1. The problem describes a rhombus with vertices approximately at $(-5,5)$, $(5,5)$, $(0,10)$, and $(0,0)$ on the coordinate plane. We are asked to find the side length $a$, the height $h$, and the area $S$ of the rhombus. 2. First, calculate the side length $a$. The rhombus sides are the distances between consecutive vertices. Calculate the distance between $(-5,5)$ and $(0,10)$ using the distance formula: $$a = \sqrt{(0 + 5)^2 + (10 - 5)^2} = \sqrt{5^2 + 5^2} = \sqrt{50} = 5\sqrt{2}$$ 3. The height $h$ is the perpendicular distance between parallel sides. We can find $h$ using the area formula of a rhombus once $S$ is known or by calculating the vertical height between points that lie opposite each other in the rhombus. 4. Calculate the area $S$. The rhombus area formula is: $$S = \text{base} \times \text{height} = a \times h$$ Alternatively, since the rhombus is on a coordinate grid, we can use the formula for the area using vertices: $$S = \frac{1}{2} \times |(x_1y_2 + x_2y_3 + x_3y_4 + x_4y_1) - (y_1x_2 + y_2x_3 + y_3x_4 + y_4x_1)|$$ Substituting the points $(-5,5)$, $(5,5)$, $(0,10)$, and $(0,0)$: $$S = \frac{1}{2} \times |((-5)\times5 + 5\times10 + 0\times0 + 0\times5) - (5\times5 + 5\times0 + 10\times0 + 0\times(-5))|$$ $$= \frac{1}{2} \times |(-25 + 50 + 0 + 0) - (25 + 0 + 0 + 0)| = \frac{1}{2} \times |25 - 25| = \frac{1}{2} \times 0 = 0$$ Looks like an error; let's reorder points properly to match the polygon traversal: $(-5,5)$, $(0,10)$, $(5,5)$, $(0,0)$: $$S = \frac{1}{2} \times |(-5 \times 10 + 0 \times 5 + 5 \times 0 + 0 \times 5) - (5 \times 0 + 10 \times 5 + 5 \times 0 + 0 \times (-5))|$$ $$= \frac{1}{2} \times |(-50 + 0 + 0 + 0) - (0 + 50 + 0 + 0)| = \frac{1}{2} \times |-50 - 50| = \frac{1}{2} \times 100 = 50$$ 5. Now, find the height $h$ using $S = a \times h$: $$50 = 5\sqrt{2} \times h \implies h = \frac{50}{5\sqrt{2}} = \frac{10}{\sqrt{2}} = 5\sqrt{2}$$ 6. Summary: - Side length $a = 5\sqrt{2}$ - Height $h = 5\sqrt{2}$ - Area $S = 50$ Therefore, the rhombus has side length $5\sqrt{2}$, height $5\sqrt{2}$, and area $50$.