Question

is $\triangle pmr$ similar to $\triangle smn$? if so, which postulate or theorem proves these two triangles are similar?

• $\triangle pmr$ is similar to $\triangle smn$ by the ssa similarity theorem.
• $\triangle pmr$ is similar to $\triangle smn$ by the sas similarity theorem.
• $\triangle pmr$ is similar to $\triangle smn$ by the sss similarity theorem.
• $\triangle pmr$ is not similar to $\triangle smn$.

(image of triangle with points m, s, p, n, r and side lengths: ms = 14 ft, mp = 14 + 8 = 22? wait, no, ms is 14 ft, sp is 8 ft; mn is 21 ft, nr is 12 ft. so pm is ps + sm = 8 + 14 = 22? wait, no, looking at the image: s is on pm, n is on mr. so pm is from p to m, with s between p and m: ps = 8 ft, sm = 14 ft. mr is from m to r, with n between m and r: mn = 21 ft, nr = 12 ft. so pm = ps + sm = 8 + 14 = 22? wait, no, maybe sm is 14, sp is 8, so pm = sp + sm = 8 + 14 = 22. mr = mn + nr = 21 + 12 = 33. then sm = 14, pm = 22; mn = 21, mr = 33. also, angle at m is common. so check ratios: sm/pm = 14/22 = 7/11? wait, no, maybe i got the triangles reversed. $\triangle pmr$ and $\triangle smn$. so vertices: p, m, r and s, m, n. so angle at m is common. then sides: pm and sm, mr and mn. so pm = 8 + 14 = 22? wait, no, sp is 8, sm is 14, so pm = sp + sm = 22. sm is 14. mr is mn + nr = 21 + 12 = 33. mn is 21. so sm/pm = 14/22 = 7/11, mn/mr = 21/33 = 7/11. so sm/pm = mn/mr = 7/11, and angle at m is common. so by sas similarity (since two sides in proportion and included angle equal). wait, but lets check the options. the options are ssa (not a valid similarity theorem), sas, sss, or not similar. so first, check if the triangles are similar. lets compute the ratios. sm = 14, pm = 8 + 14 = 22? wait, no, maybe sp is 8, so pm is from p to m, with s between p and m, so ps = 8, sm = 14, so pm = ps + sm = 22. mr is from m to r, with n between m and r, so mn = 21, nr = 12, so mr = mn + nr = 33. then sm/pm = 14/22 = 7/11, mn/mr = 21/33 = 7/11. and angle at m is common. so by sas similarity (since two sides in proportion and included angle equal). so the correct option should be sas similarity theorem. but the ocr text is the question: is $\triangle pmr$ similar to $\triangle smn$? if so, which postulate or theorem proves these two triangles are similar? then the options are as listed. the image shows the triangle with side lengths: ps = 8 ft, sm = 14 ft, mn = 21 ft, nr = 12 ft. so pm = ps + sm = 24? wait, 8 + 14 = 22? no, 8 + 14 is 22? wait, 8 + 14 is 22? wait, 8 + 14 = 22? yes. 21 + 12 = 33. then 14/22 = 7/11, 21/33 = 7/11. so ratio of sm to pm is 14/22 = 7/11, ratio of mn to mr is 21/33 = 7/11. and angle at m is common. so sas similarity (since two sides in proportion and included angle). so the question is a multiple-choice with an image, about triangle similarity.

Explanation:

Step1: Calculate ratios of corresponding sides

First, find the ratio of \( MS \) to \( MP \) and \( MN \) to \( MR \).
\( MP = MS + SP = 14 + 8 = 22 \)? Wait, no, wait: \( MS = 14 \), \( SP = 8 \), so \( MP = MS + SP = 14 + 8 = 22 \)? Wait, no, looking at the diagram: \( M \) to \( S \) is 14, \( S \) to \( P \) is 8, so \( MP = 14 + 8 = 22 \)? Wait, no, maybe \( M \) to \( S \) is 14, \( S \) to \( P \) is 8, so \( MP = MS + SP = 14 + 8 = 22 \)? Wait, no, the other side: \( M \) to \( N \) is 21, \( N \) to \( R \) is 12, so \( MR = MN + NR = 21 + 12 = 33 \).

Now, ratio of \( MS \) to \( MP \): \( \frac{MS}{MP} = \frac{14}{14 + 8} = \frac{14}{22} = \frac{7}{11} \)? Wait, no, wait, maybe I got the sides wrong. Wait, \( \triangle PMR \) and \( \triangle SMN \): angle at \( M \) is common. So we check the sides around angle \( M \).

For \( \triangle SMN \) and \( \triangle PMR \), angle \( M \) is common. So we need to check \( \frac{SM}{PM} \) and \( \frac{NM}{RM} \).

\( SM = 14 \), \( PM = SM + SP = 14 + 8 = 22 \)? Wait, no, maybe \( SP = 8 \), \( SM = 14 \), so \( PM = 14 + 8 = 22 \). \( NM = 21 \), \( RM = NM + NR = 21 + 12 = 33 \).

Now, \( \frac{SM}{PM} = \frac{14}{22} = \frac{7}{11} \)? Wait, no, that can't be. Wait, maybe I mixed up the sides. Wait, \( SM = 14 \), \( PM = 14 + 8 = 22 \)? No, wait, maybe \( M \) to \( S \) is 14, \( S \) to \( P \) is 8, so \( PM = 14 + 8 = 22 \). \( M \) to \( N \) is 21, \( N \) to \( R \) is 12, so \( MR = 21 + 12 = 33 \).

Wait, no, maybe the sides are \( SM = 14 \), \( PM = 14 + 8 = 22 \)? No, wait, \( \frac{SM}{PM} = \frac{14}{22} = \frac{7}{11} \), and \( \frac{NM}{RM} = \frac{21}{33} = \frac{7}{11} \). Oh! So both ratios are equal, and the included angle \( \angle M \) is common. So by SAS Similarity (since two sides in proportion and included angle equal), the triangles are similar.

Wait, let's recalculate:

\( SM = 14 \), \( PM = SM + SP = 14 + 8 = 22 \). So \( \frac{SM}{PM} = \frac{14}{22} = \frac{7}{11} \).

\( NM = 21 \), \( RM = NM + NR = 21 + 12 = 33 \). So \( \frac{NM}{RM} = \frac{21}{33} = \frac{7}{11} \).

So \( \frac{SM}{PM} = \frac{NM}{RM} = \frac{7}{11} \), and \( \angle M \) is common (included angle). Therefore, by SAS Similarity Theorem, \( \triangle PMR \sim \triangle SMN \).

Wait, but let's check the options. The options are:

1. SSA (not a valid similarity theorem)
2. SAS Similarity
3. SSS Similarity
4. Not similar

So the correct one is the second option: \( \triangle PMR \) is similar to \( \triangle SMN \) by the SAS Similarity Theorem.

Step2: Verify the similarity condition

Since the ratio of two sides around the common angle \( M \) are equal (\( \frac{SM}{PM} = \frac{NM}{RM} = \frac{7}{11} \)) and the included angle \( \angle M \) is congruent, the SAS Similarity Theorem applies.

Answer:

\( \triangle PMR \) is similar to \( \triangle SMN \) by the SAS Similarity Theorem. (Corresponding option: the one with SAS Similarity Theorem)