Special right triangles, similarity and congruence

Two right triangles with fixed shapes

Two right triangles come with side ratios you should know without working them out. Both follow from the Pythagorean theorem, and both turn up whenever a square is cut along its diagonal or an equilateral triangle is cut by its height.

Triangle	Sides, in order (short to long)	Where it comes from
$45^\circ$-$45^\circ$-$90^\circ$	$s, \; s, \; s\sqrt{2}$	half a square, cut along a diagonal
$30^\circ$-$60^\circ$-$90^\circ$	$s, \; s\sqrt{3}, \; 2s$	half an equilateral triangle, cut by its height

The 45-45-90 triangle is isosceles, so its legs are equal. If each leg is $s$, the hypotenuse is $\sqrt{s^2 + s^2} = s\sqrt{2}$. Going the other way, a hypotenuse $h$ gives legs of $\frac{h}{\sqrt{2}} = \frac{h\sqrt{2}}{2}$.

The 30-60-90 triangle is half of an equilateral triangle with side $2s$. The cut runs from one vertex to the midpoint of the opposite side, so the short leg is $s$, the hypotenuse is the full side $2s$, and the long leg is $\sqrt{(2s)^2 - s^2} = s\sqrt{3}$. Keep the order straight with angles: the shortest side faces the smallest angle, $30^\circ$; the hypotenuse faces $90^\circ$; and the middle side, $s\sqrt{3}$, faces $60^\circ$. Since $\sqrt{3} \approx 1.73$ lies between 1 and 2, the long leg sits between the other two sides, as it must.

Always work through the short leg. If you are given the long leg, divide by $\sqrt{3}$ to get the short leg; if you are given the hypotenuse, halve it. Then build the rest.

Where they hide

• A square with side $s$ has diagonal $s\sqrt{2}$. A square with diagonal $d$ has area $\frac{d^2}{2}$.
• An equilateral triangle with side $s$ has height $\frac{s\sqrt{3}}{2}$ and area $\frac{\sqrt{3}}{4}s^2$.
• A $30^\circ$ or $60^\circ$ angle in a parallelogram, trapezoid or rhombus usually means a height can be found from a 30-60-90 triangle. Trigonometry is not tested on the GRE, so when such an angle appears, the special triangle is the route.

Similar triangles

Two triangles are similar when they have the same shape: equal angles, and every side of one equal to the matching side of the other multiplied by the same factor $k$. To prove triangles similar you need only two pairs of equal angles, because the third angles are then equal too (all three add to $180^\circ$).

Two set-ups produce similar triangles again and again.

1. A line parallel to one side. In triangle $ABC$, draw segment $DE$ with $D$ on $AB$, $E$ on $AC$ and $DE$ parallel to $BC$. Then triangle $ADE$ is similar to triangle $ABC$: they share angle $A$, and the parallels make angle $ADE$ equal to angle $ABC$ (corresponding angles).
2. Two triangles sharing a vertex angle with a pair of parallel or perpendicular sides, such as two right triangles that share an acute angle.

Match sides by angles, not by position on the page. The sides opposite equal angles correspond. Write the correspondence in order (triangle $ADE$ matches triangle $ABC$, so $AD$ matches $AB$, $DE$ matches $BC$, $AE$ matches $AC$) and the proportions write themselves.

Scaling lengths, areas and volumes

If two figures are similar with length ratio $k$, then every length (sides, heights, perimeters) scales by $k$, every area by $k^2$ and every volume by $k^3$. The reason: area is length times length, so both factors grow by $k$. A triangle whose sides are three times as long has $9$ times the area, not $3$ times.

Congruent triangles

Congruent triangles are similar with $k = 1$: same shape and same size. Three conditions guarantee it: three pairs of equal sides (SSS); two sides and the angle between them (SAS); two angles and any side in matching position (ASA or AAS). Two sides with an angle that is not between them do not settle it in general, because two different triangles can share those three facts. On the test, congruence is mostly a reason to copy a known length or angle from one triangle to another.