https://file.aiquickdraw.com/mathbot/file/1729061438734ey8fp8c6.jpg

The image you provided shows a quadrilateral, specifically a trapezoid, with two right angles and labeled side lengths of 4 km, 9 km, and 13 km.

To find the area of this quadrilateral, we can treat it as the sum of two right triangles. The right angles allow us to split the trapezoid into two triangles by dropping a perpendicular line from the top vertex to the base.

### Step-by-Step Process:

1. **Identify the two right triangles:**
   - One triangle has sides 4 km (vertical height) and 13 km (the base).
   - The other triangle has sides 9 km (vertical height) and 13 km (the base).

2. **Calculate the area of each triangle:**
   The formula for the area of a right triangle is:
   \[
   A = \frac{1}{2} \times \text{base} \times \text{height}
   \]

   - For the left triangle with height 4 km and base 13 km:
     \[
     A_1 = \frac{1}{2} \times 4 \times 13 = 26 \, \text{km}^2
     \]

   - For the right triangle with height 9 km and base 13 km:
     \[
     A_2 = \frac{1}{2} \times 9 \times 13 = 58.5 \, \text{km}^2
     \]

3. **Sum the areas:**
   \[
   \text{Total Area} = A_1 + A_2 = 26 + 58.5 = 84.5 \, \text{km}^2
   \]

Thus, the area of the quadrilateral is **84.5 km²**.

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

Here are some related questions:
1. How can we verify the solution using another method?
2. What is the perimeter of the quadrilateral?
3. How does the Pythagorean theorem apply to this problem?
4. How would we find the height of the trapezoid if the diagonal was unknown?
5. What would happen if this shape was not a right-angled quadrilateral?

**Tip:** When solving for areas involving irregular shapes, breaking the shape into simpler components, like triangles, makes the calculations more manageable.

Find the area of a quadrilateral with right angles and side lengths 4 km, 9 km, and 13 km as shown in the image.

Learn how to calculate the area of a quadrilateral with right angles and side lengths 4 km, 9 km, and 13 km. This problem involves using triangle area formulas and applying geometry concepts suitable for students in Grades 7-9.

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