Triangles and Polygons

The Densest Topic in GRE Geometry

Triangles are where the GRE concentrates its geometry difficulty. Almost every hard quant figure resolves into one or more triangles, and the polygon questions reduce to triangles too. The good news is that a compact toolkit — five facts about triangles, two special right-triangle ratios, one similarity idea, and one polygon formula — covers essentially the entire domain. And a reminder that saves points on every figure: the GRE tests special right triangles, but it does not test trigonometry, so you will never need sine, cosine, or tangent. The side ratios below do all the work those functions would.

Angles Inside a Triangle

The interior angles of every triangle sum to $180^\circ$. Know two angles and the third is forced. The exterior-angle theorem is a fast corollary: an exterior angle (formed by extending one side) equals the sum of the two remote (non-adjacent) interior angles. Instead of finding an interior angle and subtracting from $180^\circ$, you add the two far angles directly.

Special triangles carry built-in angle facts. An isosceles triangle has two equal sides and, opposite them, two equal base angles. An equilateral triangle has three equal sides and three $60^\circ$ angles. So an isosceles triangle with a $40^\circ$ vertex angle has base angles of $\dfrac{180^\circ - 40^\circ}{2} = 70^\circ$ each.

The Triangle Inequality

Three lengths form a triangle only if each side is shorter than the sum of the other two. Equivalently, the third side of a triangle with sides $a$ and $b$ must satisfy

$|a - b| < \text{third side} < a + b.$

If two sides are $5$ and $9$, the third side lies strictly between $4$ and $14$. This range is a favorite GRE device, especially in "how many integer values are possible?" items and in Quantitative Comparisons where the third side is deliberately left unpinned so the answer is "cannot be determined."

Area

The workhorse area formula is

$A = \tfrac{1}{2} b h,$

where the height $h$ is measured perpendicular to the chosen base $b$ — not along a slanted side. Any side may serve as the base as long as you pair it with the matching perpendicular height. For an equilateral triangle of side $s$, a useful shortcut is $A = \dfrac{\sqrt{3}}{4}s^2$.

The Pythagorean Theorem and the Special Right Triangles

In a right triangle with legs $a$, $b$ and hypotenuse $c$ (the side opposite the right angle, always the longest),

$a^2 + b^2 = c^2.$

Two families of whole-number right triangles recur so often they should be automatic: the $3$-$4$-$5$ and the $5$-$12$-$13$ triples, along with all their multiples (a $6$-$8$-$10$ triangle is just $3$-$4$-$5$ doubled). Spotting a scaled triple lets you skip the arithmetic entirely.

Two special right triangles are defined by their angles and have fixed side ratios:

In a $30$-$60$-$90$ triangle the side opposite $30^\circ$ is half the hypotenuse, and the side opposite $60^\circ$ is $\sqrt{3}$ times the short side. In a $45$-$45$-$90$ triangle the two legs are equal and the hypotenuse is a leg times $\sqrt{2}$. These ratios appear inside squares (a diagonal splits a square into two $45$-$45$-$90$ triangles) and equilateral triangles (an altitude splits it into two $30$-$60$-$90$ triangles), so they show up far beyond "right triangle" problems.

Similar Triangles

Two triangles are similar when their corresponding angles are equal; their corresponding sides are then in a constant ratio. Similarity is the engine behind most "find the missing length" problems. Once you establish similarity — often because two triangles share an angle and each contains a right angle, or because a line is drawn parallel to one side — set up a proportion and cross-multiply:

$\frac{\text{side in triangle 1}}{\text{matching side in triangle 2}} = \frac{\text{another side in 1}}{\text{its match in 2}}.$

Shadow problems are pure similarity: the sun's parallel rays make a person and a tree cast proportional shadows. If a $6$-foot person casts a $4$-foot shadow while a tree casts a $20$-foot shadow, then $\dfrac{6}{4} = \dfrac{h}{20}$, giving $h = 30$ feet. The one discipline that matters is matching each side to its true corresponding side before writing the ratio.

Polygons: Reduce to Triangles

Any convex polygon with $n$ sides can be cut into $n - 2$ triangles from a single vertex, so its interior angles sum to

$(n - 2)\times 180^\circ.$

A quadrilateral sums to $360^\circ$, a pentagon to $540^\circ$, a hexagon to $720^\circ$. If the polygon is regular (all angles equal), each interior angle is $\dfrac{(n-2)\times 180^\circ}{n}$. Setting that equal to a given angle lets you solve for the number of sides — a $150^\circ$ interior angle forces $n = 12$. The exterior angles of any convex polygon always sum to $360^\circ$, so each exterior angle of a regular $n$-gon is $\dfrac{360^\circ}{n}$.

The Habits That Matter

Look for hidden right triangles and scaled triples before reaching for the calculator. When a figure hands you a $30^\circ$, $45^\circ$, or $60^\circ$ angle in a right triangle, use the ratio table instead of trig. Confirm that a claimed set of sides actually satisfies the triangle inequality. And in comparison problems, ask whether a length or angle is truly pinned down — the GRE loves configurations that look determined but are not.