Daniel finds a rectangular index card and measures its diagonal to be $2\sqrt{34}$ centimeters. He cuts out equal squares of side $1$ cm at two opposite corners and measures the distance between the two closest vertices of these squares to be $4\sqrt{5}$ centimeters. What is the area of the original index card?

1. Let the width and height of the index card be $a$ and $b$ centimeters. The diagonal gives $a^2+b^2=(2\sqrt{34})^2=136$. The distance between the closest vertices gives $(a-2)^2+(b-2)^2=(4\sqrt5)^2=80$.
2. Expanding the second equation and using the first gives $136-4(a+b)+8=80$, so $a+b=16$.
3. Therefore, the area is
   $$ab=\frac{(a+b)^2-(a^2+b^2)}2=\frac{256-136}{2}=60$$
   square centimeters.