Question

state if the triangles in each pair are similar. if so, state how you know they are similar and complete the similarity statement.
12)
13)
\\( \triangle u v w \sim \\)
a) similar; sss similarity; \\( \triangle k m l \\)
b) similar; aa similarity; \\( \triangle k m l \\)
c) not similar
d) similar; sss similarity; \\( \triangle k l m \\)
\\( \triangle a b c \sim \\)
a) similar; sss similarity; \\( \triangle k l m \\)
b) similar; sas similarity; \\( \triangle m k l \\)
c) similar; sas similarity; \\( \triangle k l m \\)
d) not similar
14)
15)
\\( \triangle f g h \sim \\)
a) similar; sas similarity; \\( \triangle c e d \\)
b) similar; sss similarity; \\( \triangle c d e \\)
c) similar; sss similarity; \\( \triangle c e d \\)
d) not similar
\\( \triangle a b c \sim \\)
a) similar; sas similarity; \\( \triangle a m n \\)
b) similar; sas similarity; \\( \triangle n a m \\)
c) similar; sas similarity; \\( \triangle m a n \\)
d) not similar

Explanation:

12)

Step1: Check SSS similarity

For SSS similarity, the ratios of corresponding sides should be equal.
In \(\triangle UVW\), sides are \(UV = 42\), \(VW=42\), \(WU = 42\) (assuming it's an isosceles triangle with two sides \(42\) and included angle \(60^{\circ}\), so it's equilateral).
In \(\triangle KLM\), sides are \(KL = 18\), \(LM=17\), \(MK\) (not equal ratios).
But wait, check the problem again. Wait, no:
Wait, for \(\triangle UVW\): \(UV = 42\), \(UW = 42\), \(\angle U=60^{\circ}\) (so it's equilateral, all sides \(42\)).
For \(\triangle KLM\): check ratios. Wait no, wait the problem is about similarity.
Wait, check SSS:
\(\frac{UV}{KL}\), \(\frac{VW}{LM}\), \(\frac{WU}{MK}\). But \(UV = 42\), \(KL\) - no, wait the lower triangle \(\triangle MLK\): \(ML = 17\), \(LK = 18\), \(MK\) - no. Wait no, the problem is wrong. Wait no, wait the first triangle \(\triangle UVW\): \(UV = 42\), \(UW = 42\), \(\angle U = 60^{\circ}\) (so it's equilateral, all sides \(42\)). The second triangle \(\triangle MLK\): sides \(ML = 17\), \(LK = 18\), \(MK\) - no. Wait no, wait the options:
Option D: \(\triangle UVW\sim\triangle KLM\) by SSS. But \(42/22
eq42/18
eq42/20\). No. Wait no, wait the first triangle \(\triangle UVW\): \(UV = 42\), \(UW = 42\), \(\angle U = 60^{\circ}\). The second triangle \(\triangle MLK\): \(ML = 17\), \(LK = 18\), \(\angle K=60^{\circ}\). No, sides not in proportion. Wait no - wait the problem is mis - written. Wait no, check the ratios:
If \(\triangle UVW\) and \(\triangle KLM\):
\(\frac{UV}{KL}=\frac{42}{18}=\frac{7}{3}\), \(\frac{UW}{ML}=\frac{42}{17}\) (not equal). So not SSS.
But wait - no, wait the first triangle \(\triangle UVW\): two sides \(42\), included angle \(60^{\circ}\). The second triangle \(\triangle MLK\): two sides \(17\) and \(18\), included angle \(60^{\circ}\). Sides not in proportion (\(42/17
eq42/18\)). So not similar. But wait the options:
Option C: not similar.

13)

Step1: Check SAS similarity

For \(\triangle ABC\) and \(\triangle KLM\):
In \(\triangle ABC\): \(AB = 117\), \(BC = 130\), \(AC = 143\)
In \(\triangle KLM\): \(KL = 18\), \(LM = 20\), \(MK = 22\)
Check ratios:
\(\frac{AB}{MK}=\frac{117}{22}\), \(\frac{BC}{LM}=\frac{130}{20}=\frac{13}{2}\), \(\frac{AC}{KL}=\frac{143}{18}\) (not equal).
Check another way:
For SAS similarity: two sides in proportion and included angle equal.
But no information about angles.
Alternatively, check SSS:
\(\frac{117}{22}
eq\frac{130}{20}
eq\frac{143}{18}\). So not similar. Option D.

14)

Step1: Check SSS similarity

In \(\triangle FGH\): \(FG = 24\), \(GH = 12\), \(FH = 16\)
In \(\triangle CDE\): \(CD = 18\), \(DE = 9\), \(CE = 12\)
Check ratios:
\(\frac{FG}{CD}=\frac{24}{18}=\frac{4}{3}\), \(\frac{GH}{DE}=\frac{12}{9}=\frac{4}{3}\), \(\frac{FH}{CE}=\frac{16}{12}=\frac{4}{3}\)
Since \(\frac{FG}{CD}=\frac{GH}{DE}=\frac{FH}{CE}\), by SSS similarity \(\triangle FGH\sim\triangle CDE\) (Option B: \(\triangle CDE\) - note the order. \(\triangle FGH\sim\triangle CDE\) as per SSS)

15)

Step1: Check SAS similarity

In \(\triangle ABC\) and \(\triangle AMN\):
\(\frac{AB}{AM}=\frac{16}{8} = 2\), \(\frac{AC}{AN}=\frac{20}{10}=2\)
\(\angle A\) is common.
By SAS similarity (\(AB/AM=AC/AN\) and \(\angle A=\angle A\)), \(\triangle ABC\sim\triangle AMN\) (Option A)

Answer:

1. C. not similar
2. D. not similar
3. B. similar; SSS similarity; \(\triangle CDE\)
4. A. similar; SAS similarity; \(\triangle AMN\)