Finding the Longer Diagonal of a Parallelogram I was studying for some exams when a wild question appeared. It looks like this:

A certain city block is in the form of parallelogram. Two of its sides measure 32 feet and 41 feet. If the area of the land in the block is 656 square feet, what is the length of the longer diagonal?

My work:
I imagined the problem like this:

This might seem easy for me, but when I think about it, the things I could think of were the areas of parallelogram, which is $Area = base \times height$ and $Area = \frac{1}{2}d_1 d_2 \sin \theta$.
This might be a silly question, but how do you get the length of longer diagonal?
 A: Let the parallelogram be ABCD, where A and C are the acute angles. 
Extend AB, and draw the height of the parallelogram, from C down perpendicular onto that extended AB. Let P be that other end point of the height.
From the area and the length of base AB, you can get the height CP.
Using Pythagoras in the triangle BPC you get the length BP.
You then also know length of AP = AB+BP.
Now in the triangle APC you can apply Pythagoras again to get the length of the long diagonal AC.
A: Let the parallelogram b e $ABCD$, where $A$ is on the lower left corner and the letters go in counter-clockwise fashion $($so $AB$ has length $41)$.
Let $H$ be the point on $AD$ such that $BH$ is perpendicular to $AD$; in particular, the length of $BH$ is the height of your parallelogram.
Since you have the area and the base, you can infer $BH$ from your formula.
Now, consider the right triangle $\triangle ABH$, and let $\alpha=\angle BAH$. We have that $\sin\alpha=\frac{BH}{AB}$.
Now, consider the angle $\beta=\angle ADC$ We have $\beta=180^\circ-\alpha$, so $\cos\beta=-\cos\alpha$. Since you know $\sin\alpha$, you can infer the value of $\cos\beta$ from $(\sin\alpha)^2+(\cos\alpha)^2=1$.
Finally, apply the law of cosines to $\triangle ADC$ to find that the length of $AC$ (the longer diagonal) satisfies
$$AC^2=AD^2+CD^2-2AD\cdot CD\cdot\cos\beta$$
