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We are given that Reynaldo's original rectangular design is 6 inches wide and 8 inches tall. After enlarging, the new width is 10 inches, and we need to find the height of the enlarged design, maintaining the same proportions.

We can set up a proportion based on the fact that the width and height are proportional before and after enlargement. The original ratio of width to height is:

\[
\frac{6}{8} = \frac{10}{\text{new height}}
\]

Now, we solve for the new height:

\[
\frac{6}{8} = \frac{10}{\text{new height}}
\]

First, we simplify \(\frac{6}{8}\) to \(\frac{3}{4}\):

\[
\frac{3}{4} = \frac{10}{\text{new height}}
\]

Cross-multiply to find the new height:

\[
3 \times \text{new height} = 4 \times 10
\]

\[
3 \times \text{new height} = 40
\]

Now, divide both sides by 3:

\[
\text{new height} = \frac{40}{3} \approx 13.33
\]

Rounding to the nearest tenth, the new height is approximately **13.3 inches**.

Would you like any more details or have any questions?

Here are 5 related questions to expand on this concept:
1. How would you solve this if the width was doubled instead of being changed to 10 inches?
2. What is the scale factor in this problem?
3. How can we use the concept of proportions in other geometric shapes?
4. How would the problem change if the width was reduced instead of increased?
5. How can this concept be applied in real-life design or architecture?

**Tip**: When working with proportions, simplifying the ratio first can make solving easier!

Reynaldo drew a rectangular design that was 6 inches wide and 8 inches tall. He used a copy machine to enlarge the rectangular design so that the width was 10 inches. What was the height of the enlarged design? Round your answer to the nearest tenth (one decimal place).

Learn how to solve a proportion problem involving the enlargement of a rectangular design. Reynaldo enlarges a 6x8 inch rectangle to a width of 10 inches, and we calculate the new height using ratios and proportions. Suitable for Grades 6-8.

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