Question

1. find the length of \\( \overline { e n } \\) if \\( \triangle s a m \\) and \\( \triangle b e n \\) are similar triangles.

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answers:

Explanation:

First Problem (Top Triangle Problem)

Step1: Identify similar triangles ratio

Since the triangles are similar, the ratios of corresponding sides are equal. Let the missing side (let's say the side corresponding to 8 in in Triangle 1 and the side we need to find in Triangle 2, but wait, actually looking at the sides: Triangle 1 has sides 12 in, 10 in, 8 in. Triangle 2 has sides 6 in, 5 in, and the missing side (let's call it \( x \)). The ratio of 12/6 = 2, 10/5 = 2, so the scale factor is 2. So the side corresponding to 8 in in Triangle 1 (length 8) should correspond to \( x \) in Triangle 2, so \( 8 / x = 2 \) (wait, no, actually Triangle 2 is smaller, so scale factor from Triangle 1 to Triangle 2 is 6/12 = 1/2. So the side corresponding to 8 in (Triangle 1) in Triangle 2 is \( 8 \times \frac{1}{2} = 4 \)? Wait, no, let's check the sides. Triangle 1: 12 in (one side), 10 in (base), 8 in (other side). Triangle 2: 6 in (corresponding to 12 in), 5 in (corresponding to 10 in), so the third side (corresponding to 8 in) should be \( 8 \times \frac{6}{12} = 4 \)? Wait, the answer choices have A. 4 in, so maybe that's it. Wait, let's do it properly. Let the sides of Triangle 1 be \( a = 12 \), \( b = 10 \), \( c = 8 \). Triangle 2: \( a' = 6 \), \( b' = 5 \), \( c' =? \). Since similar, \( \frac{a}{a'} = \frac{b}{b'} = \frac{c}{c'} \). \( \frac{12}{6} = 2 \), \( \frac{10}{5} = 2 \), so \( \frac{8}{c'} = 2 \) → \( c' = 4 \). So the missing side is 4 in.

Step2: Confirm the ratio

Check the ratios: 12/6 = 2, 10/5 = 2, so 8/c' = 2 → c' = 4. So the answer is 4 in.

Step1: Identify corresponding sides

Since \( \triangle SAM \sim \triangle BEN \), the ratios of corresponding sides are equal. Let's find the corresponding sides. In \( \triangle SAM \): \( SA = 22 \) m, \( SM = 24 \) m, \( AM = 28 \) m. In \( \triangle BEN \): \( EB = 11 \) m, \( BN = 12 \) m, \( EN =? \). First, find the scale factor. \( SA = 22 \) corresponds to \( EB = 11 \), so the scale factor from \( \triangle SAM \) to \( \triangle BEN \) is \( \frac{11}{22} = \frac{1}{2} \).

Step2: Find \( EN \) using the scale factor

The side \( AM = 28 \) m in \( \triangle SAM \) corresponds to \( EN \) in \( \triangle BEN \). So \( EN = AM \times \) scale factor \( = 28 \times \frac{1}{2} = 14 \)? Wait, no, wait. Wait, maybe the correspondence is different. Wait, \( \triangle SAM \) and \( \triangle BEN \): let's check the sides. \( SA = 22 \), \( EB = 11 \) (so \( SA \) corresponds to \( EB \)), \( SM = 24 \), \( BN = 12 \) (so \( SM \) corresponds to \( BN \)), so the scale factor is \( \frac{11}{22} = \frac{1}{2} \), \( \frac{12}{24} = \frac{1}{2} \). Then \( AM = 28 \) corresponds to \( EN \), so \( EN = 28 \times \frac{1}{2} = 14 \)? Wait, but let's check the other sides. Wait, maybe the sides are \( SA = 22 \), \( EB = 11 \); \( SM = 24 \), \( BN = 12 \); so the third side \( AM \) (28) corresponds to \( EN \). So \( EN = 28 \times \frac{11}{22} = 14 \). Wait, but let's verify with the other ratio. \( SM = 24 \), \( BN = 12 \), so \( 24/12 = 2 \), so scale factor from \( \triangle BEN \) to \( \triangle SAM \) is 2. So \( EN \) corresponds to \( AM \), so \( AM = EN \times 2 \) → \( 28 = EN \times 2 \) → \( EN = 14 \). So the length of \( \overline{EN} \) is 14 m.

Answer:

A. 4 in

Second Problem (Triangle SAM and BEN)