Question
To find the area of the original rectangle, we can use the formula for the area of a rectangle:
\[
\text{Area} = \text{length} \times \text{width}
\]
The sides of the original rectangle are 6 inches and 10 inches, so we can calculate its area:
\[
\text{Area}_{\text{original}} = 6 \, \text{inches} \times 10 \, \text{inches} = 60 \, \text{square inches}
\]
Now, in order to determine the area of the dilated rectangle, we need to determine the dimensions of the dilated rectangle with scale factor \( \frac{1}{4} \).
We apply the scale factor to both dimensions:
\[
\text{Length}_{\text{dilated}} = 10 \, \text{inches} \times \frac{1}{4} = 2.5 \, \text{inches}
\]
\[
\text{Width}_{\text{dilated}} = 6 \, \text{inches} \times \frac{1}{4} = 1.5 \, \text{inches}
\]
We now determine the area of the dilated rectangle:
[]
\text{Area}_{\text{dilated}} = \text{Length}_{\text{dilated}} \times \text{Width}_{\text{dilated}} = 2.5 \, \text{inches} \times 1.5 \, \text{inches} = 3.75 \, \text{square inches}
We now find out just how much bigger the area of the original rectangle is compared to the area of the dilated rectangle. We do so by taking the area of the original rectangle and dividing it by the area of the dilated rectangle:
\\text{Ratio} = \frac{\text{Area}_{\text{original}}}{\text{Area}_{\text{dilated}}} = \frac{60 \\, \text{square inches}}{3.75 \, \text{square inches}} = 16
Thus, the original rectangle has an area **16 times greater** than that of the dilated rectangle.
### Summary of Results:
- Original rectangle area: \( 60 \, \text{square inches } \)
- Dilated rectangle area: \( 3.75 \, \text{square inches } \)
- The area of the original rectangle is \( 16 \) times greater than the area of the dilated rectangle.
\[
\text{Area} = \text{length} \times \text{width}
\]
The sides of the original rectangle are 6 inches and 10 inches, so we can calculate its area:
\[
\text{Area}_{\text{original}} = 6 \, \text{inches} \times 10 \, \text{inches} = 60 \, \text{square inches}
\]
Now, in order to determine the area of the dilated rectangle, we need to determine the dimensions of the dilated rectangle with scale factor \( \frac{1}{4} \).
We apply the scale factor to both dimensions:
\[
\text{Length}_{\text{dilated}} = 10 \, \text{inches} \times \frac{1}{4} = 2.5 \, \text{inches}
\]
\[
\text{Width}_{\text{dilated}} = 6 \, \text{inches} \times \frac{1}{4} = 1.5 \, \text{inches}
\]
We now determine the area of the dilated rectangle:
[]
\text{Area}_{\text{dilated}} = \text{Length}_{\text{dilated}} \times \text{Width}_{\text{dilated}} = 2.5 \, \text{inches} \times 1.5 \, \text{inches} = 3.75 \, \text{square inches}
We now find out just how much bigger the area of the original rectangle is compared to the area of the dilated rectangle. We do so by taking the area of the original rectangle and dividing it by the area of the dilated rectangle:
\\text{Ratio} = \frac{\text{Area}_{\text{original}}}{\text{Area}_{\text{dilated}}} = \frac{60 \\, \text{square inches}}{3.75 \, \text{square inches}} = 16
Thus, the original rectangle has an area **16 times greater** than that of the dilated rectangle.
### Summary of Results:
- Original rectangle area: \( 60 \, \text{square inches } \)
- Dilated rectangle area: \( 3.75 \, \text{square inches } \)
- The area of the original rectangle is \( 16 \) times greater than the area of the dilated rectangle.
Answers
GPT-4o mini
Your calculations and reasoning about the area of the original rectangle and the dilated rectangle appear to be accurate. Here's a concise summary of the important steps and results:
1. **Calculating Area of Original Rectangle**:
\[
\text{Area}_{\text{original}} = \text{length} \times \text{width} = 6 \, \text{inches} \times 10 \, \text{inches} = 60 \, \text{square inches}
\]
2. **Calculating Dimensions of Dilated Rectangle** with a scale factor of \( \frac{1}{4} \):
\[
\text{Length}_{\text{dilated}} = 10 \, \text{inches} \times \frac{1}{4} = 2.5 \, \text{inches}
\]
\[
\text{Width}_{\text{dilated}} = 6 \, \text{inches} \times \frac{1}{4} = 1.5 \, \text{inches}
\]
3. **Calculating Area of Dilated Rectangle**:
\[
\text{Area}_{\text{dilated}} = \text{Length}_{\text{dilated}} \times \text{Width}_{\text{dilated}} = 2.5 \, \text{inches} \times 1.5 \, \text{inches} = 3.75 \, \text{square inches}
\]
4. **Finding the Ratio of Areas**:
\[
\text{Ratio} = \frac{\text{Area}_{\text{original}}}{\text{Area}_{\text{dilated}}} = \frac{60 \, \text{square inches}}{3.75 \, \text{square inches}} = 16
\]
### Summary of Results:
- Area of the original rectangle: \( 60 \, \text{square inches} \)
- Area of the dilated rectangle: \( 3.75 \, \text{square inches} \)
- The area of the original rectangle is **16 times greater** than that of the dilated rectangle.
Everything is correct, and your final conclusion clearly states the relationship between the areas of the two rectangles. Well done!
1. **Calculating Area of Original Rectangle**:
\[
\text{Area}_{\text{original}} = \text{length} \times \text{width} = 6 \, \text{inches} \times 10 \, \text{inches} = 60 \, \text{square inches}
\]
2. **Calculating Dimensions of Dilated Rectangle** with a scale factor of \( \frac{1}{4} \):
\[
\text{Length}_{\text{dilated}} = 10 \, \text{inches} \times \frac{1}{4} = 2.5 \, \text{inches}
\]
\[
\text{Width}_{\text{dilated}} = 6 \, \text{inches} \times \frac{1}{4} = 1.5 \, \text{inches}
\]
3. **Calculating Area of Dilated Rectangle**:
\[
\text{Area}_{\text{dilated}} = \text{Length}_{\text{dilated}} \times \text{Width}_{\text{dilated}} = 2.5 \, \text{inches} \times 1.5 \, \text{inches} = 3.75 \, \text{square inches}
\]
4. **Finding the Ratio of Areas**:
\[
\text{Ratio} = \frac{\text{Area}_{\text{original}}}{\text{Area}_{\text{dilated}}} = \frac{60 \, \text{square inches}}{3.75 \, \text{square inches}} = 16
\]
### Summary of Results:
- Area of the original rectangle: \( 60 \, \text{square inches} \)
- Area of the dilated rectangle: \( 3.75 \, \text{square inches} \)
- The area of the original rectangle is **16 times greater** than that of the dilated rectangle.
Everything is correct, and your final conclusion clearly states the relationship between the areas of the two rectangles. Well done!