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Answers:
11. B) Acute
12. D) BA = QP
13. C) 36
15. C) BC = ST
16. Hypontenuse is roughly 23.6 millimeters
17. C) 17 ft

Work Shown

Problem 11)
T+U+V = 180
38+U+72 = 180
38+72+U = 180
110+U = 180
110+U-110 = 180-110
U = 70
All three angles (38, 72, 70) are less than 90 degrees. This triangle is acute

Problem 12)
If triangle ABC = triangle PQR, then the corresponding parts must be congruent. So,
AB = PQ
BC = QR
AC = PR
angle A = angle P
angle B = angle Q
angle C = angle R
The only choice that fits with the equations mentioned above is choice D, which is why it's the answer
Keep in mind BA is the same as AB. The order doesn't matter

Problem 13)
FP is twice as long as PY
FP = 2*PY
FY = FP+PY
FY = 2*PY+PY
FY = 3*PY
Because FP = 24, we can find PY
FP = 2*PY
24 = 2*PY
PY = 12
Which is then used to find FY
FY = 3*PY
FY = 3*12
FY = 36

Problem 15)
HL stands for "hypotenuse leg". We already have a pair of congruent legs which is RS = AB, given by the tick marks. We just need a pair of hypotenuses. In this case, BC and ST are the hypotenuse of the triangles. So if we knew BC = ST, then we'd have enough for HL to be used. 

Problem 16)
a^2 + b^2 = c^2
14^2 + 19^2 = c^2
557 = c^2
c^2 = 557
c = sqrt(557) <<--- "sqrt" stands for "square root"
c = 23.60084
c = 23.6
The hypotenuse is roughly 23.6 millimeters long

Problem 17)
Let a,b,c be the sides of the triangle. The given sides are a=6 and b=17. The unknown side is c. We don't know the exact value of c, but we can figure out how small and how large it can get.
It turns out that c is restricted through this compound inequality
b-a < c < b+a
Plug in the given values and simplify
b-a < c < b+a
17-6 < c < 17+6
11 < c < 23
So the third side can be between 11 and 23, but not equal to either endpoint. The only choice that fits this inequality is 17, so that's why the answer is C.