In the largest triangle, the missing side has length
√((11 + 5)² - <em>x</em> ²) = √(256 - <em>x</em> ²)
But it's also the hypotenuse of the triangle with side lengths 11 and <em>y</em>, so that its length can also be written as
√(11² + <em>y</em> ²) = √(121 + <em>y</em> ²)
so that
√(256 - <em>x</em> ²) = √(121 + <em>y</em> ²)
or, by taking the squares of both sides,
256 - <em>x</em> ² = 121 + <em>y</em> ²
<em>y</em> ² = 135 - <em>x</em> ²
In the smallest triangle, you have
5² + <em>y</em> ² = <em>x</em> ² ==> <em>x</em> ² = 25 + <em>y</em> ²
Substitute this into the previous equation and solve for <em>y</em> :
<em>y</em> ² = 135 - (25 + <em>y</em> ²)
<em>y</em> ² = 110 - <em>y</em> ²
2<em>y</em> ² = 110
<em>y</em> ² = 55
<em>y</em> = √55