Question

1. similar proofs: decide if each pair of triangles below is similar. use a flowchart to organize your facts and conclusion for each pair of triangles. show why they are or are not similar triangles.

a.
triangle with vertices y, r, n: ry=9, yn=12, rn=8; triangle with vertices e, d, a: de=15, ea=10, da=11.25
b.
triangle with vertices m, h, t: mh=32, mt=30, angle at m; triangle with vertices s, w, q: sq=48, wq=44, angle at q

Explanation:

Step1: Analyze Triangle Pair a

First, we need to check the ratios of corresponding sides. Let's label the sides:
- Triangle \( RYN \): \( RY = 9 \), \( YN = 12 \), \( RN = 8 \)
- Triangle \( DEA \): \( DE = 15 \), \( EA = 10 \), \( DA = 11.25 \)

Now, find the ratios of corresponding sides:
- \( \frac{RY}{EA} = \frac{9}{10} = 0.9 \)
- \( \frac{YN}{DE} = \frac{12}{15} = 0.8 \) Wait, that's not right. Maybe we need to re - check the correspondence. Let's try to match the sides in order. Let's sort the sides of each triangle.

For triangle \( RYN \): sides are \( 8, 9, 12 \) (sorted: \( 8, 9, 12 \))
For triangle \( DEA \): sides are \( 10, 11.25, 15 \) (sorted: \( 10, 11.25, 15 \))

Now, check the ratios of corresponding sorted sides:
- \( \frac{8}{10}=\frac{4}{5} = 0.8 \)
- \( \frac{9}{11.25}=\frac{9}{\frac{45}{4}}=9\times\frac{4}{45}=\frac{36}{45}=\frac{4}{5}=0.8 \)
- \( \frac{12}{15}=\frac{4}{5}=0.8 \)

Since all three ratios of corresponding sides are equal (\( \frac{4}{5} \)), by the Side - Side - Side (SSS) similarity criterion, the triangles \( RYN \) and \( DEA \) are similar.

Step2: Analyze Triangle Pair b

For triangle \( MTH \): sides are \( MT = 30 \), \( MH = 32 \), \( HT \) (we can assume it's a right triangle? Wait, the angle at \( M \) is marked as equal to the angle at \( Q \). Wait, first, check the given sides.

Triangle \( MTH \): \( MT = 30 \), \( MH = 32 \)
Triangle \( WQS \): \( WQ = 44 \), \( QS = 48 \)

First, check the ratio of the sides adjacent to the equal angles. The angle at \( M \) and angle at \( Q \) are equal (marked with the same angle symbol). So, if the triangles are similar by the Side - Angle - Side (SAS) similarity criterion, the ratio of the sides adjacent to the equal angle should be equal, and the ratio of the other pair of sides should also be equal.

The sides adjacent to the equal angle:
- In triangle \( MTH \): \( MT = 30 \), \( MH = 32 \)
- In triangle \( WQS \): \( WQ = 44 \), \( QS = 48 \)

Check the ratio of \( MT/WQ=\frac{30}{44}=\frac{15}{22}\approx0.6818 \)
Check the ratio of \( MH/QS=\frac{32}{48}=\frac{2}{3}\approx0.6667 \)

Since \( \frac{15}{22} eq\frac{2}{3} \), the triangles \( MTH \) and \( WQS \) are not similar by SAS. Also, let's check the SSS ratio. The sides of triangle \( MTH \): let's assume it's a right triangle? Wait, if we consider the sides, for triangle \( MTH \), if it's a right triangle (assuming \( \angle T = 90^{\circ} \)), then \( HT=\sqrt{32^{2}-30^{2}}=\sqrt{(32 + 30)(32 - 30)}=\sqrt{62\times2}=\sqrt{124}\approx11.14 \)

For triangle \( WQS \), if \( \angle W = 90^{\circ} \), then \( SW=\sqrt{48^{2}-44^{2}}=\sqrt{(48 + 44)(48 - 44)}=\sqrt{92\times4}=\sqrt{368}\approx19.18 \)

Now, check the ratios of sides:
- \( \frac{MT}{WQ}=\frac{30}{44}=\frac{15}{22}\approx0.6818 \)
- \( \frac{HT}{SW}=\frac{\sqrt{124}}{\sqrt{368}}=\sqrt{\frac{124}{368}}=\sqrt{\frac{31}{92}}\approx\sqrt{0.337}\approx0.581 \)
- \( \frac{MH}{QS}=\frac{32}{48}=\frac{2}{3}\approx0.6667 \)

Since the ratios are not equal, the triangles in pair b are not similar.

Step1 (Revised for Pair a): Correct Correspondence

Wait, maybe I made a mistake in the first correspondence. Let's try to find the correct corresponding sides. Let's check \( \frac{RN}{DA}=\frac{8}{11.25}=\frac{8}{\frac{45}{4}}=\frac{32}{45}\approx0.711 \), no. Wait, let's use the Law of Sines or check the ratios again.

Wait, \( \frac{RY}{DE}=\frac{9}{15}=\frac{3}{5} \), \( \frac{YN}{EA}=\frac{12}{10}=\frac{6}{5} \), no. Wait, maybe the triangles are labeled differently. Let's check \( \frac{RN}{EA}=\frac{8}{10}=\frac{4}{5} \), \( \frac{RY}{DA}=\frac{9}{11.25}=\frac{4}{5} \), \( \frac{YN}{DE}=\frac{12}{15}=\frac{4}{5} \). Ah! So \( RN \) corresponds to \( EA \), \( RY \) corresponds to \( DA \), and \( YN \) corresponds to \( DE \). So the ratios are:

\( \frac{RN}{EA}=\frac{8}{10}=\frac{4}{5} \), \( \frac{RY}{DA}=\frac{9}{11.25}=\frac{4}{5} \), \( \frac{YN}{DE}=\frac{12}{15}=\frac{4}{5} \)

Since all three ratios of corresponding sides are equal, by the SSS (Side - Side - Side) similarity criterion, triangle \( RYN \sim \) triangle \( DAE \) (wait, maybe the correspondence is \( RYN \) and \( DAE \) with \( R ightarrow D \), \( Y ightarrow A \), \( N ightarrow E \)? No, the side lengths: \( RN = 8 \), \( DA = 11.25 \); \( RY = 9 \), \( DE = 15 \); \( YN = 12 \), \( EA = 10 \). Wait, no, \( \frac{RY}{DE}=\frac{9}{15}=\frac{3}{5} \), \( \frac{YN}{EA}=\frac{12}{10}=\frac{6}{5} \), that's not equal. Wait, I think I messed up the side labels. Let's list the sides of each triangle:

Triangle 1 (RYN): \( RY = 9 \), \( YN = 12 \), \( RN = 8 \)

Triangle 2 (DEA): \( DE = 15 \), \( EA = 10 \), \( DA = 11.25 \)

Now, let's find the ratios of \( RY/EA=\frac{9}{10}=0.9 \), \( YN/DE=\frac{12}{15}=0.8 \), \( RN/DA=\frac{8}{11.25}\approx0.711 \). No, that's not right. Wait, maybe the triangles are similar by a different correspondence. Let's check the ratios of \( RN/EA=\frac{8}{10}=0.8 \), \( RY/DA=\frac{9}{11.25}=0.8 \), \( YN/DE=\frac{12}{15}=0.8 \). Ah! So \( RN \) corresponds to \( EA \), \( RY \) corresponds to \( DA \), and \( YN \) corresponds to \( DE \). So the sides:

- \( RN = 8 \), \( EA = 10 \), ratio \( 8/10 = 0.8 \)
- \( RY = 9 \), \( DA = 11.25 \), ratio \( 9/11.25=0.8 \) (since \( 11.25\times0.8 = 9 \))
- \( YN = 12 \), \( DE = 15 \), ratio \( 12/15 = 0.8 \)

Yes! So all three ratios are equal (\( 0.8 \) or \( \frac{4}{5} \)). So by the SSS similarity criterion, triangle \( RYN \) and triangle \( DAE \) (wait, \( DAE \) has sides \( DA = 11.25 \), \( AE = 10 \), \( DE = 15 \)) are similar.

Step2 (Revised for Pair b): Check SAS

For triangle \( MTH \) and \( WQS \), we have a common angle (the marked angle at \( M \) and \( Q \)). Let's check the ratios of the sides around the common angle.

In triangle \( MTH \): sides adjacent to the marked angle are \( MT = 30 \) and \( MH = 32 \)

In triangle \( WQS \): sides adjacent to the marked angle are \( WQ = 44 \) and \( QS = 48 \)

The ratio of the first pair of sides: \( \frac{MT}{WQ}=\frac{30}{44}=\frac{15}{22}\approx0.6818 \)

The ratio of the second pair of sides: \( \frac{MH}{QS}=\frac{32}{48}=\frac{2}{3}\approx0.6667 \)

Since \( \frac{15}{22} eq\frac{2}{3} \), the SAS similarity criterion is not satisfied. Also, if we check the other sides, the ratio of the non - included sides (the sides opposite the marked angle) would be \( \frac{HT}{SW} \). For triangle \( MTH \), using Pythagoras (assuming \( \angle T = 90^{\circ} \)), \( HT=\sqrt{32^{2}-30^{2}}=\sqrt{1024 - 900}=\sqrt{124}\approx11.14 \). For triangle \( WQS \), assuming \( \angle W = 90^{\circ} \), \( SW=\sqrt{48^{2}-44^{2}}=\sqrt{2304 - 1936}=\sqrt{368}\approx19.18 \). The ratio \( \frac{HT}{SW}=\frac{\sqrt{124}}{\sqrt{368}}\approx\frac{11.14}{19.18}\approx0.581 \), which is not equal to the other ratios. So the triangles in pair b are not similar.

Answer:

a. The triangles are similar (by SSS similarity, as the ratios of corresponding sides are equal: \( \frac{8}{10}=\frac{9}{11.25}=\frac{12}{15}=\frac{4}{5} \)).
b. The triangles are not similar (the ratios of sides around the common angle are not equal, so SAS similarity is not satisfied).