Rhombus Problem

For a rhombus with diagonals $a,b$ the area is $\dfrac{ab}{2}$ and the side length is $\dfrac{1}{2}\sqrt{a^{2} + b^{2}}$ .

We are thus given that $\dfrac{ab}{2} + 4 \times \dfrac{1}{2}\sqrt{a^{2} + b^{2}} = 200 \Longrightarrow \dfrac{ab}{2} + 2 \times \sqrt{625} = 200 \Longrightarrow ab = 300$ .

So then $(a + b)^{2} = a^{2} + b^{2} + 2ab = 625 + 2 \times 300 = 1225 \Longrightarrow a + b = 35 \Longrightarrow$

$a + \dfrac{300}{a} = 35 \Longrightarrow a^{2} - 35a + 300 = 0 \Longrightarrow (a - 15)(a - 20) = 0$ .

As $a \lt b$ we thus have that $(a,b) = (15,20)$ , and so $x = b - a = \boxed{5}$ .

Let $a$ be the shorter diagonal and $b$ be the longer diagonal. By applying the Pythagorean Theorem, $s = \sqrt{(\frac{a}{2})^2 + (\frac{b}{2})^2} = \sqrt{\frac{a^2}{4} + \frac{b^2}{4}} = \frac{1}{2}\sqrt{a^2 + b^2} = \frac{1}{2}\sqrt{625} = 12.5$

Solving for the perimeter

$P = 4s = 4(12.5) = 50$

Solving for the area

$A = \frac{1}{2}ab = \frac{1}{2}(a)(a + x) = \frac{1}{2}a^2 + \frac{1}{2}ax$

Adding the area and the perimeter

$A + P = 200$

$\frac{1}{2}a^2 + \frac{1}{2}ax + 50 = 200$

After simplifying, we get

$\frac{1}{2}a^2 + \frac{1}{2}ax = 150$ (equation 1)

We know that $a^2 + b^2 = 625$ , so

$a^2 + (a + x)^2 = 625$

After simplifying, we get

$2a^2 + 2ax + x^2 = 625$ (equation 2)

We multiply equation 1 by -4 then add to equation 2 to solve for x,

$(\frac{1}{2}a^2 + \frac{1}{2}ax = 150)(-4)$

It becomes

$-2a^2 -2ax = -600$

Add the above equation to equation 2, we get

$x^2 = 25$

Finally,

$x = 5$ $inches$