Question
Diana draws a square with a side length of x. She dilates the square with a scale factor of 3.
What is the area of the original square in terms of x?
What is the area of the dilated square?
How many times greater is the area of the dilated square compared to the area of the original square?
Answers
GPT-4o mini
To solve this problem, we will first determine the area of the original square, then the area of the dilated square, and finally, we will compare the two areas.
1. **Area of the original square**:
The area \( A \) of a square is given by the formula:
\[
A = \text{side length}^2
\]
For the original square with side length \( x \):
\[
\text{Area of the original square} = x^2
\]
2. **Area of the dilated square**:
When the square is dilated with a scale factor of 3, the new side length becomes:
\[
3x
\]
The area of the dilated square is then:
\[
\text{Area of the dilated square} = (3x)^2 = 9x^2
\]
3. **Comparison of the areas**:
To find how many times greater the area of the dilated square is compared to the area of the original square, we take the ratio of the area of the dilated square to the area of the original square:
\[
\text{Ratio} = \frac{\text{Area of the dilated square}}{\text{Area of the original square}} = \frac{9x^2}{x^2} = 9
\]
Thus, the area of the dilated square is 9 times greater than the area of the original square.
**Summary**:
- Area of the original square: \( x^2 \)
- Area of the dilated square: \( 9x^2 \)
- The area of the dilated square is 9 times greater than the area of the original square.
1. **Area of the original square**:
The area \( A \) of a square is given by the formula:
\[
A = \text{side length}^2
\]
For the original square with side length \( x \):
\[
\text{Area of the original square} = x^2
\]
2. **Area of the dilated square**:
When the square is dilated with a scale factor of 3, the new side length becomes:
\[
3x
\]
The area of the dilated square is then:
\[
\text{Area of the dilated square} = (3x)^2 = 9x^2
\]
3. **Comparison of the areas**:
To find how many times greater the area of the dilated square is compared to the area of the original square, we take the ratio of the area of the dilated square to the area of the original square:
\[
\text{Ratio} = \frac{\text{Area of the dilated square}}{\text{Area of the original square}} = \frac{9x^2}{x^2} = 9
\]
Thus, the area of the dilated square is 9 times greater than the area of the original square.
**Summary**:
- Area of the original square: \( x^2 \)
- Area of the dilated square: \( 9x^2 \)
- The area of the dilated square is 9 times greater than the area of the original square.