We can compare these numbers by comparing their squares.

By the Pythagorean Theorem, 

$\displaystyle (AC) ^{2}= (AB) ^{2}+ (BC)^{2} = 100 ^{2}+ 40 ^{2} = 10,000 + 1,600 = 11,600$

Also,

$\displaystyle 108^{2} = 11,664$

$\displaystyle 108^{2} >(AC) ^{2}$, so $\displaystyle 108 > AC$.

Suppose $\displaystyle AC = 55$.

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 $\displaystyle (AC)^{2}$

$\displaystyle (AB)^{2}+ ( BC )^{2} = 33^{2} + 44 ^{2} = 1,089+ 1,936= 3,025$

$\displaystyle (AC)^{2} = 55^{2} = 3,025$

Therefore, if $\displaystyle AC = 55$

$\displaystyle (AB)^{2}+ ( BC )^{2} =(AC)^{2}$, 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.

However, $\displaystyle \angle B$ has degree measure greater than 90, so, as a consequence of the Converse of the Pythagorean Theorem and the SAS Inequality Theorem, it holds that $\displaystyle AC > 55$.

No information is given about the legs of either triangle; therefore, no information about their perimeters can be deduced.

By the Pythagorean Theorem, the distance from B to C, which we will call $\displaystyle D$, is equal to 

$\displaystyle D = \sqrt{13^{2}- 5^{2}} = \sqrt{169- 25 }= \sqrt{144} = 12$.

The perimeter of the triangle is 

$\displaystyle 5+12 + 13 = 30$.

The insect traveled the entirety of the hypotenuse, which is 13 units long, and 75% of the longer leg, which adds 75% of 12, or $\displaystyle 0.75 \times 12 = 9$ units. Therefore, the insect has traveled 22 out of the 30 units perimeter, or 

$\displaystyle \frac{22}{30} \times 100 = 73 \frac{1}{3} \%$ of the perimeter.

The altitude of a right triangle from the vertex of its right angle - which, here, is $\displaystyle \overline{BX}$ - divides the triangle into two triangles similar to each other. The ratio of the hypotenuse of $\displaystyle \bigtriangleup CXB$ to that of $\displaystyle \bigtriangleup AXB$ (which are corresponding sides) is 

$\displaystyle \frac{BC}{AB} = \frac{40}{30} =\frac{4}{3}$ ,

making this the similarity ratio. The ratio of the perimeters of two similar triangles is the same as their similarity ratio; therefore, if $\displaystyle X$ is the perimeter of $\displaystyle \bigtriangleup AXB$ and $\displaystyle Y$ is the perimeter of $\displaystyle \bigtriangleup CXB$, it follows that

$\displaystyle \frac{Y}{X} = \frac{4}{3}$

$\displaystyle \frac{Y}{X} \cdot X = \frac{4}{3} \cdot X$

$\displaystyle Y = \frac{4}{3} X$

Multiply both sides by 3:

$\displaystyle 3 \cdot Y =3 \cdot \frac{4}{3} X$

$\displaystyle 3Y = 4X$

Three times the perimeter of $\displaystyle \bigtriangleup CXB$ is therefore equal to four times that of $\displaystyle \bigtriangleup AXB$.

Quantity A:  This is the special $\displaystyle 3$-$\displaystyle 4$-$\displaystyle 5$ triangle, where the two sides have lengths of $\displaystyle 3$ and $\displaystyle 4$ and the hypotenuse is $\displaystyle 5$.

Quantity B: This triangle has an area of $\displaystyle 10$ and base of $\displaystyle 4$. The area of a triangle is $\displaystyle \frac{bh}{2}$, so that height must be $\displaystyle 5$. 

Quantity A and Quantity B are equal.

Since the shorter leg of the right triangle is half the hypotenuse, the triangle is a $\displaystyle 30 ^{\circ } -60^{\circ } -90^{\circ }$ triangle, with the $\displaystyle 30 ^{\circ }$ angle opposite the shorter leg. That makes $\displaystyle x = 30$.

The degree measures of the acute angles of a right triangle total 90, so we solve for $\displaystyle x$ in the following equation:

$\displaystyle \left ( x + 30 \right )+ \left ( 2x- 57\right ) = 90$

$\displaystyle 3x-27 = 90$

$\displaystyle 3x-27+ 27 = 90 + 27$

$\displaystyle 3x = 117$

$\displaystyle 3x \div 3 = 117 \div 3$

$\displaystyle x= 39$

(a) $\displaystyle m\angle A = \left ( x + 30 \right )^ {\circ} = \left ( 39 + 30 \right )^ {\circ} = 69^ {\circ}$

(b) $\displaystyle m\angle C =\left ( 90 - 69 \right ) ^ {\circ} = 21^ {\circ}$

$\displaystyle m\angle A > m\angle C$

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, $\displaystyle m \angle A = 45 ^{\circ } = m \angle C$.