The points $(4, 0)$ and $(-4, 0)$ are two non-consecutive vertices of a rhombus with an area of $80$ square units. One of the other vertices is $(0, K)$ where $K > 0$. What is the value of $K$?
I know that a rhombus is a quadrilateral with four congruent sides, so the distance between any two adjacent vertices is the same. I also know that the area of a rhombus is half the product of the lengths of its diagonals, so I can use that formula to find $K$. The diagonals of the rhombus are the segments joining $(4, 0)$ and $(-4, 0)$, and $(0, K)$ and the other unknown vertex. The length of the first diagonal is just $8$, since it is the difference of the $x$-coordinates of the endpoints. The length of the second diagonal is $2K$, since it is twice the $y$-coordinate of the midpoint. So the area of the rhombus is $\frac{1}{2} \cdot 8 \cdot 2K = 8K$. But I also know that the area is $80$, so I can set up an equation and solve for $K$. I get $8K = 80$, which implies $K = 10$. 