The perimeter of $\displaystyle \bigtriangleup DEF$ is actually irrelevant to this problem. Corresponding sides of similar triangles are in proportion, so use this to calculate $\displaystyle DF \div DE$, or $\displaystyle \frac{DF}{DE}$:

$\displaystyle \frac{DF}{DE} = \frac{AC}{AB} = \frac{36}{24} = 1.5$

The ratio of the perimeters of two similar triangles is equal to the ratio of the lengths of a pair of corresponding sides. Therefore, 

$\displaystyle \frac{DE}{AB}= \frac{P_{\bigtriangleup DEF}}{P_{\bigtriangleup ABC }}$ and $\displaystyle \frac{EF}{BC}= \frac{P_{\bigtriangleup DEF}}{P_{\bigtriangleup ABC }}$, or

$\displaystyle \frac{DE}{AB}=\frac{EF}{BC} = \frac{P_{\bigtriangleup DEF}}{P_{\bigtriangleup ABC }}$

By one of the properties of proportions, it follows that

$\displaystyle \frac{DE+EF}{AB+BC} = \frac{P_{\bigtriangleup DEF}}{P_{\bigtriangleup ABC }}$

The perimeter of $\displaystyle \bigtriangleup ABC$ is

$\displaystyle AB+BC+ AC = 24+ 20+ 36 = 80$, so

$\displaystyle \frac{DE+EF}{24+20} = \frac{300}{80}$

$\displaystyle \frac{DE+EF}{44} = \frac{300}{80}$

$\displaystyle \frac{DE+EF}{44} \cdot 44 = \frac{300}{80} \cdot 44$

$\displaystyle DE+EF=165$

The ratio of the perimeters of two similar triangles is the same as the ratio of the lengths of a pair of corresponding sides. Therefore, 

$\displaystyle \frac{P_{\bigtriangleup ABC }}{P_{\bigtriangleup DEF}} = \frac{AB}{DE}$

$\displaystyle \frac{P_{\bigtriangleup ABC }}{90} = \frac{12}{40}$

$\displaystyle \frac{P_{\bigtriangleup ABC }}{90}\cdot 90 = \frac{12}{40} \cdot 90$

$\displaystyle P_{\bigtriangleup ABC} = 27$

The similarity of the triangles is actually extraneous information here. The sum of the measures of a triangle is $\displaystyle 180 ^{\circ }$, so:

$\displaystyle m \angle A + m \angle B + m \angle C = 180^{\circ }$

$\displaystyle m \angle D + m \angle E + m \angle F = 180^{\circ }$

$\displaystyle m \angle A + m \angle B + m \angle C +m \angle D + m \angle E + m \angle F = 180^{\circ }+ 180^{\circ }$

$\displaystyle m \angle C + m \angle E + (m \angle A +m \angle D) + (m \angle B + m \angle F )= 360^{\circ }$

$\displaystyle m \angle C + m \angle E +94 ^{\circ } +94 ^{\circ }= 360^{\circ }$

$\displaystyle m \angle C + m \angle E +188 ^{\circ }= 360^{\circ }$

$\displaystyle m \angle C + m \angle E +188 ^{\circ }- 188^{\circ } = 360^{\circ } - 188^{\circ }$

$\displaystyle m \angle C + m \angle E = 172^{\circ }$

Two pairs of corresponding angles are stated to be congruent in the main body of the problem; it follows from the Angle-Angle Similarity Postulate that the triangles are similar. No further information is needed.

 The similarity ratio of $\displaystyle \bigtriangleup ABC$ to $\displaystyle \bigtriangleup DEF$ is equal to the ratio of two corresponding sidelengths, which is given as $\displaystyle \frac{25}{36}$; the similarity ratio of $\displaystyle \bigtriangleup DEF$ to $\displaystyle \bigtriangleup ABC$ is the reciprocal of this, or $\displaystyle \frac{36}{25}$.

The ratio of the area of a figure to that of one to which it is similar is the square of the similarity ratio, so the ratio of the area of $\displaystyle \bigtriangleup DEF$ to that of $\displaystyle \bigtriangleup ABC$ is 

$\displaystyle \left (\frac{36}{25} \right ) ^{2}= \frac{1,296}{625}$

Corresponding angles of similar triangles are congruent, so the measures of the angles of $\displaystyle \bigtriangleup ABC$ are equal to those of $\displaystyle \bigtriangleup DEF$.

Two of the angles of $\displaystyle \bigtriangleup DEF$ have measures $\displaystyle 44 ^{\circ }$ and $\displaystyle 26^{\circ }$; its third angle measures 

$\displaystyle 180^{\circ } - (44^{\circ }+ 26^{\circ } ) = 180^{\circ } - 70 ^{\circ } = 110^{\circ }$.

One of the angles having measure greater than $\displaystyle 90^{\circ }$ makes $\displaystyle \bigtriangleup DEF$ - and, consequently, $\displaystyle \bigtriangleup ABC$ - an obtuse triangle. Also, the three angles have different measures, so the sides do as well, making $\displaystyle \bigtriangleup ABC$ scalene.

$\displaystyle m \angle A = 66^{\circ }$

Corresponding angles of similar triangles are congruent, so $\displaystyle m \angle C =m \angle F = 63^{\circ }$.

Consequently, 

$\displaystyle m \angle B = 180^{\circ } - (m \angle A +m \angle C)$

$\displaystyle m \angle B = 180^{\circ } - ( 66^{\circ }+ 63^{\circ })$

$\displaystyle m \angle B = 180^{\circ } - 129^{\circ } = 51^{\circ }$

Therefore,

$\displaystyle m \angle B < m \angle C < m \angle A$.

Corresponding sides of similar triangles are in proportion; since $\displaystyle \bigtriangleup ABC \sim \bigtriangleup DEF$,

$\displaystyle \frac{AC}{DF} = \frac{AB}{DE}$

$\displaystyle \frac{AC}{100} = \frac{50}{75}$

$\displaystyle \frac{AC}{100} \cdot 100 = \frac{50}{75} \cdot 100$

$\displaystyle AC = 66\frac{2}{3}$

Therefore, $\displaystyle AB < AC < BC$. 

The angle opposite the longest (shortest) side of a triangle is the angle of greatest (least) measure, so

$\displaystyle m \angle C < m \angle B < m \angle A$.

$\displaystyle \overline{DF}$ and $\displaystyle \overline{AC}$ are corresponding sides of their respective triangles, and $\displaystyle DF < AC$, so it easily follows from proportionality that each side of $\displaystyle \bigtriangleup DEF$ is shorter than its corresponding side in $\displaystyle \bigtriangleup ABC$. Therefore, $\displaystyle \bigtriangleup DEF$ is of lesser perimeter than $\displaystyle \bigtriangleup ABC$. By the same reasoning, since $\displaystyle AB < GH$,  $\displaystyle \bigtriangleup ABC$ is of lesser perimeter than $\displaystyle \bigtriangleup GHI$.

The correct response is

$\displaystyle \bigtriangleup DEF, \bigtriangleup ABC,\bigtriangleup GHI$