\displaystyle \angle E\), so its legs measure \(\displaystyle ME\) and \(\displaystyle EC\); its area is \(\displaystyle \frac{1}{2}\left ( ME \right ) \left (EC \right )\). Since \(\displaystyle M\) is the midpoint of \(\displaystyle \overline{RE}\), \(\displaystyle ME = \frac{1}{2} \left ( RE \right )\), making the area of the triangle

\(\displaystyle \frac{1}{2}\left ( ME \right ) \left (EC \right ) = \frac{1}{2} \cdot \frac{1}{2} \left ( RE \right ) \left (EC \right ) = \frac{1}{4} \left ( RE \right ) \left (EC \right )\)

Rectangle \(\displaystyle RECT\) has area \(\displaystyle \left ( RE \right ) \left (EC \right )\). 

Quadrilateral \(\displaystyle RMCT\),  which is the portion of \(\displaystyle RECT\) not in \(\displaystyle \Delta MEC\), has as its area

\(\displaystyle \left ( RE \right ) \left (EC \right ) - \frac{1}{4} \left ( RE \right ) \left (EC \right ) = \frac{3}{4} \left ( RE \right ) \left (EC \right )\)

Therefore, the area of Quadrilateral \(\displaystyle RMCT\) is three times that of \(\displaystyle \Delta MEC\), making (a) and (b) equal.

Each triangle is a right triangle, and each has its two legs as its base and height.

(a) \(\displaystyle M\) is the midpoint of \(\displaystyle \overline{RE}\), so \(\displaystyle ME = \frac{1}{2} RE\).

The area of \(\displaystyle \Delta MEC\) is \(\displaystyle A = \frac{1}{2}bh= \frac{1}{2} \left ( ME \right ) \left ( EC\right )= \frac{1}{2} \left ( \frac{1}{2} RE \right ) \left ( EC\right ) =\frac{1}{4} \left ( RE \right ) \left ( EC\right )\).

(b) \(\displaystyle N\) is the midpoint of \(\displaystyle \overline{EC}\), so \(\displaystyle NE = \frac{1}{2} EC\).

 The area of \(\displaystyle \Delta NER\) is

\(\displaystyle A = \frac{1}{2}bh= \frac{1}{2} \left ( RE \right ) \left ( NE \right )= \frac{1}{2} \left ( \frac{1}{2} RE \right ) \left ( \frac{1}{2} EC\right ) =\frac{1}{4} \left ( RE \right ) \left ( EC\right )\).

The triangles have equal area.

(a) Let \(\displaystyle s\) be the sidelength of the square. Then its area is \(\displaystyle A = s^{2}\).

(b) In a \(\displaystyle 30 ^{\circ }- 60 ^{\circ }-90 ^{\circ }\) triangle, the shorter leg is one-half as long as the hypotenuse. The hypotenuse has length \(\displaystyle 2s\), so the shorter leg has length \(\displaystyle s\). The longer leg is \(\displaystyle \sqrt{3}\) times as long as the shorter leg, so the longer leg will have length \(\displaystyle s\sqrt{3}\). The area of the triangle is

\(\displaystyle A = \frac{1}{2} bh = \frac{1}{2} s \cdot s\sqrt{ 3} = \frac{\sqrt{ 3}}{2} s^{2}\).

\(\displaystyle \frac{\sqrt{3} }{2}\approx \frac{1.7 }{2} < 1\), so \(\displaystyle \frac{\sqrt{ 3}}{2} s^{2} < s^{2}\);  the square has the greater area.

The area of a triangle is half the product of its base and its height; for a right triangle, the legs, which are perpendicular, serve as base and height.

\(\displaystyle A = \frac{1}{2} (k+2) (k-2)\)

\(\displaystyle A = \frac{1}{2} (k^{2}-2^{2})\)

\(\displaystyle A = \frac{1}{2} (k^{2}-4)\)

\(\displaystyle A = \frac{1}{2} k^{2}-2\)

The area of the triangle is 

\(\displaystyle A = \frac{1}{2} \cdot 18 \cdot 30 = 270\) square inches.

The sidelength of the square is \(\displaystyle 2 \times 12 = 24\) inches, so the area of the square is 

\(\displaystyle A = 24 ^{2} = 24 \times 24 = 576\).

The question becomes "what percent of 576 is 270", which is answered as follows:

\(\displaystyle P = \frac{270}{576} \times 100 = 46 \frac{7}{8}\)

The correct answer is \(\displaystyle 46 \frac{7}{8} \%\).