https://file.aiquickdraw.com/mathbot/file/17266913184496kj63vca.jpg

We have an irregular quadrilateral in the image with the following dimensions:

- A vertical side of 8 units,
- A horizontal base of 15 units,
- A slanted top side of 3 units,
- Another vertical side of 3 units.

### Step 1: Break the shape into simpler components

We can split the figure into two parts:
1. A rectangle with sides 15 units and 8 units.
2. A right triangle with a height of 5 units (since the total height is 8 and the right vertical side is 3, the remaining height is \(8 - 3 = 5\)) and a base of 3 units.

### Step 2: Calculate the area

**Area of the rectangle**:
\[
A_{\text{rect}} = 15 \times 8 = 120 \text{ square units}
\]

**Area of the triangle**:
\[
A_{\text{tri}} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 3 \times 5 = 7.5 \text{ square units}
\]

**Total area**:
\[
A_{\text{total}} = A_{\text{rect}} + A_{\text{tri}} = 120 + 7.5 = 127.5 \text{ square units}
\]

### Step 3: Calculate the perimeter

The perimeter is the sum of all sides:
\[
P = 15 + 8 + 3 + 3 + \text{slanted side}
\]

We need to find the length of the slanted side using the Pythagorean theorem for the right triangle:
\[
\text{slanted side} = \sqrt{(3^2 + 5^2)} = \sqrt{9 + 25} = \sqrt{34} \approx 5.83
\]

So, the perimeter is:
\[
P = 15 + 8 + 3 + 3 + 5.83 \approx 34.83 \text{ units}
\]

### Final Answer:
- **Area**: 127.5 square units
- **Perimeter**: 34.83 units

Would you like more details on any of the steps?

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Here are five related questions you can explore:
1. How do you find the area of a trapezoid?
2. What is the formula for the perimeter of a triangle?
3. How does the Pythagorean theorem help in finding diagonal lengths?
4. How would the area change if the slanted side were longer?
5. What if the figure was 3D? How would you calculate its volume?

**Tip**: When breaking complex shapes, always look for common geometric figures like rectangles or triangles.

Find the area and perimeter of the adjoining figure.

Learn how to find the area and perimeter of an irregular quadrilateral by breaking it into simpler shapes, such as rectangles and right triangles. This problem involves using key geometry concepts like area and perimeter calculations, along with the Pythagorean theorem. Suitable for students in Grades 7-9.

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