Rhombus

Area of a rhombus = (side)^2 - (difference of diagonals)^2/4

Relevant wiki: Pythagorean Theorem

Let $a$ be the longer diagonal and $b$ be the shorter diagonal. Using pythagorean theorem, we have

$68^2=\left(\dfrac{b}{2}\right)^2+\left(\dfrac{a}{2}\right)^2$

However, $a=b+56$ ,

$4624=\dfrac{b^2}{4}+\left(\dfrac{b+56}{2}\right)^2$

$4624=\dfrac{b^2}{4}+\dfrac{b^2+112b+3136}{4}$

$18496=2b^2+112b+3136$

$2b^2+112b-15360=0$

$b^2+56b-7680=0$

$(b+120)(b-64)=0$

$b=-120$ or $b=64$ (reject the negative value)

It follows that $a=64+56=120$ .

$A=\dfrac{1}{2}ab=\dfrac{1}{2}(64)(120)=$ $\boxed{3840}$

Note:

The area of a rhombus is half the product of the diagonals.