| 模型 | 模型答案 | 正确答案 |
|---|---|---|
| 金声玉亮2.0 | Given that polygon ACDF is similar to polygon VWYZ, we can set up a proportion based on the corresponding sides. Similar polygons have corresponding sides that are proportional. Let's set up the proportion based on the given side lengths: \[ \frac{AF}{YZ} = \frac{CD}{YW} = \frac{AC}{VW} = \frac{DF}{ZY} = \frac{AD}{VZ} \] We are given that: - AF = 12 - CD = 9 - YZ = 10 - YW = 6 - ZV = 3y-1 We can use the first two pieces of information to set up a proportion: \[ \frac{AF}{YZ} = \frac{CD}{YW} \] Substituting the given values: \[ \frac{12}{10} = \frac{9}{6} \] Simplifying both sides: \[ \frac{6}{5} = \frac{3}{2} \] This equation is true, confirming that the polygons are similar. However, we need to find y, which is part of the expression for ZV. We can use the proportion involving ZV: \[ \frac{AD}{VZ} = \frac{AF}{YZ} \] We do not have the value for AD, but we can use the proportion involving CD and YW since we have all the necessary values: \[ \frac{CD}{YW} = \frac{ZV}{AF} \] Substituting the given values: \[ \frac{9}{6} = \frac{3y-1}{12} \] Simplifying the left side: \[ \frac{3}{2} = \frac{3y-1}{12} \] Cross multiplying to solve for y: \[ 3 \times 12 = 2 \times (3y-1) \] \[ 36 = 6y - 2 \] \[ 38 = 6y \] \[ y = \frac{38}{6} = \frac{19}{3} \] Thus, the value of y is \(\frac{19}{3}\). |
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| deepseekr1 | Given that polygon ACDF is similar to polygon VWYZ, we know that the corresponding sides are proportional. The order of the vertices indicates the correspondence between the sides: AC corresponds to VW, CD corresponds to WY, DF corresponds to YZ, and FA corresponds to ZV. We are given the following side lengths: - \( AF = 12 \) - \( CD = 9 \) - \( YZ = 10 \) - \( YW = 6 \) - \( ZV = 3y - 1 \) First, we determine the similarity ratio by comparing the corresponding sides CD and WY: \[ \frac{CD}{WY} = \frac{9}{6} = \frac{3}{2} \] This ratio applies to all corresponding sides. Therefore, the ratio of FA to ZV should also be \(\frac{3}{2}\). Given \( FA = 12 \) and \( ZV = 3y - 1 \), we set up the equation: \[ \frac{12}{3y - 1} = \frac{3}{2} \] Solving for \( y \): \[ 12 = \frac{3}{2}(3y - 1) \] Multiply both sides by 2: \[ 24 = 3(3y - 1) \] Divide both sides by 3: \[ 8 = 3y - 1 \] Add 1 to both sides: \[ 9 = 3y \] Divide by 3: \[ y = 3 \] Thus, the value of \( y \) is \(\boxed{3}\). |
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| gpto1 | To solve this problem, we'll use the property of similar polygons that corresponding sides are proportional. We'll first identify the corresponding sides and then set up a proportion to solve for \( y \). **Step 1: Identify Corresponding Sides** Given that polygons \( ACDF \) and \( VWYZ \) are similar, we need to establish the correspondence between their vertices: - \( A \leftrightarrow V \) - \( C \leftrightarrow W \) - \( D \leftrightarrow Y \) - \( F \leftrightarrow Z \) Therefore, the corresponding sides are: - \( CD \leftrightarrow WY \) - \( AF \leftrightarrow VZ \) (since \( A \leftrightarrow V \) and \( F \leftrightarrow Z \)) **Step 2: Write the Proportions** Using the lengths given: 1. \( CD = 9 \), \( WY = YW = 6 \) 2. \( AF = 12 \), \( VZ = ZV = 3y - 1 \) Set up the proportion between the corresponding sides: \[ \frac{AF}{VZ} = \frac{CD}{WY} \] Substitute the known values: \[ \frac{12}{3y - 1} = \frac{9}{6} \] **Step 3: Simplify the Proportion** Simplify the right side of the equation: \[ \frac{9}{6} = \frac{3}{2} \] So the proportion becomes: \[ \frac{12}{3y - 1} = \frac{3}{2} \] **Step 4: Solve for \( y \)** Cross-multiply to solve for \( y \): \[ 12 \times 2 = (3y - 1) \times 3 \] Simplify: \[ 24 = 9y - 3 \] Add 3 to both sides: \[ 24 + 3 = 9y \] \[ 27 = 9y \] Divide both sides by 9: \[ y = \frac{27}{9} \] \[ y = 3 \] **Answer:** \[ y = 3 \] |
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| GPT-4o | To solve the problem, we need to use the property of similar polygons, which states that corresponding sides of similar polygons are proportional. Given: - Polygon ACDF is similar to polygon VWYZ. - AF = 12, CD = 9, YZ = 10, YW = 6, and ZV = 3y - 1. Since the polygons are similar, the ratios of corresponding sides are equal. We can use the given lengths to set up a proportion. First, let's identify the corresponding sides: - AF corresponds to VW (since the first letters A and V, and the last letters F and W). - CD corresponds to YZ. - Since we have YZ and ZV (which is 3y - 1), we can use these to find the value of y. Given: - CD = 9 - YZ = 10 - ZV = 3y - 1 The ratio of the sides of the similar polygons should be the same, so: \[ \frac{CD}{YZ} = \frac{9}{10} \] Now we can use this ratio to find y. Since ZV is the corresponding side to CD and YZ, the proportion should also hold: \[ \frac{CD}{YZ} = \frac{ZV}{CD} \] Given ZV = 3y - 1 and CD = 9, we can set up the proportion: \[ \frac{9}{10} = \frac{3y - 1}{9} \] Cross-multiply to solve for y: \[ 9 \cdot 9 = 10 \cdot (3y - 1) \] \[ 81 = 30y - 10 \] Add 10 to both sides: \[ 81 + 10 = 30y \] \[ 91 = 30y \] Divide both sides by 30: \[ y = \frac{91}{30} \] So the value of y is: \[ y \approx 3.033 \] Therefore, the value of y is \(\frac{91}{30}\) or approximately 3.033. |
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| llama3.1-70B-Instruct | Since the two polygons are similar, the corresponding sides are proportional. We can set up a proportion using the given side lengths: AF / CD = YW / YZ 12 / 9 = 6 / 10 To solve for y, we can use the fact that the ratio of the corresponding sides is equal to the ratio of the corresponding altitudes. Since ZV is the altitude of triangle VWZ, we can set up the following proportion: ZV / YZ = AF / CD (3y - 1) / 10 = 12 / 9 Cross-multiplying, we get: 9(3y - 1) = 12(10) 27y - 9 = 120 Adding 9 to both sides, we get: 27y = 129 Dividing both sides by 27, we get: y = 129/27 y = 4.78 (approximately) So, the value of y is approximately 4.78. |
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| llama3.1-405B-Instruct | Since the polygons are similar, we can set up a proportion using the corresponding sides: AF / CD = YW / YZ Substituting the given values, we get: 12 / 9 = 6 / 10 Simplifying, we get: 4/3 = 3/5 Cross-multiplying, we get: 4(5) = 3(3) 20 = 9 Now, we can solve for y: ZV = 3y - 1 Since ZV corresponds to CD, we can set up another proportion: CD / AF = ZV / YW Substituting the given values, we get: 9 / 12 = (3y - 1) / 6 Simplifying, we get: 3/4 = (3y - 1) / 6 Cross-multiplying, we get: 3(6) = 4(3y - 1) 18 = 12y - 4 Adding 4 to both sides, we get: 22 = 12y Dividing both sides by 12, we get: y = 22/12 y = 11/6 Therefore, the value of y is 11/6. |
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| DeepSeek-Coder-V2 | To solve the problem, we need to use the property of similar polygons, which states that corresponding sides of similar polygons are proportional. Given: - Polygon ACDF is similar to polygon VWYZ. - Corresponding sides: - AF corresponds to YZ - CD corresponds to YW - AC corresponds to ZV We are given the following lengths: - AF = 12 - CD = 9 - YZ = 10 - YW = 6 - ZV = 3y - 1 Since the polygons are similar, the ratios of the corresponding sides are equal. Therefore, we can set up the following proportion: \[ \frac{AF}{YZ} = \frac{CD}{YW} = \frac{AC}{ZV} \] First, let's find the ratio of the given sides: \[ \frac{AF}{YZ} = \frac{12}{10} = \frac{6}{5} \] \[ \frac{CD}{YW} = \frac{9}{6} = \frac{3}{2} \] Since the ratios must be equal, we need to find the common ratio. Notice that: \[ \frac{6}{5} \neq \frac{3}{2} \] This suggests that there might be an error in the problem statement or the given lengths. However, let's proceed with the assumption that the common ratio is \(\frac{6}{5}\) and use it to find \(y\). Using the ratio \(\frac{6}{5}\) for the sides involving \(y\): \[ \frac{AC}{ZV} = \frac{6}{5} \] Since \(AC\) corresponds to \(ZV\), we can set up the equation: \[ \frac{AC}{3y - 1} = \frac{6}{5} \] To solve for \(y\), we need to know the length of \(AC\). However, since \(AC\) is not directly given, we assume it follows the same ratio as the other sides. Let's solve the equation: \[ \frac{6}{5} = \frac{AC}{3y - 1} \] Cross-multiplying to solve for \(y\): \[ 6(3y - 1) = 5 \cdot AC \] Since we don't have the exact length of \(AC\), we assume it follows the same proportionality. Let's solve for \(y\) using the given lengths: \[ 6(3y - 1) = 5 \cdot 12 \] \[ 6(3y - 1) = 60 \] \[ 18y - 6 = 60 \] \[ 18y = 66 \] \[ y = \frac{66}{18} \] \[ y = \frac{11}{3} \] Thus, the value of \(y\) is: \[ \boxed{\frac{11}{3}} \] |
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| Qwen2-72B-Instruct | Given that polygon ACDF is similar to polygon VWYZ, we can set up a proportion based on the corresponding sides. Similar polygons have corresponding sides that are proportional. Let's set up the proportion based on the given side lengths: \[ \frac{AF}{YZ} = \frac{CD}{YW} = \frac{AC}{VW} = \frac{DF}{ZY} = \frac{AD}{VZ} \] We are given that: - AF = 12 - CD = 9 - YZ = 10 - YW = 6 - ZV = 3y-1 We can use the first two pieces of information to set up a proportion: \[ \frac{AF}{YZ} = \frac{CD}{YW} \] Substituting the given values: \[ \frac{12}{10} = \frac{9}{6} \] Simplifying both sides: \[ \frac{6}{5} = \frac{3}{2} \] This equation is true, confirming that the polygons are similar. However, we need to find y, which is part of the expression for ZV. We can use the proportion involving ZV: \[ \frac{AD}{VZ} = \frac{AF}{YZ} \] We do not have the value for AD, but we can use the proportion involving CD and YW since we have all the necessary values: \[ \frac{CD}{YW} = \frac{ZV}{AF} \] Substituting the given values: \[ \frac{9}{6} = \frac{3y-1}{12} \] Simplifying the left side: \[ \frac{3}{2} = \frac{3y-1}{12} \] Cross multiplying to solve for y: \[ 3 \times 12 = 2 \times (3y-1) \] \[ 36 = 6y - 2 \] \[ 38 = 6y \] \[ y = \frac{38}{6} = \frac{19}{3} \] Thus, the value of y is \(\frac{19}{3}\). |
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| baichuan4 | Since polygons ACDF and VWYZ are similar, their corresponding sides are proportional. That is, the ratio of any two corresponding sides are equal. Let's set up a proportion using the given sides: (AF/CD) = (YZ/YW) = (ZV/DF) We know that AF = 12, CD = 9, YZ = 10, YW = 6, and ZV = 3y - 1. We need to find DF. Since we don't know DF directly, we can use the Pythagorean theorem on triangle ACD to find it. In right triangle ACD, we have: AC^2 + CD^2 = AD^2 We are given AC = 9 and CD = 9, so: 9^2 + 9^2 = AD^2 81 + 81 = AD^2 162 = AD^2 AD = √162 Now, in right triangle ADF, we have: DF^2 + AF^2 = AD^2 DF^2 + 12^2 = (√162)^2 DF^2 + 144 = 162 DF^2 = 162 - 144 DF^2 = 18 DF = √18 Now, let's set up our proportion using the known values: (12/9) = (10/6) = (3y - 1)/√18 Cross multiply the last fraction to find y: (3y - 1) * √18 = 12 * 9 3y√18 - √18 = 108 3y√18 = 108 + √18 y = (108 + √18) / (3√18) Simplify: y = (108 + √18) / (3√18) y = (36√18 + √18) / (3√18) y = (37√18) / (3√18) y = 37 / 3 y = 12.333... So, y is approximately 12.333. |
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| doubao | Since the polygons are similar, the corresponding sides are proportional. The ratio of corresponding sides AF to YZ is 12/10. The ratio of corresponding sides CD to YW is 9/6. So we can set up the proportion: 12/10 = 9/6. Cross multiplying gives: 12*6 = 10*9. 72 = 90 (not true), so there seems to be an error in the problem. If the correct proportion is used, then we can find the value of y using the ratio of corresponding sides AF to ZV. 12/(3y-1) = 12/10. Cross multiplying gives: 12*10 = 12*(3y-1). 120 = 36y - 12. 36y = 132. y = 132/36 = 11/3. |
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| minimax | To solve this problem, we need to use the property that corresponding sides of similar polygons are proportional. This means that the ratio of any two sides in one polygon is equal to the ratio of the corresponding sides in the other polygon. Given that polygon ACDF is similar to polygon VWYZ, we can set up the following proportions: \[ \frac{AF}{VW} = \frac{CD}{YZ} = \frac{DF}{WZ} \] We are given the following lengths: \[ AF = 12, \quad CD = 9, \quad YZ = 10, \quad YW = 6 \] We are also given that \( ZV = 3y - 1 \), and since \( WZ = YZ - ZV \), we can write: \[ WZ = 10 - (3y - 1) = 11 - 3y \] Now, let's use the proportions to find the value of \( y \). We'll start with the proportion involving \( AF \) and \( VW \): \[ \frac{AF}{VW} = \frac{CD}{YZ} \] Substituting the given lengths: \[ \frac{12}{6} = \frac{9}{10} \] This simplifies to: \[ 2 = \frac{9}{10} \] This is not true, which means we've made a mistake in setting up our proportions. We should instead compare corresponding sides that are not adjacent. Let's use \( AF \) and \( YZ \): \[ \frac{AF}{VW} = \frac{CD}{YZ} \] We don't know \( VW \), but we can use \( CD \) and \( YZ \) to find \( VW \): \[ \frac{CD}{YZ} = \frac{9}{10} \] Since \( VW \) and \( YW \) are adjacent sides in the similar polygons, we can use them to find \( VW \): \[ \frac{VW}{YW} = \frac{CD}{YZ} \] Substituting the given lengths: \[ \frac{VW}{6} = \frac{9}{10} \] Solving for \( VW \): \[ VW = \frac{9}{10} \times 6 = \frac{54}{10} = 5.4 \] Now we have \( VW \), and we can use the proportion involving \( AF \) and \( VW \): \[ \frac{AF}{VW} = \frac{CD}{YZ} \] Substituting the known lengths: \[ \frac{12}{5.4} = \frac{9}{10} \] Solving for \( VW \) again: \[ VW = \frac{12 \times 10}{9} = \frac{120}{9} = \frac{40}{3} = 13.\overline{3} \] This is a contradiction, as we previously found \( VW \) to be 5.4. This suggests that there is a mistake in our approach. Let's re-evaluate the problem. We should compare the sides that are not adjacent in the similar polygons. Let's use \( AF \) and \( YZ \) and \( CD \) and \( YW \): \[ \frac{AF}{YZ} = \frac{CD}{YW} \] Substituting the given lengths: \[ \frac{12}{10} = \frac{9}{6} \] Solving for \( YW \): \[ YW = \frac{9 \times 10}{12} = \frac{90}{12} = 7.5 \] This is incorrect because we already know \( YW = 6 \). We need to use the correct sides for the proportion. Let's use \( AF \) and \( YZ \) and \( CD \) and \( YW \) correctly: \[ \frac{AF}{YZ} = \frac{CD}{YW} \] Substituting the given lengths: \[ \frac{12}{10} = \frac{9}{6} \] Solving for \( YW \): \[ YW = \frac{9 \times 10}{12} = \frac{90}{12} = 7.5 \] This is incorrect because we already know \( YW = 6 \). We need to use the correct sides for the proportion. Let's use \( AF \) and \( YZ \) and \( CD \) and \( YW \) correctly: \[ \frac{AF}{YZ} = \frac{CD}{YW} \] Substituting the given lengths: \[ \frac{12}{10} = \frac{9}{6} \] Solving for \( YW \): \[ YW = \frac{9 \times 10}{12} = \frac{90}{12} = 7.5 \] This is incorrect because we already know \( YW = 6 \). We need to use the correct sides for the proportion. Let's use \( AF \) and \( YZ \) and \( CD \) and \( WZ \) (since \( WZ = YZ - ZV \)): \[ \frac{AF}{YZ} = \frac{CD}{WZ} \] Substituting the given lengths: \[ \frac{12}{10} = \frac{9}{11 - 3y} \] Cross-multiplying to solve for \( y \): \[ 12 \times (11 - 3y) = 10 \times 9 \] \[ 132 - 36y = 90 \] \[ 132 - 90 = 36y \] \[ 42 = 36y \] \[ y = \frac{42}{36} = \frac{7}{6} \] So the value of \( y \) is \( \frac{7}{6} \) or approximately 1.1667. |
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| kimi | Since polygons ACDF and VWYZ are similar, their corresponding sides are proportional. This means that the ratio of the lengths of corresponding sides is the same for both polygons. We can set up the following proportion based on the given side lengths: \[ \frac{AF}{YZ} = \frac{CD}{YW} = \frac{AD}{VW} \] Given: - \( AF = 12 \) - \( CD = 9 \) - \( YZ = 10 \) - \( YW = 6 \) - \( ZV = 3y - 1 \) First, let's find the ratio of \( AF \) to \( YZ \): \[ \frac{AF}{YZ} = \frac{12}{10} = \frac{6}{5} \] This ratio should be the same for the other pairs of corresponding sides. So, we can write: \[ \frac{CD}{YW} = \frac{6}{5} \] Now, we can solve for \( VW \) using the given length of \( YW \): \[ \frac{9}{YW} = \frac{6}{5} \Rightarrow YW = \frac{5 \times 9}{6} = \frac{45}{6} = 7.5 \] However, we are given that \( YW = 6 \), which contradicts our calculation. This means that the sides \( CD \) and \( YW \) are not corresponding sides. Instead, \( CD \) must correspond to \( ZV \). Let's correct our approach. Since \( CD \) corresponds to \( ZV \), we have: \[ \frac{CD}{ZV} = \frac{6}{5} \] Now we can solve for \( ZV \): \[ \frac{9}{ZV} = \frac{6}{5} \Rightarrow ZV = \frac{5 \times 9}{6} = \frac{45}{6} = 7.5 \] But \( ZV \) is given as \( 3y - 1 \), so we can set up the equation: \[ 3y - 1 = 7.5 \] Solving for \( y \): \[ 3y = 7.5 + 1 \Rightarrow 3y = 8.5 \Rightarrow y = \frac{8.5}{3} \Rightarrow y = \frac{17}{6} \Rightarrow y = 2.8333... \] So, \( y \) is approximately \( 2.8333 \). |
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| wenxinyiyan | To solve this problem, we can use the properties of similar polygons. Since polygons ACDF and VWYZ are similar, the corresponding sides are in proportion. We can set up a proportion equation using the given side lengths. Let's compare the sides AF and YW, as well as CD and YZ, to set up our proportion: $$\frac{AF}{YW} = \frac{CD}{YZ}$$Substitute the given values: $$\frac{12}{6} = \frac{9}{10}$$However, this proportion does not directly involve the unknown variable y. To find y, we need to use another set of corresponding sides that include the unknown side ZV. Let's compare the sides AF and VY (which is equal to YW + ZV): $$\frac{AF}{VY} = \frac{AF}{YW + ZV} = \frac{CD}{YZ}$$Again, substitute the given values, and express ZV in terms of y: $$\frac{12}{6 + (3y - 1)} = \frac{9}{10}$$Now we can solve for y: $$12 \times 10 = 9 \times (6 + 3y - 1)$$Expand and simplify: $$120 = 9 \times (5 + 3y)$$$$120 = 45 + 27y$$$$27y = 75$$$$y = \frac{75}{27} = \frac{25}{9}$$Therefore, the value of y is $\frac{25}{9}$. |
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| zhipu | Since the polygons ACDF and VWYZ are similar, their corresponding sides are proportional. This means that the ratios of the lengths of corresponding sides are equal. Let's match the corresponding sides: - AF corresponds to YZ - CD corresponds to YW - AC (or DF) would correspond to VW (or ZY) We are given the lengths of AF, CD, YZ, and YW, and we need to find the value of y from the length of ZV. First, let's set up the proportionality equations based on the given side lengths: For AF and YZ: \( \frac{AF}{YZ} = \frac{12}{10} = \frac{6}{5} \) For CD and YW: \( \frac{CD}{YW} = \frac{9}{6} = \frac{3}{2} \) Since the polygons are similar, the ratio of any two corresponding sides should be the same. Therefore, the ratio of CD to YW should also be equal to the ratio of AF to YZ. However, in this case, they are not directly equal (6/5 is not equal to 3/2), which suggests that we may have mismatched the corresponding sides or there may be an error in the given information. But let's proceed with the information given and assume that the ratios should be consistent for the sides we are considering. We are tasked with finding the value of y from ZV, which corresponds to either AC or DF (it doesn't matter which one we choose since the polygons are similar and the ratios will be the same). So, we can set up the proportionality equation for ZV and CD (or AF, but CD is used here for consistency): \( \frac{CD}{YZ} = \frac{ZV}{VW} \) We know CD = 9, YZ = 10, and ZV = 3y - 1. We need to find y. \( \frac{9}{10} = \frac{3y - 1}{VW} \) Since VW is not given, we cannot use this direct approach. Instead, we should use the consistent ratio we have from AF and YZ: \( \frac{AF}{YZ} = \frac{12}{10} = \frac{6}{5} \) Now we apply this ratio to find ZV: \( \frac{AF}{YZ} = \frac{CD}{ZV} \) \( \frac{6}{5} = \frac{9}{3y - 1} \) Cross-multiply to solve for y: \( 6(3y - 1) = 5 \cdot 9 \) \( 18y - 6 = 45 \) Add 6 to both sides: \( 18y = 51 \) Divide both sides by 18: \( y = \frac{51}{18} \) Simplify the fraction: \( y = \frac{17}{6} \) So, the value of y is \( \frac{17}{6} \) or approximately 2.8333 when expressed as a decimal. |
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| gpto1 | To solve this problem, we'll use the property of similar polygons that corresponding sides are proportional. We'll first identify the corresponding sides and then set up a proportion to solve for \( y \). **Step 1: Identify Corresponding Sides** Given that polygons \( ACDF \) and \( VWYZ \) are similar, we need to establish the correspondence between their vertices: - \( A \leftrightarrow V \) - \( C \leftrightarrow W \) - \( D \leftrightarrow Y \) - \( F \leftrightarrow Z \) Therefore, the corresponding sides are: - \( CD \leftrightarrow WY \) - \( AF \leftrightarrow VZ \) (since \( A \leftrightarrow V \) and \( F \leftrightarrow Z \)) **Step 2: Write the Proportions** Using the lengths given: 1. \( CD = 9 \), \( WY = YW = 6 \) 2. \( AF = 12 \), \( VZ = ZV = 3y - 1 \) Set up the proportion between the corresponding sides: \[ \frac{AF}{VZ} = \frac{CD}{WY} \] Substitute the known values: \[ \frac{12}{3y - 1} = \frac{9}{6} \] **Step 3: Simplify the Proportion** Simplify the right side of the equation: \[ \frac{9}{6} = \frac{3}{2} \] So the proportion becomes: \[ \frac{12}{3y - 1} = \frac{3}{2} \] **Step 4: Solve for \( y \)** Cross-multiply to solve for \( y \): \[ 12 \times 2 = (3y - 1) \times 3 \] Simplify: \[ 24 = 9y - 3 \] Add 3 to both sides: \[ 24 + 3 = 9y \] \[ 27 = 9y \] Divide both sides by 9: \[ y = \frac{27}{9} \] \[ y = 3 \] **Answer:** \[ y = 3 \] |
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| deepseekv3 | To solve for \( y \), we'll use the fact that polygons **ACDF** and **VWYZ** are similar. This means their corresponding sides are proportional. ### Given: - \( AF = 12 \) - \( CD = 9 \) - \( YZ = 10 \) - \( YW = 6 \) - \( ZV = 3y - 1 \) ### Step 1: Identify the corresponding sides From the similarity of the polygons: - \( AF \) corresponds to \( VW \) - \( CD \) corresponds to \( YZ \) - \( DF \) corresponds to \( ZV \) ### Step 2: Set up the proportion The ratio of the corresponding sides is constant. Using sides \( CD \) and \( YZ \): \[ \frac{CD}{YZ} = \frac{AF}{VW} \] Substitute the known values: \[ \frac{9}{10} = \frac{12}{VW} \] Solve for \( VW \): \[ VW = \frac{12 \cdot 10}{9} = \frac{120}{9} = \frac{40}{3} \] ### Step 3: Use the proportion to find \( ZV \) Now, using the proportion: \[ \frac{DF}{ZV} = \frac{AF}{VW} \] Substitute the known values: \[ \frac{DF}{3y - 1} = \frac{12}{\frac{40}{3}} \] Simplify the right-hand side: \[ \frac{12}{\frac{40}{3}} = \frac{12 \cdot 3}{40} = \frac{36}{40} = \frac{9}{10} \] So: \[ \frac{DF}{3y - 1} = \frac{9}{10} \] To solve for \( DF \), note that \( DF \) corresponds to \( ZV \) in the similarity, so: \[ DF = ZV = 3y - 1 \] Thus: \[ \frac{3y - 1}{3y - 1} = \frac{9}{10} \] This simplifies to: \[ 1 = \frac{9}{10} \] This is a contradiction, so we need to reconsider the approach. ### Step 4: Re-evaluate the proportion Instead, use the proportion: \[ \frac{AF}{VW} = \frac{CD}{YZ} \] Substitute the known values: \[ \frac{12}{VW} = \frac{9}{10} \] Solve for \( VW \): \[ VW = \frac{12 \cdot 10}{9} = \frac{120}{9} = \frac{40}{3} \] ### Step 5: Find \( ZV \) Since \( DF \) corresponds to \( ZV \), and \( DF = AF - CD = 12 - 9 = 3 \), we have: \[ DF = ZV = 3 \] Substitute \( ZV = 3y - 1 \): \[ 3y - 1 = 3 \] Solve for \( y \): \[ 3y = 4 \] \[ y = \frac{4}{3} \] ### Final Answer: \[ y = \frac{4}{3} \] |
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