Question

A composite of two rectangles. The top rectangle is a smaller rectangle and is centered on the top width of the bottom larger rectangle. the measurements ar 2m, 3m, 4m, and 8m.
The area of the composite figure is 52 square meters. What is the value in meters of the height of the larger rectangle?

Answer 1

To solve for the height of the larger rectangle in the composite figure, let's denote the dimensions of the rectangles clearly.

Let:
- The height of the smaller rectangle (top rectangle) = \( h_1 = 2 \) m (based on the best assumption given the measurements).
- The width of the smaller rectangle (top rectangle) = \( w_1 = 3 \) m.
- The height of the larger rectangle (bottom rectangle) = \( h_2 \).
- The width of the larger rectangle (bottom rectangle) = \( w_2 = 4 \) m.

Now, the area of the composite figure is given as 52 square meters.

The area of the smaller rectangle is:
\[
\text{Area}_{\text{small}} = h_1 \times w_1 = 2 \times 3 = 6 \text{ square meters}
\]

Let’s denote the area of the larger rectangle as:
\[
\text{Area}_{\text{large}} = h_2 \times w_2 = h_2 \times 4
\]

Now, the total area of the composite figure is:
\[
\text{Area}_{\text{composite}} = \text{Area}_{\text{small}} + \text{Area}_{\text{large}} = 6 + (h_2 \times 4)
\]

Given that the total area is 52 square meters, we can write:
\[
6 + (h_2 \times 4) = 52
\]

Now, we can solve for \( h_2 \):
\[
h_2 \times 4 = 52 - 6
\]
\[
h_2 \times 4 = 46
\]
\[
h_2 = \frac{46}{4} = 11.5 \text{ meters}
\]

Thus, the height of the larger rectangle is **11.5 meters**.