problem: Kite $ABCD$ (a quadrilateral with two pairs of adjacent equal sides) has coordinates $A\ (0,7),\ B\ (1,0),\ C\ (12,-2),$ and $D\ (7,8).$ What is the area of $ABCD,$ given that the area of a kite is equal to half the product of its diagonals?
  
  [asy]
  string sp(pair P, string P2){return "$" + P2 + "\,(" + string(P.x) + "," + string(P.y) + ")$";}
  size(150); defaultpen(fontsize(10)); draw((-5,0)--(15,0),Arrows(4)); draw((0,-5)--(0,10),Arrows(4)); pair A=(0,7),B=(1,0),C=(12,-2),D=(7,8); draw(A--B--C--D--cycle, linewidth(0.7)); draw(A--C, dashed); draw(B--D, dashed); label(sp(A,"A"),A,W); label(sp(B,"B"),B,S); label(sp(C,"C"),C,E); label(sp(D,"D"),D,N);
  [/asy]
solution: As the problem suggests, we need to compute the lengths of the diagonals $\overline{AC}$ and $\overline{BD}$. By the distance formula,
  
  \begin{align*}
  AC &= \sqrt{(12 -0)^2 + (-2-7)^2} = \sqrt{12^2 + 9^2} = 15\\
  BD &= \sqrt{(7-1)^2 + (8-0)^2} = \sqrt{6^2 + 8^2} = 10\\
  \end{align*}Thus, the answer is $\frac 12 \cdot 10 \cdot 15 = \boxed{75}$.
  
  As an extra challenge, can you figure out why the area of a kite equals half the product of the lengths of its diagonals?