1. **Problem statement:** Triangles $\triangle ABC$ and $\triangle XYZ$ are similar right-angled isosceles triangles. Squares $KLM B$ and $PQRS$ inside them have equal areas. Given the area of $\triangle ABC$ is 200, find the area of $\triangle XYZ$. 2. **Formula and rules:** For similar triangles, areas scale by the square of the similarity ratio. If the scale factor of sides is $k$, then the area scale factor is $k^2$. 3. **Step 1: Express areas of triangles and squares.** - Let the leg of $\triangle ABC$ be $a$. Since it is right-angled isosceles, area is $\frac{1}{2}a^2 = 200 \implies a^2 = 400 \implies a = 20$. 4. **Step 2: Relate square side to triangle leg.** - Square $KLM B$ is inside $\triangle ABC$ with vertices on sides. Let the square side length be $s$. - Since $K$ and $M$ lie on legs $AB$ and $BC$, and $L$ lies on hypotenuse $AC$, the square side $s$ relates to $a$ by geometry of the right isosceles triangle. 5. **Step 3: Find $s$ in terms of $a$.** - The square inside a right isosceles triangle with legs $a$ has side length $s = a(\sqrt{2} - 1)$ (known geometric result or by coordinate geometry). 6. **Step 4: Area of square $KLM B$ is $s^2 = a^2(\sqrt{2} - 1)^2 = 400(3 - 2\sqrt{2})$. 7. **Step 5: Since $PQRS$ has equal area, its side length $t$ satisfies $t^2 = s^2$. 8. **Step 6: Let the leg of $\triangle XYZ$ be $b$. Similarly, the square inside it has side length $t = b(\sqrt{2} - 1)$. 9. **Step 7: Equate square areas:** $$ s^2 = t^2 \implies a^2(3 - 2\sqrt{2}) = b^2(3 - 2\sqrt{2}) \implies a^2 = b^2 \implies b = a = 20 $$ 10. **Step 8: But the problem implies different triangles, so check carefully.** - The squares have equal areas, so $$ s^2 = t^2 \implies a^2(3 - 2\sqrt{2}) = b^2(3 - 2\sqrt{2}) \implies a^2 = b^2$$ - This means $b = a$ or $b = -a$ (discard negative). 11. **Step 9: Therefore, the triangles are congruent, so area of $\triangle XYZ$ is also 200.** **Final answer:** The area of $\triangle XYZ$ is $\boxed{200}$.