QUESTION IMAGE
Question
pablo was asked to find the original dimension of an enlarged pentagon. his solution is shown next to the pentagons.
now drawn to scale
what error did pablo make in his solution?
o pablo set up an incorrect proportion.
o pablo made an error when he cross multiplied the proportion.
o pablo divided incorrectly when he solved for x.
o pablo multiplied instead of divided when he solved for x.
write a proportion. \\( \frac { 3 } { x } = \frac { 1 2 } { 8 } \\)
cross multiply. \\( 8 x = 3 6 \\)
solve for x. \\( x = 4 5 \\)
To determine the error, we analyze the proportion setup. The original pentagon has a side of 3 cm, and the enlarged one has a corresponding side of 12 cm. Another side of the original is \( x \) cm, and the enlarged is 8 cm. The correct proportion should be \(\frac{3}{12}=\frac{x}{8}\) (original/enlarged for corresponding sides). But Pablo set up \(\frac{3}{12}=\frac{x}{8}\) incorrectly? Wait, no—wait, when setting up proportions for similar figures, the ratio of corresponding sides should be consistent. The original small pentagon's side is 3, enlarged is 12 (so scale factor from small to large is \( \frac{12}{3}=4 \)). The other side of small is \( x \), large is 8. So correct proportion is \( \frac{3}{12}=\frac{x}{8} \)? Wait, no—if small to large, then \( \frac{\text{small side}}{\text{large side}}=\frac{3}{12}=\frac{x}{8} \)? Wait, no, if the small pentagon has side 3 and \( x \), and large has 12 and 8, then the ratio of small to large should be \( \frac{3}{12}=\frac{x}{8} \), but when cross - multiplying, it should be \( 12x = 3\times8=24 \), so \( x = 2 \). But Pablo set up \( \frac{3}{12}=\frac{x}{8} \) and then cross - multiplied as \( 8x = 36 \) (wait, 3×12 = 36). Wait, no, the proportion Pablo used was \( \frac{3}{12}=\frac{x}{8} \), cross - multiplying should be \( 12x=3\times8 \), but Pablo did \( 8x = 3\times12 \) (since 3×12 = 36). Wait, no, the problem says Pablo's cross - multiply was \( 8x = 36 \). Wait, the correct proportion setup: if the two pentagons are similar, the ratio of corresponding sides should be equal. The small pentagon has side 3 and \( x \), the large has 12 and 8. So the ratio of small to large for the first pair is \( \frac{3}{12} \), and for the second pair should be \( \frac{x}{8} \) (if \( x \) is small and 8 is large) or \( \frac{8}{x} \) (if 8 is small and \( x \) is large). Wait, the diagram: the small pentagon has sides 3 and \( x \), large has 12 and 8. So the correct proportion is \( \frac{3}{12}=\frac{x}{8} \) (small side 3 corresponds to large side 12, small side \( x \) corresponds to large side 8). So cross - multiplying: \( 12x=3\times8 = 24 \), so \( x = 2 \). But Pablo set up \( \frac{3}{12}=\frac{x}{8} \) and cross - multiplied as \( 8x=3\times12 = 36 \) (since 3×12 = 36). Wait, no, the proportion Pablo wrote was \( \frac{3}{12}=\frac{x}{8} \), then cross - multiplied to get \( 8x = 36 \). But the correct cross - multiplication for \( \frac{3}{12}=\frac{x}{8} \) is \( 12x=3\times8 \), not \( 8x = 3\times12 \). So Pablo set up an incorrect proportion? Wait, no—maybe the correspondence of sides is wrong. The small pentagon: one side 3, another \( x \); large pentagon: one side 12, another 8. If the scale is from small to large, then \( \frac{3}{12}=\frac{x}{8} \), but if it's from large to small, \( \frac{12}{3}=\frac{8}{x} \), which would be \( 12x = 24 \), \( x = 2 \). But Pablo's proportion is \( \frac{3}{12}=\frac{x}{8} \), which is a wrong setup because the sides are not corresponding correctly. Wait, the first option is "Pablo set up an incorrect proportion." Let's check the other options:
- Option 2: "Pablo made an error when he cross multiplied the proportion." If the proportion was correct, cross - multiplying error? But if the proportion is wrong, the setup is wrong.
- Option 3: "Pablo divided incorrectly when he solved for x." If we take Pablo's equation \( 8x = 36 \), solving for \( x \) would be \( x=\frac{36}{8}=4.5 \), but if the proportion was wrong, the setup is wrong.
- Option 4: "Pablo multiplied instead of divided when he solved for x." No, solving \( 8x = 36 \…
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A. Pablo set up an incorrect proportion.