1. **State the problem:** We have a semicircle with diameter 10 and a rectangle inscribed inside it. One corner of the rectangle is at the midpoint of the diameter, and the opposite corners lie on the semicircle and the diameter line. We want to find the area of this shaded rectangle. 2. **Set up the semicircle equation:** The semicircle has diameter 10, so radius $r = 5$. Place the diameter along the x-axis from $-5$ to $5$ with the center at the origin $(0,0)$. The semicircle equation (top half circle) is: $$y = \sqrt{25 - x^2}$$ 3. **Define the rectangle vertices:** Let the rectangle have vertices at - midpoint of diameter: $(0,0)$, - at some positive $x$ along the base: $(x,0)$, - the corresponding point on the semicircle above: $(x,y)$ where $y = \sqrt{25 - x^2}$, - and the symmetric point at $(-x,y)$ (due to equal distances on the semicircle). 4. **Calculate rectangle dimensions:** - Base width = distance between $(-x,0)$ and $(x,0)$ = $2x$ - Height = $y = \sqrt{25 - x^2}$ Since one corner is at midpoint $(0,0)$, the rectangle is formed by $(0,0)$, $(x,0)$, $(x,y)$, and $(0,y)$. 5. **Find the area $A$ of rectangle:** $$A = \text{width} \times \text{height} = x \times y = x \sqrt{25 - x^2}$$ 6. **Maximize the area:** To find $x$ that maximizes $A$, treat $A$ as a function: $$A(x) = x \sqrt{25 - x^2}$$ 7. **Use derivative:** $$A'(x) = \sqrt{25 - x^2} + x \times \frac{1}{2 \sqrt{25 - x^2}} (-2x) = \sqrt{25 - x^2} - \frac{x^2}{\sqrt{25 - x^2}} = \frac{25 - x^2 - x^2}{\sqrt{25 - x^2}} = \frac{25 - 2x^2}{\sqrt{25 - x^2}}$$ 8. **Set derivative to zero:** $$25 - 2x^2 = 0 \implies 2x^2 = 25 \implies x^2 = \frac{25}{2} \implies x = \frac{5}{\sqrt{2}} = \frac{5 \sqrt{2}}{2}$$ 9. **Calculate height at this $x$:** $$y = \sqrt{25 - x^2} = \sqrt{25 - \frac{25}{2}} = \sqrt{\frac{25}{2}} = \frac{5 \sqrt{2}}{2}$$ 10. **Calculate maximum area:** $$A_{max} = x y = \frac{5 \sqrt{2}}{2} \times \frac{5 \sqrt{2}}{2} = \frac{25 \times 2}{4} = \frac{50}{4} = 12.5$$ 11. **Conclusion:** The maximum area of the rectangle inside the semicircle under the described constraints is 12.5, not 20 or 25. If your teacher said 20, possibly the rectangle is positioned differently or using a diameter other than 10. **Final answer:** $$\boxed{12.5}$$