https://file.aiquickdraw.com/mathbot/file/1729823642277ypjtoi4r.jpg

The figure shows a shaded region in the shape of a large rectangle with two smaller rectangles removed. Let's break this down step by step to find the area of the shaded region.

### Step 1: Calculate the area of the large outer rectangle.
- Length: 27 m
- Height: 12 m

The area of the large rectangle is:
\[
\text{Area of large rectangle} = 27 \, \text{m} \times 12 \, \text{m} = 324 \, \text{m}^2
\]

### Step 2: Calculate the area of the cut-out smaller rectangle.
The smaller rectangle has the following dimensions:
- Length: 17 m
- Height: 5 m

The area of this smaller rectangle is:
\[
\text{Area of small rectangle} = 17 \, \text{m} \times 5 \, \text{m} = 85 \, \text{m}^2
\]

### Step 3: Calculate the total shaded area.
The shaded area is the area of the large rectangle minus the area of the smaller rectangle. Thus:
\[
\text{Shaded area} = 324 \, \text{m}^2 - 85 \, \text{m}^2 = 239 \, \text{m}^2
\]

Therefore, the area of the shaded region is **239 m²**.

Would you like further details on any part of the solution?

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Here are 5 questions to expand your understanding:
1. How do you find the perimeter of the shaded region?
2. What would happen if the smaller cut-out rectangle was in a different position?
3. How could you use this method to calculate more complex shapes?
4. If the dimensions of the larger rectangle changed, how would it affect the shaded area?
5. How can you check your calculations for accuracy?

**Tip**: When subtracting areas, make sure both shapes have the same units and check dimensions carefully!

Find the area of the shaded region. All angles are right angles. The dimensions provided are: Outer rectangle: 27 m (length) and 12 m (height), Inner rectangle: 17 m (length) and 5 m (height).

Learn how to find the area of a shaded region by subtracting the area of a smaller cut-out rectangle from a larger outer rectangle. This example involves an outer rectangle of 27 m × 12 m and a smaller rectangle of 17 m × 5 m. Suitable for middle school students.

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