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DeeDee draws a rectangle with sides measuring x inches and y inches. She dilates the rectangle using a scale factor of 14 . Wha...Question
DeeDee draws a rectangle with sides measuring x inches and y inches. She dilates the rectangle using a scale factor of 1/4 . What is the area of the original rectangle in terms of x and y? What is the area of the dilated rectangle? How many times greater is the area of the original rectangle compared to the area of the dilated rectangle? Explain your answers and show all your work.(4 points)
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Answered by
GPT-4o mini
To find the area of the original rectangle in terms of \( x \) and \( y \), we use the formula for the area of a rectangle, which is:
\[
\text{Area} = \text{length} \times \text{width}
\]
For DeeDee's original rectangle, the dimensions are \( x \) inches and \( y \) inches. Thus, the area \( A_{\text{original}} \) is:
\[
A_{\text{original}} = x \times y
\]
Now, DeeDee dilates the rectangle using a scale factor of \( \frac{1}{4} \). When we dilate a shape, each dimension is multiplied by the scale factor. Therefore, the dimensions of the dilated rectangle are:
\[
\text{New length} = x \times \frac{1}{4} = \frac{x}{4}
\]
\[
\text{New width} = y \times \frac{1}{4} = \frac{y}{4}
\]
Next, we can find the area of the dilated rectangle \( A_{\text{dilated}} \):
\[
A_{\text{dilated}} = \left(\frac{x}{4}\right) \times \left(\frac{y}{4}\right)
\]
Calculating this gives:
\[
A_{\text{dilated}} = \frac{x}{4} \times \frac{y}{4} = \frac{xy}{16}
\]
Now we will compare the area of the original rectangle to the area of the dilated rectangle. We can do this by calculating the ratio of the areas:
\[
\text{Ratio of areas} = \frac{A_{\text{original}}}{A_{\text{dilated}}} = \frac{xy}{\frac{xy}{16}}
\]
This simplifies to:
\[
\text{Ratio of areas} = \frac{xy \cdot 16}{xy} = 16
\]
Thus, the area of the original rectangle is **16 times greater** than the area of the dilated rectangle.
**Summary:**
1. Area of the original rectangle: \( A_{\text{original}} = xy \)
2. Area of the dilated rectangle: \( A_{\text{dilated}} = \frac{xy}{16} \)
3. The area of the original rectangle is **16 times greater** than the area of the dilated rectangle.
\[
\text{Area} = \text{length} \times \text{width}
\]
For DeeDee's original rectangle, the dimensions are \( x \) inches and \( y \) inches. Thus, the area \( A_{\text{original}} \) is:
\[
A_{\text{original}} = x \times y
\]
Now, DeeDee dilates the rectangle using a scale factor of \( \frac{1}{4} \). When we dilate a shape, each dimension is multiplied by the scale factor. Therefore, the dimensions of the dilated rectangle are:
\[
\text{New length} = x \times \frac{1}{4} = \frac{x}{4}
\]
\[
\text{New width} = y \times \frac{1}{4} = \frac{y}{4}
\]
Next, we can find the area of the dilated rectangle \( A_{\text{dilated}} \):
\[
A_{\text{dilated}} = \left(\frac{x}{4}\right) \times \left(\frac{y}{4}\right)
\]
Calculating this gives:
\[
A_{\text{dilated}} = \frac{x}{4} \times \frac{y}{4} = \frac{xy}{16}
\]
Now we will compare the area of the original rectangle to the area of the dilated rectangle. We can do this by calculating the ratio of the areas:
\[
\text{Ratio of areas} = \frac{A_{\text{original}}}{A_{\text{dilated}}} = \frac{xy}{\frac{xy}{16}}
\]
This simplifies to:
\[
\text{Ratio of areas} = \frac{xy \cdot 16}{xy} = 16
\]
Thus, the area of the original rectangle is **16 times greater** than the area of the dilated rectangle.
**Summary:**
1. Area of the original rectangle: \( A_{\text{original}} = xy \)
2. Area of the dilated rectangle: \( A_{\text{dilated}} = \frac{xy}{16} \)
3. The area of the original rectangle is **16 times greater** than the area of the dilated rectangle.
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