| 模型 | 模型答案 | 正确答案 |
|---|---|---|
| 金声玉亮2.0 | C. 50 |
C |
| deepseekr1 | The problem involves a rectangle with length twice its width and a diagonal of \(5\sqrt{5}\). Let the width be \(w\), so the length is \(2w\). Using the Pythagorean theorem for the diagonal: \[ (5\sqrt{5})^2 = (2w)^2 + w^2 \] Simplifying: \[ 125 = 4w^2 + w^2 \implies 125 = 5w^2 \implies w^2 = 25 \implies w = 5 \] The length is \(2w = 10\). The area is: \[ \text{Area} = \text{length} \times \text{width} = 10 \times 5 = 50 \] Verification confirms the diagonal calculation is correct. The answer is **C**. **Answer: C** |
C |
| gpto1 | C | C |
| GPT-4o | To find the area of the rectangle, we start by defining the width of the rectangle as \( w \) and the length as \( 2w \). Given that the length of the diagonal is \( 5\sqrt{5} \), we can use the Pythagorean theorem to relate the width, length, and diagonal: \[ \text{Diagonal}^2 = \text{Length}^2 + \text{Width}^2 \] Substituting the given values: \[ (5\sqrt{5})^2 = (2w)^2 + w^2 \] \[ 25 \cdot 5 = 4w^2 + w^2 \] \[ 125 = 5w^2 \] \[ w^2 = 25 \] \[ w = 5 \] Now, the length \( 2w \) is: \[ 2w = 2 \times 5 = 10 \] The area of the rectangle is: \[ \text{Area} = \text{Length} \times \text{Width} = 10 \times 5 = 50 \] Therefore, the correct answer is: C. 50 |
C |
| llama3.1-70B-Instruct | D |
C |
| llama3.1-405B-Instruct | We can use the Pythagorean Theorem to find the length of the rectangle. Let $x$ be the width of the rectangle. Then, the length is $2x$. We have: $$x^2 + (2x)^2 = (5\sqrt{5})^2$$ Simplifying, we get: $$x^2 + 4x^2 = 125$$ Combine like terms: $$5x^2 = 125$$ Divide by 5: $$x^2 = 25$$ Take the square root: $$x = 5$$ So, the width of the rectangle is 5. Since the length is twice the width, the length is 10. The area of the rectangle is: $$\text{Area} = \text{length} \times \text{width} = 10 \times 5 = 50$$ Therefore, the area of the rectangle is 50. The correct answer is C. |
C |
| DeepSeek-Coder-V2 | D. 25 | C |
| Qwen2-72B-Instruct | Let's denote the width of the rectangle as \(w\) and the length as \(l\). According to the problem, \(l = 2w\). The length of the diagonal, by the Pythagorean theorem, is \(\sqrt{l^2 + w^2}\). Given that the diagonal is \(5\sqrt{5}\), we can set up the equation: \[ \sqrt{(2w)^2 + w^2} = 5\sqrt{5} \] Simplifying inside the square root: \[ \sqrt{4w^2 + w^2} = 5\sqrt{5} \] \[ \sqrt{5w^2} = 5\sqrt{5} \] Squaring both sides: \[ 5w^2 = 25 \cdot 5 \] \[ w^2 = 25 \] \[ w = 5 \] Thus, the width \(w = 5\) and the length \(l = 2w = 10\). The area of the rectangle is \(l \times w = 10 \times 5 = 50\). Answer: C. 50 |
C |
| baichuan4 | C | C |
| doubao | Let the width of the rectangle be $x$, then the length is $2x$. Using the Pythagorean theorem, $(2x)^2+x^2=(5\sqrt{5})^2$, $4x^2+x^2=125$, $5x^2=125$, $x^2=25$, $x=5$ or $x=-5$ (reject the negative value). So the width is 5 and the length is 2*5=10. The area is $5\times10=50$. So the answer is C. |
C |
| minimax | Let's denote the width of the rectangle as \( w \) and the length as \( l \). According to the problem, the length is twice the width, so we can write \( l = 2w \). We are also given that the diagonal \( d \) is \( 5\sqrt{5} \). In a rectangle, the diagonal, width, and length form a right-angled triangle, so we can use the Pythagorean theorem to relate these quantities: \[ d^2 = l^2 + w^2 \] Substituting \( l = 2w \) into the equation, we get: \[ (5\sqrt{5})^2 = (2w)^2 + w^2 \] \[ 25 \cdot 5 = 4w^2 + w^2 \] \[ 125 = 5w^2 \] \[ w^2 = \frac{125}{5} \] \[ w^2 = 25 \] \[ w = 5 \] Since \( l = 2w \), we have: \[ l = 2 \cdot 5 = 10 \] Now, the area \( A \) of the rectangle is given by the product of its length and width: \[ A = l \cdot w \] \[ A = 10 \cdot 5 \] \[ A = 50 \] So the area of the rectangle is 50 square units. The correct answer is: C. 50 |
C |
| kimi | Let's denote the width of the rectangle as \( w \) and the length as \( l \). According to the problem, the length is twice the width, so \( l = 2w \). The diagonal \( d \) of a rectangle can be found using the Pythagorean theorem: \[ d = \sqrt{l^2 + w^2} \] Given that \( d = 5\sqrt{5} \), we can substitute \( l \) with \( 2w \) to get: \[ 5\sqrt{5} = \sqrt{(2w)^2 + w^2} \] \[ 5\sqrt{5} = \sqrt{4w^2 + w^2} \] \[ 5\sqrt{5} = \sqrt{5w^2} \] \[ (5\sqrt{5})^2 = 5w^2 \] \[ 25 \cdot 5 = 5w^2 \] \[ 125 = 5w^2 \] \[ w^2 = 25 \] \[ w = 5 \] Now that we have the width, we can find the length: \[ l = 2w = 2 \cdot 5 = 10 \] The area \( A \) of the rectangle is given by the product of its length and width: \[ A = l \cdot w \] \[ A = 10 \cdot 5 \] \[ A = 50 \] So the correct answer is C. 50. |
C |
| wenxinyiyan | C Explanation: Let the width of the rectangle be $x$ units. Then, the length is $2x$ units. By the Pythagorean theorem, the diagonal $d$ satisfies: $d^2 = \text{length}^2 + \text{width}^2$ Given $d = 5\sqrt{5}$, we have: $(5\sqrt{5})^2 = (2x)^2 + x^2$ $25 \times 5 = 4x^2 + x^2$ $125 = 5x^2$ $x^2 = 25$ $x = 5$ (since width cannot be negative, we take the positive root) The length is then $2 \times 5 = 10$ units. The area of the rectangle is $\text{length} \times \text{width} = 10 \times 5 = 50$ square units. |
C |
| zhipu | To solve this problem, let's denote the width of the rectangle as \( w \) and the length as \( l \). According to the problem, the length is twice the width, so we have: \[ l = 2w \] We are also given the length of the diagonal, which is \( 5\sqrt{5} \). We can use the Pythagorean theorem to relate the length, width, and diagonal of the rectangle. The Pythagorean theorem states: \[ l^2 + w^2 = d^2 \] where \( d \) is the diagonal. Substituting \( l = 2w \) and \( d = 5\sqrt{5} \) into the equation, we get: \[ (2w)^2 + w^2 = (5\sqrt{5})^2 \] Simplifying each term, we have: \[ 4w^2 + w^2 = 25 \times 5 \] \[ 5w^2 = 125 \] Dividing both sides by 5, we find: \[ w^2 = 25 \] \[ w = 5 \] Now, substituting \( w = 5 \) back into \( l = 2w \), we get: \[ l = 2 \times 5 = 10 \] The area of the rectangle is given by the product of its length and width: \[ \text{Area} = l \times w = 10 \times 5 = 50 \] Therefore, the area of the rectangle is 50. The correct answer is: C. 50 |
C |
| gpto1 | C | C |
| deepseekv3 | Let’s solve the problem step by step. 1. Let the width of the rectangle be \( w \). Then, the length of the rectangle is \( 2w \). 2. The diagonal of the rectangle forms a right triangle with the length and width. Using the Pythagorean theorem: \[ \text{Diagonal}^2 = \text{Length}^2 + \text{Width}^2 \] Substituting the given diagonal length \( 5\sqrt{5} \): \[ (5\sqrt{5})^2 = (2w)^2 + w^2 \] \[ 125 = 4w^2 + w^2 \] \[ 125 = 5w^2 \] \[ w^2 = 25 \] \[ w = 5 \] 3. The length of the rectangle is: \[ 2w = 2 \times 5 = 10 \] 4. The area of the rectangle is: \[ \text{Area} = \text{Length} \times \text{Width} = 10 \times 5 = 50 \] So, the correct answer is: \boxed{C} |
C |