# If an ice rink is sqrt(404) across then what is its perimeter?

Oct 21, 2017

It could be $44$, but I'm not particularly convinced.

#### Explanation:

Since there is not enough information in the question, let us work with the following assumptions:

• The ice rink is rectangular.

• The length of each side is an integer.

• The diagonal measurement of the ice rink is $\sqrt{404}$.

• The rectangle is as close as possible to being square given the above conditions.

By Pythagoras, we can tell that we need a pair of positive integers, the sum of whose squares is $404$.

Note that:

$14 = \sqrt{196} < \sqrt{\frac{404}{2}} = \sqrt{202} < \sqrt{225} = 15$

So one of the sides of the rectangle is $\ge 15$ and the other $\le 14$.

Trying each possible larger side from $15$ upwards, we find:

• $404 - {15}^{2} = 404 - 225 = \textcolor{red}{\cancel{\textcolor{b l a c k}{179}}}$

• $404 - {16}^{2} = 404 - 256 = \textcolor{red}{\cancel{\textcolor{b l a c k}{148}}}$

• $404 - {17}^{2} = 404 - 289 = \textcolor{red}{\cancel{\textcolor{b l a c k}{115}}}$

• $404 - {18}^{2} = 404 - 324 = \textcolor{red}{\cancel{\textcolor{b l a c k}{80}}}$

• $404 - {19}^{2} = 404 - 361 = \textcolor{red}{\cancel{\textcolor{b l a c k}{43}}}$

• $404 - {20}^{2} = 404 - 400 = 4 = {2}^{2}$

So with the above assumptions, the ice rink is $20 \times 2$, with perimeter $20 + 2 + 20 + 2 = 44$.

Well there's an answer, but $20 \times 2$ seems strange proportions for a real ice rink.