Question

the pentagons abcde and jklmn are similar.
find the length x of \\(\overline{nj}\\).
(there are two pentagon figures, one labeled abcde with side lengths: ab=1, bc=4, cd=1, de=2, ae=3; the other labeled jklmn with side lengths: jk=1.6, kl=6.4, lm=1.6, mn=3.2, and nj=x.)
\\(x = \square\\)

Explanation:

Step1: Identify corresponding sides

In similar pentagons, corresponding sides are proportional. Let's find the ratio of corresponding sides. For example, side \( AB = 1 \) in pentagon \( ABCDE \) corresponds to side \( JK = 1.6 \) in pentagon \( JKLMN \)? Wait, no, looking at the labels: \( A \) corresponds to \( J \), \( B \) to \( K \), \( C \) to \( L \), \( D \) to \( M \), \( E \) to \( N \). So side \( AB = 1 \) (from \( A \) to \( B \)) corresponds to \( JK = 1.6 \)? Wait, no, \( A \) to \( J \), \( B \) to \( K \), \( C \) to \( L \), \( D \) to \( M \), \( E \) to \( N \). So side \( AE = 3 \) (from \( A \) to \( E \)) corresponds to \( JN = x \) (from \( J \) to \( N \)), and side \( DE = 2 \) (from \( D \) to \( E \)) corresponds to \( MN = 3.2 \) (from \( M \) to \( N \)). Let's check the ratio of \( DE \) to \( MN \): \( \frac{DE}{MN} = \frac{2}{3.2} \). Wait, or maybe \( AB = 1 \) (from \( A \) to \( B \)) corresponds to \( JK = 1.6 \)? Wait, \( A \) to \( B \) is length 1, \( J \) to \( K \) is length 1.6? Wait, no, the first pentagon: \( A \) to \( B \) is 1, \( B \) to \( C \) is 4, \( C \) to \( D \) is 1, \( D \) to \( E \) is 2, \( E \) to \( A \) is 3. The second pentagon: \( J \) to \( K \) is 1.6, \( K \) to \( L \) is 6.4, \( L \) to \( M \) is 1.6, \( M \) to \( N \) is 3.2, \( N \) to \( J \) is \( x \). So corresponding sides: \( AB \) (length 1) corresponds to \( JK \) (length 1.6)? Wait, no, \( A \) to \( B \) is 1, \( J \) to \( K \) is 1.6? Wait, \( A \) to \( E \) is 3, \( J \) to \( N \) is \( x \); \( D \) to \( E \) is 2, \( M \) to \( N \) is 3.2. Let's find the scale factor. Let's take \( DE = 2 \) and \( MN = 3.2 \). The ratio of \( MN \) to \( DE \) is \( \frac{3.2}{2} = 1.6 \). Let's check another pair: \( CD = 1 \) and \( LM = 1.6 \). \( \frac{LM}{CD} = \frac{1.6}{1} = 1.6 \). Good, so the scale factor is 1.6. Now, \( AE = 3 \) (from \( E \) to \( A \)) corresponds to \( JN = x \) (from \( N \) to \( J \))? Wait, no, \( E \) to \( A \) is 3, \( N \) to \( J \) is \( x \). Wait, \( E \) to \( A \) is length 3, \( N \) to \( J \) is length \( x \). Since the scale factor is 1.6 (from first pentagon to second), then \( x = 3 \times 1.6 \)? Wait, no, wait: first pentagon is \( ABCDE \), second is \( JKLMN \). So \( ABCDE \sim JKLMN \), so the ratio of corresponding sides is \( \frac{JKLMN}{ABCDE} = 1.6 \). So \( AE \) (length 3) in \( ABCDE \) corresponds to \( JN \) (length \( x \)) in \( JKLMN \). So \( x = 3 \times 1.6 \)? Wait, but let's check with \( DE = 2 \) and \( MN = 3.2 \): \( 2 \times 1.6 = 3.2 \), which matches. \( CD = 1 \) and \( LM = 1.6 \): \( 1 \times 1.6 = 1.6 \), which matches. \( AB = 1 \) and \( JK = 1.6 \): \( 1 \times 1.6 = 1.6 \), which matches. \( BC = 4 \) and \( KL = 6.4 \): \( 4 \times 1.6 = 6.4 \), which matches. So yes, the scale factor is 1.6. Therefore, \( AE = 3 \) corresponds to \( JN = x \), so \( x = 3 \times 1.6 = 4.8 \)? Wait, no, wait: \( AE \) is from \( E \) to \( A \), length 3. \( JN \) is from \( J \) to \( N \), length \( x \). So since the pentagons are similar, the ratio of \( JN \) to \( AE \) should be equal to the scale factor. The scale factor is \( \frac{MN}{DE} = \frac{3.2}{2} = 1.6 \). So \( \frac{x}{3} = 1.6 \), so \( x = 3 \times 1.6 = 4.8 \). Wait, but let's check another way. Let's use \( AB = 1 \) and \( JK = 1.6 \). So ratio is \( 1.6/1 = 1.6 \). Then \( AE = 3 \), so \( JN = 3 \times 1.6 = 4.8 \). Yes, that works.

Step2: Calculate x

We know that in similar figures, the ratio of corresponding sides is equal. Let's take side \( DE = 2 \) (in \( ABCDE \)) and side \( MN = 3.2 \) (in \( JKLMN \)). The ratio of similarity is \( \frac{MN}{DE} = \frac{3.2}{2} = 1.6 \). Now, side \( AE = 3 \) (in \( ABCDE \)) corresponds to side \( JN = x \) (in \( JKLMN \)). So we set up the proportion: \( \frac{x}{3} = 1.6 \). Solving for \( x \), we multiply both sides by 3: \( x = 3 \times 1.6 = 4.8 \).

Answer:

\( x = 4.8 \)