Diagonals Part 1

Nice problem! To start, let's set up a simple system of equations, where $x$ and $y$ are the two side lengths of the triangle created by splitting the rectangle in half. Then:

$xy=500$

and

$\frac{x}{y}=\tan{47}$

Solving, we find that the two side lengths are approximately $21.5929845552$ and $23.1556688474$ . Therefore the hypotenuse is approximately $31.66$ which rounds to $\boxed{32}$ .

Without any explicit calculation. Since 47° is more or less 45°, the rectangle is actually near to a square.

Now let x , be the side of our hypotethical square, and d her diagonal. By pithagoras $2x^2=d^2$ , also we are given that the area is 500, therefore $x^2=500$ . So $d \approx \sqrt{1000}$ .

For our last approximation. Notice that $31^{2}=961$ and $32^{2}=1024$ , the nearest one to 1000 is 32.

We conclude that the hypotenuse rounds 32.

BC=BD Cos47, .......DC=BD Sin47.
So area rect. 500=BD * DC=BD * BD * 1/2 * Sin94.
So BD= $\dfrac{\sqrt{500}}{1/2*Sin94.}=31.66\approx\large \color{#D61F06}{32}.$

By the sine rule, BD = $\frac{CD}{sin 47}$ = $\frac{BC}{sin 43}$

Area of the rectangle: CD*BC=500

BD^2= $\frac{CD}{sin 47}$ x $\frac{BC}{sin 43}$ = $\frac{500}{sin 47 * sin 43}$

BD = 31.661...

For me I thought about the factors of 500. Also I considered that in a right isosceles triangle, the angle 2 angles are 45 degrees. As such with this angle being 47 the lengths of the sides of the triangle (ie the factors of 500) must be relatively close to each other. 25 and 20 seemed the only logical choice.