1. **State the problem:** We have a square ABCD with side length $2\sqrt{15}$. Points E and F are midpoints of sides AB and BC respectively. We need to find the area of triangle $\triangle DMN$ inside the square, where M and N lie on segments EF and FD respectively. 2. **Determine coordinates of points:** - Let A be at origin $(0,0)$. - Then B is at $(2\sqrt{15},0)$. - C is at $(2\sqrt{15},2\sqrt{15})$. - D is at $(0,2\sqrt{15})$. 3. **Find midpoints E and F:** - E midpoint of AB is $\left(\frac{0+2\sqrt{15}}{2}, \frac{0+0}{2}\right) = (\sqrt{15},0)$. - F midpoint of BC is $\left(\frac{2\sqrt{15}+2\sqrt{15}}{2}, \frac{0+2\sqrt{15}}{2}\right) = (2\sqrt{15}, \sqrt{15})$. 4. **Identify points M and N:** - Given description, M is on EF, N is on FD. - Line EF is between E $(\sqrt{15},0)$ and F $(2\sqrt{15}, \sqrt{15})$. - Line FD is between F $(2\sqrt{15}, \sqrt{15})$ and D $(0, 2\sqrt{15})$. Since points M and N are located such that DMN forms a triangle inside the square, and with no further info, we assume M and N are E and F respectively. 5. **Coordinates:** - D: $(0, 2\sqrt{15})$ - M (E): $(\sqrt{15}, 0)$ - N (F): $(2\sqrt{15}, \sqrt{15})$ 6. **Area of $\triangle DMN$ using coordinates:** Area = $\frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$ Substitute: - $x_1 = 0$, $y_1 = 2\sqrt{15}$ - $x_2 = \sqrt{15}$, $y_2 = 0$ - $x_3 = 2\sqrt{15}$, $y_3 = \sqrt{15}$ Calculate: $$ \frac{1}{2} |0(0 - \sqrt{15}) + \sqrt{15}(\sqrt{15} - 2\sqrt{15}) + 2\sqrt{15}(2\sqrt{15} - 0)| = \frac{1}{2} |0 + \sqrt{15}(-\sqrt{15}) + 2\sqrt{15}(2\sqrt{15})| = \frac{1}{2} |0 - 15 + 4 \times 15| = \frac{1}{2} | -15 + 60| = \frac{1}{2} \times 45 = 22.5 $$ **Final answer:** The area of $\triangle DMN$ is $22.5$ square units.