Compute the exact area or length as directed. Square roots do not need to be simplified, but arithmetic does.
\( \text{square's area: } 5^2 = 25 \)
\( \text{triangle's area: } \frac{1}{2} 5 \cdot 5 = \frac{25}{2} \) → Student used slant height \((5)\), not vertical height \( \left( \sqrt{5^2 - (5/2)^2} \right) \)
\( \text{total area: } 25 + \frac{25}{2} = \frac{75}{2} \)
correct area: \( 25 + \frac{5}{2} \sqrt{18.75} = 25 + \dfrac{5\sqrt{75}}{4} = 25 + \dfrac{25\sqrt{3}}{4} \)