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We are given the quadrilateral \( ABCD \), where:

- \( \angle ABC = \angle ADC = 90^\circ \),
- \( AD = DC \),
- \( AB + BC = 20 \).

We need to find the area of quadrilateral \( ABCD \).

### Step-by-Step Breakdown:

1. **Identify Right Triangles**:
   - Both \( \triangle ABC \) and \( \triangle ADC \) are right-angled triangles, as given by the right angles at \( B \) and \( D \).

2. **Let \( AB = x \) and \( BC = y \)**:
   - We are told that \( AB + BC = 20 \), so \( x + y = 20 \).
   - Since \( AD = DC \), let \( AD = DC = z \).

3. **Pythagorean Theorem in \( \triangle ABC \)**:
   Since \( \triangle ABC \) is a right-angled triangle at \( B \), we apply the Pythagorean theorem:
   \[
   AC^2 = AB^2 + BC^2 \quad \Rightarrow \quad AC^2 = x^2 + y^2.
   \]

4. **Pythagorean Theorem in \( \triangle ADC \)**:
   Similarly, for \( \triangle ADC \), we apply the Pythagorean theorem:
   \[
   AC^2 = AD^2 + DC^2 \quad \Rightarrow \quad AC^2 = z^2 + z^2 = 2z^2.
   \]

5. **Equating Both Expressions for \( AC^2 \)**:
   From both triangles, we have two expressions for \( AC^2 \). Equating them:
   \[
   x^2 + y^2 = 2z^2.
   \]

6. **Solving for the Area**:
   The area of quadrilateral \( ABCD \) is the sum of the areas of the two right-angled triangles:
   \[
   \text{Area of } ABCD = \text{Area of } \triangle ABC + \text{Area of } \triangle ADC.
   \]
   The area of \( \triangle ABC \) is:
   \[
   \text{Area of } \triangle ABC = \frac{1}{2} \times AB \times BC = \frac{1}{2} \times x \times y.
   \]
   The area of \( \triangle ADC \) is:
   \[
   \text{Area of } \triangle ADC = \frac{1}{2} \times AD \times DC = \frac{1}{2} \times z \times z = \frac{1}{2} z^2.
   \]
   Therefore, the total area is:
   \[
   \text{Area of } ABCD = \frac{1}{2} \times x \times y + \frac{1}{2} \times z^2.
   \]

Now let's solve for \( z \) using the earlier equations. I'll calculate this.We have two possible solutions for \( x \) and \( y \), both dependent on \( z \). The solutions are:

1. \( x = 10 - \sqrt{(z - 10)(z + 10)} \) and \( y = 10 + \sqrt{(z - 10)(z + 10)} \),
2. \( x = 10 + \sqrt{(z - 10)(z + 10)} \) and \( y = 10 - \sqrt{(z - 10)(z + 10)} \).

Now, we need to find a valid value for \( z \) that satisfies \( AD = DC = z \). To do this, we substitute different values of \( z \) into the system and find the best fit for the geometric constraints. I'll compute the area next.The area of quadrilateral \( ABCD \) is \( 100 \, \text{square units} \).

Would you like further details on this solution or any other clarifications?

Here are 5 related questions for you:
1. How is the Pythagorean theorem applied in solving problems involving right-angled triangles?
2. What is the significance of the condition \( AD = DC \) in this problem?
3. How do the properties of quadrilaterals affect their area calculations?
4. What would happen if the sum of \( AB + BC \) was a different value?
5. How can we generalize the approach used here for other quadrilateral problems?

**Tip:** When solving problems involving multiple right-angled triangles, always look for shared sides or angles to simplify calculations using geometric relationships.

In quadrilateral ABCD, ∠ABC = ∠ADC = 90°, AD = DC, and AB + BC = 20. What is the area of quadrilateral ABCD?

Learn how to solve the area of quadrilateral ABCD where ∠ABC = ∠ADC = 90°, AD = DC, and AB + BC = 20. The problem involves applying the Pythagorean theorem and calculating the areas of two right triangles.

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