Question

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triangles d and e are scaled copies of each other. what is the measurement of the missing side length in triangle d?
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triangle d (right triangle) has a base of \\(\frac{3}{2}\\) and a hypotenuse marked with ?. triangle e (right triangle) has a base of 6 and a hypotenuse of 10. answer options: 2, \\(\left(\frac{5}{2}\
ight)\\), 8, \\(\left(\frac{10}{2}\
ight)\\).
triangles d and e are scaled copies of each other.

Explanation:

Step1: Find the scale factor

Since triangles \( D \) and \( E \) are scaled copies, the ratio of corresponding sides is equal. Let's take the horizontal sides: in triangle \( E \), the horizontal side is \( 6 \), and in triangle \( D \), it is \( \frac{3}{2} \). The scale factor from \( E \) to \( D \) is \( \frac{\frac{3}{2}}{6}=\frac{3}{2}\times\frac{1}{6}=\frac{1}{4} \)? Wait, no, wait. Wait, maybe from \( D \) to \( E \). Wait, triangle \( D \) has horizontal side \( \frac{3}{2} \), triangle \( E \) has horizontal side \( 6 \). So the scale factor (scaling \( D \) to \( E \)) is \( \frac{6}{\frac{3}{2}} = 6\times\frac{2}{3}=4 \)? Wait, no, wait the hypotenuse of \( E \) is \( 10 \), so let's check the ratio of horizontal sides: \( 6 \) (E) and \( \frac{3}{2} \) (D). So \( 6\div\frac{3}{2}=6\times\frac{2}{3} = 4 \). Wait, but then the hypotenuse of \( E \) is \( 10 \), so hypotenuse of \( D \) should be \( 10\div4=\frac{10}{4}=\frac{5}{2} \)? Wait, no, maybe I mixed up. Wait, triangle \( D \) is smaller, triangle \( E \) is larger. So the ratio of corresponding sides (D to E) is \( \frac{\frac{3}{2}}{6}=\frac{1}{4} \)? No, that can't be. Wait, let's do it properly. Let the missing side in \( D \) be \( x \). Then, since they are similar triangles, the ratios of corresponding sides are equal. So \( \frac{x}{10}=\frac{\frac{3}{2}}{6} \). Let's solve for \( x \).

Step2: Solve for the missing side

We have the proportion \( \frac{x}{10}=\frac{\frac{3}{2}}{6} \). Simplify the right - hand side: \( \frac{\frac{3}{2}}{6}=\frac{3}{2}\times\frac{1}{6}=\frac{3}{12}=\frac{1}{4} \)? Wait, no, \( \frac{3}{2}\div6=\frac{3}{2}\times\frac{1}{6}=\frac{1}{4} \)? Wait, but \( \frac{3}{2} \) is the horizontal side of \( D \), \( 6 \) is the horizontal side of \( E \). So \( \frac{\text{side of } D}{\text{side of } E}=\frac{\frac{3}{2}}{6}=\frac{1}{4} \). Then, the hypotenuse of \( D \) (x) over hypotenuse of \( E \) (10) should be equal to the same ratio. So \( \frac{x}{10}=\frac{1}{4} \)? No, that would give \( x = \frac{10}{4}=\frac{5}{2} \). Wait, let's check again. Wait, maybe the ratio is \( \frac{\text{side of } E}{\text{side of } D}=\frac{6}{\frac{3}{2}} = 4 \), so the scale factor from \( D \) to \( E \) is \( 4 \). So hypotenuse of \( E \) is \( 10 \), so hypotenuse of \( D \) is \( 10\div4=\frac{10}{4}=\frac{5}{2} \). Yes, that makes sense. So the missing side length in triangle \( D \) is \( \frac{5}{2} \)? Wait, no, wait the options are \( 2 \), \( \frac{5}{2} \), \( 8 \), \( \frac{10}{2} \). Wait, \( \frac{10}{2}=5 \), no. Wait, maybe I made a mistake in the corresponding sides. Let's look at the triangles: both are right - angled triangles. So in triangle \( E \), horizontal side is \( 6 \), hypotenuse is \( 10 \), so the vertical side can be found by Pythagoras: \( \sqrt{10^{2}-6^{2}}=\sqrt{100 - 36}=\sqrt{64}=8 \). So triangle \( E \) has sides \( 6 \), \( 8 \), \( 10 \) (a 3 - 4 - 5 triangle scaled by 2, since \( 6 = 3\times2 \), \( 8 = 4\times2 \), \( 10 = 5\times2 \)). Then triangle \( D \) has horizontal side \( \frac{3}{2} \), which is \( 3\times\frac{1}{2} \). So the scale factor from \( E \) to \( D \) is \( \frac{1}{2}\div2=\frac{1}{4} \)? No, wait \( 6 \) (E) corresponds to \( \frac{3}{2} \) (D). So \( 6\times k=\frac{3}{2} \), so \( k=\frac{3}{2}\div6=\frac{1}{4} \)? No, \( 6\times\frac{1}{4}=\frac{3}{2} \), yes. Then the hypotenuse of \( D \) is \( 10\times\frac{1}{4}=\frac{10}{4}=\frac{5}{2} \). Yes, that's correct.

Answer:

\(\frac{5}{2}\)