Quadrilaterals, polygons, area and perimeter

Angles of any polygon

Pick one vertex of a polygon with $n$ sides and draw every diagonal from it. The polygon falls into $n - 2$ triangles, and their angles together make up the polygon's angles. So

$$\text{sum of interior angles} = (n - 2) \times 180^\circ$$

Sides $n$	Interior sum	Each angle if regular
4	$360^\circ$	$90^\circ$
5	$540^\circ$	$108^\circ$
6	$720^\circ$	$120^\circ$
8	$1080^\circ$	$135^\circ$

The exterior angles (one at each vertex, made by extending a side) always add to $360^\circ$, whatever $n$ is: walking once round the polygon, you turn through one full turn. For a regular polygon (all sides and all angles equal) each exterior angle is $\frac{360}{n}$, and each interior angle is $180$ minus that. This is the fastest route from an angle to the number of sides: an interior angle of $150^\circ$ means an exterior angle of $30^\circ$, so $n = \frac{360}{30} = 12$.

What each quadrilateral guarantees

Shape	Defined by	Also guaranteed
Parallelogram	both pairs of opposite sides parallel	opposite sides equal; opposite angles equal; neighbouring angles add to $180^\circ$; diagonals cut each other in half
Rectangle	four right angles	everything a parallelogram has; diagonals equal
Rhombus	four equal sides	everything a parallelogram has; diagonals perpendicular
Square	four equal sides and four right angles	everything above
Trapezoid	one pair of parallel sides	the two angles along each leg add to $180^\circ$

Each row inherits the rows it belongs to: a square is a rectangle and a rhombus, and both are parallelograms. The reverse never holds without proof. A quadrilateral with four equal sides is a rhombus, but it is a square only if one angle is known to be $90^\circ$. A figure that "looks like" a rectangle is just a quadrilateral until the question says otherwise.

Area formulas, and why the height is perpendicular

Shape	Area
Rectangle	$\ell w$
Parallelogram	$bh$
Trapezoid	$\frac{1}{2}(b_1 + b_2)h$
Rhombus (or any quadrilateral with perpendicular diagonals)	$\frac{1}{2}d_1 d_2$
Regular hexagon with side $s$	six equilateral triangles: $\frac{3\sqrt{3}}{2}s^2$

A parallelogram is a rectangle with one end sliced off and moved to the other end: same base, same perpendicular height, same area. That is why its area is $bh$ and not the product of its two side lengths. The slanted side is longer than the height, so multiplying the two sides overstates the area. A trapezoid's area is its average base times its height. A rhombus is four right triangles whose legs are half-diagonals, so its area is half the product of the diagonals.

When the height is not given, look for a right triangle that contains it. Dropping perpendiculars from the ends of a trapezoid's shorter base splits off two right triangles at the ends. In a parallelogram with a $30^\circ$ angle, the slanted side is the hypotenuse of a 30-60-90 triangle, so the height is half the slanted side; with a $45^\circ$ angle it is the slanted side divided by $\sqrt{2}$.

Perimeter is not area

Perimeter adds the lengths round the outside; area counts the squares inside. Fixing one does not fix the other. Rectangles with perimeter 20 include $9 \times 1$ (area 9) and $5 \times 5$ (area 25). Among rectangles with a given perimeter, the square has the largest area. In a quantitative comparison that gives only a perimeter, an area is usually not determined.

Composite figures

A shaded region is usually one known shape minus another, or several known shapes side by side. Describe it to yourself in words first: "a 10 by 6 rectangle with a semicircle of diameter 6 removed from its right end" gives an area of $60 - \frac{1}{2}\pi(3)^2 = 60 - 4.5\pi$. For a perimeter, trace the actual boundary: a cut-out removes some edges and adds new ones, so perimeters are not subtracted the way areas are.