Polygons & Quadrilaterals

A polygon is a closed plane figure with $n$ straight sides. A quadrilateral is the polygon where $n=4$. The fundamental property of any convex quadrilateral is that its interior angles sum to $360^\circ$.

Why it works. Every convex quadrilateral can be split into two triangles by drawing one diagonal. Since each triangle's interior angles sum to $180^\circ$, the total sum is $180^\circ + 180^\circ = 360^\circ$.

SAT use. If you are given three angles of a quadrilateral, subtract their sum from $360^\circ$ to find the fourth. If the polygon is regular, all sides and angles are equal.

Worked example. A quadrilateral has angles $70^\circ, 110^\circ$, and $x^\circ$. If the fourth angle is $2x^\circ$, then $70+110+x+2x=360 \Rightarrow 3x=180 \Rightarrow x=60$.

A trapezoid has exactly one pair of parallel sides (bases). An isosceles trapezoid is a special case where the non-parallel sides (legs) are equal, which forces the base angles to be equal.

Key Property. In an isosceles trapezoid, the diagonals are also equal in length. The area is calculated as $A = \frac{b_1 + b_2}{2} \cdot h$.

SAT use. Drop perpendiculars from the top vertices to the base to create a rectangle and two right triangles. This is the most efficient way to solve for unknown side lengths using the Pythagorean theorem.

Worked example. Bases are $6$ and $10$, height is $3$. Area $= \frac{6+10}{2} \cdot 3 = 8 \cdot 3 = 24$.

A parallelogram has two pairs of parallel sides. Opposite sides are equal, opposite angles are equal, and consecutive angles are supplementary ($180^\circ$).

The Diagonal Rule. The diagonals bisect each other (cut each other in half). This means they share the same midpoint.

SAT use. In coordinate geometry, if you are given three vertices of a parallelogram, find the fourth by ensuring the midpoint of one diagonal equals the midpoint of the other.

Worked example. Consecutive angles are $x^\circ$ and $2x+30^\circ$. $x + 2x + 30 = 180 \Rightarrow 3x = 150 \Rightarrow x = 50$.

A rectangle is a parallelogram with four right angles. It inherits all parallelogram properties plus one critical upgrade: the diagonals are equal in length.

Geometry Insight. Because the diagonals are equal and bisect each other, the four segments created by the intersection are all equal. This creates four isosceles triangles inside the rectangle.

SAT use. If you know the length and width, the diagonal is $\sqrt{l^2 + w^2}$. Use this to find distances between opposite corners.

Worked example. A rectangle has sides $3$ and $4$. The diagonal is $\sqrt{3^2+4^2} = 5$.

A rhombus is a parallelogram with four equal sides. Its diagonals are perpendicular ($90^\circ$) and bisect the corner angles.

Geometry Insight. The diagonals divide the rhombus into four congruent right triangles. If the diagonals are $d_1$ and $d_2$, the area is $\frac{d_1 \cdot d_2}{2}$.

SAT use. Use the Pythagorean theorem on one of the four internal right triangles to find the side length: $s = \sqrt{(\frac{d_1}{2})^2 + (\frac{d_2}{2})^2}$.

Worked example. Diagonals are $6$ and $8$. Area $= (6 \cdot 8)/2 = 24$. Side length $= \sqrt{3^2+4^2} = 5$.

A square is a regular quadrilateral. It possesses every property of the rectangle and the rhombus simultaneously.

Key Properties. Diagonals are equal, perpendicular, bisect each other, and bisect the corner angles ($45^\circ-45^\circ-90^\circ$ triangles).

SAT use. The relationship between side $s$ and diagonal $d$ is $d = s\sqrt{2}$. This is a direct application of the $45-45-90$ triangle ratio.

Worked example. A square has a diagonal of $10$. $s\sqrt{2} = 10 \Rightarrow s = \frac{10}{\sqrt{2}} = 5\sqrt{2}$.

Memorize this table for the SAT to identify shapes instantly based on diagonal behavior.

SAT use. If a question states 'the diagonals are perpendicular', you immediately eliminate rectangles and parallelograms. You are left with a rhombus or square.

For any convex polygon with $n$ sides, the sum of interior angles is $S = 180(n-2)$.

Why it works. A polygon with $n$ sides can be divided into $n-2$ triangles by drawing diagonals from a single vertex. Each triangle contributes $180^\circ$.

SAT use. For a regular $n$-gon, each interior angle is $\frac{180(n-2)}{n}$.

Worked example. A regular hexagon ($n=6$): sum $= 180(6-2) = 720^\circ$. Each angle $= 720/6 = 120^\circ$.

The sum of the exterior angles of any convex polygon is always $360^\circ$, regardless of the number of sides.

Key Relationship. At any vertex, the interior angle and exterior angle are supplementary: $Interior + Exterior = 180^\circ$.

SAT use. If a problem gives you the exterior angle of a regular polygon, finding $n$ is trivial: $n = 360 / Exterior$. This is often faster than using the interior angle formula.

Worked example. A regular polygon has an exterior angle of $24^\circ$. $n = 360/24 = 15$.

A regular hexagon is composed of 6 equilateral triangles meeting at the center. This makes the distance from the center to any vertex equal to the side length $s$.

Coordinate Geometry. For parallelograms in the coordinate plane, the midpoint of diagonal $AC$ must equal the midpoint of diagonal $BD$. Use $M = (\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2})$.

SAT use. If you see a regular hexagon, draw the center point. Every internal angle is $120^\circ$, and the triangles are $60-60-60$.

Worked example. Parallelogram vertices $(0,0), (4,0), (x,y), (2,3)$. Midpoint of $(0,0)$ to $(x,y)$ is $(x/2, y/2)$. Midpoint of $(4,0)$ to $(2,3)$ is $(3, 1.5)$. Thus $x=6, y=3$.