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$\,45^\circ\text{-}45^\circ\text{-}90^\circ\,$ TRIANGLE

Lengths of Sides in a $\,45^\circ\text{-}45^\circ\text{-}90^\circ\,$ Triangle Let $\,s\,$ denote the length of the two shorter sides. Then, the hypotenuse has length $\,\sqrt{2}s\,$. Conversely, if a triangle has sides of lengths $\,s\,$, $\,s\,$, and $\,\sqrt{2}s\,$, then it is a $\,45^\circ\text{-}45^\circ\text{-}90^\circ\,$ triangle. Since $\,\sqrt{2}\approx 1.4\,$, it follows that the hypotenuse is a little less than one and a half times the shortest side.

It's easy to see that these are the correct relationships between the lengths of the sides, as follows:

Start with an equiangular (hence equilateral) triangle, where each side has length $\,2\,$.
Notice that all the angles must be $\,\frac{180^\circ}{3} = 60^\circ\,$.
Drop a perpendicular, as shown below.
The purple and green triangles are $\,30^\circ\text{-}60^\circ\text{-}90^\circ\,$ triangles, where the length of the shortest side is $\,1\,$.
A quick application of the Pythagorean Theorem shows that the remaining side has length $\,\sqrt{3}\,$.
Then, scale all sides by $\,s\,$.

Create a triangle with two sides of length $\,1\,$ and a $\,90^\circ\,$ angle, as shown below.
Since angles opposite equal sides must be equal,
and since the two remaining angles must add to $\,90^\circ\,$,
they must each be $\,45^\circ\,$.
A quick application of the Pythagorean Theorem shows that the remaining side must equal $\,\sqrt{2}\,$.
Then, scale all sides by $\,s\,$.

This is the last exercise in Topics in Geometry.

(If you went through all the topics,
then congratulations are definitely in order!)

If you haven't already done so,
you might want to take a look at these:

A First Course in Algebra

Topics in Algebra II