Quadrilaterals & Polygon Areas

What you need to know

• Area formulas. Trapezoid: $\frac{(b_1+b_2)h}{2}$, where $b_1$, $b_2$ are the parallel sides and $h$ is the distance between them. Parallelogram: $bh$. Rhombus or kite: $\frac{d_1 d_2}{2}$, half the product of the diagonals. Example: parallel sides $5$ and $9$ with height $4$ give $28$; diagonals $6$ and $8$ give a rhombus of area $24$.
• Any quadrilateral. If the diagonals $d_1$ and $d_2$ cross at angle $\theta$, the area is $\frac12 d_1 d_2\sin\theta$. Example: diagonals $6$ and $8$ crossing at $30^\circ$ give $\frac12\cdot6\cdot8\cdot\frac12 = 12$.
• Shoelace Theorem. For corners $(x_1,y_1), \dots, (x_n,y_n)$ listed in order around the shape, area $= \frac12\left|(x_1y_2 - x_2y_1) + (x_2y_3 - x_3y_2) + \dots + (x_ny_1 - x_1y_n)\right|$. Example: $(0,0)$, $(4,0)$, $(5,3)$, $(1,4)$ give $\frac12|0 + 12 + 17 + 0| = \frac{29}{2}$.
• Regular hexagon. With side $s$, it is six equilateral triangles, so its area is $\frac{3\sqrt3}{2}s^2$. Its long diagonal (through the center) is $2s$, and its short diagonal (skipping one corner) is $s\sqrt3$. Example: $s = 2$ gives area $6\sqrt3$ and diagonals $4$ and $2\sqrt3$.
• Trapezoid with its diagonals. If the diagonals cross at $P$, the two triangles on the parallel sides are similar, and the two triangles on the legs have equal area.
• Pick's Theorem. For a polygon whose corners are lattice points (both coordinates integers), area $= I + \frac{B}{2} - 1$, where $I$ counts the lattice points inside and $B$ those on the boundary. Example: the rectangle from $(0,0)$ to $(3,2)$ has $I = 2$ and $B = 10$, so its area is $2 + 5 - 1 = 6$.

How AMC 10 tests it

• Early: the area of a figure on a grid. Count squares, use Shoelace or Pick, or subtract right triangles from a rectangle around it.
• A trapezoid given by its bases and legs. Drop perpendiculars; often a $3$-$4$-$5$ triangle gives the height.
• A square with points on its sides, joined to make a quadrilateral inside; find the ratio of the areas.
• Regular hexagons: the area of a triangle on three of its corners, or one hexagon inside another. The answer contains $\sqrt3$.

Standard approaches

1. Cut along a diagonal or drop perpendiculars to make rectangles and right triangles.
2. Find the area of a big simple region and subtract the unwanted pieces.
3. For figures on a grid or given by coordinates, use Shoelace directly.
4. For a trapezoid with its diagonals, get all four small areas from one, using the ratio of the bases.

Worked example

Trapezoid $ABCD$ has $\overline{AB}\parallel\overline{CD}$, $AB = 4$, and $CD = 10$. The diagonals meet at $P$, and the area of triangle $ABP$ is $8$. What is the area of the trapezoid?

(A) $84$ (B) $90$ (C) $98$ (D) $100$ (E) $112$

Solution. Write $[XYZ]$ for the area of triangle $XYZ$.

Step 1. Since $AB\parallel CD$, triangles $ABP$ and $CDP$ have equal angles, so they are similar, with ratio $4:10 = 2:5$. Areas go by the square of the ratio: $[CDP] = 8\cdot\left(\frac52\right)^2 = 50$.

Step 2. The similarity also gives $PD:PB = 5:2$. Triangles $APD$ and $ABP$ share the corner $A$, and their bases $PD$ and $PB$ lie on line $BD$, so they have the same height. Their areas are in the ratio of their bases: $[APD] = \frac52\cdot 8 = 20$.

Step 3. The two side triangles have equal area, so $[BPC] = 20$ as well.

Step 4. Add the four pieces:
$$8 + 50 + 20 + 20 = 98.$$
The answer is $\boxed{\textbf{(C)}\ 98}$.

Pitfalls

• Using the slanted leg of a trapezoid as its height. Example: a leg of $5$ that runs $3$ sideways gives height $4$.
• Listing corners out of order in Shoelace. That measures a different, self-crossing shape.
• Using $\frac{d_1 d_2}{2}$ when the diagonals are not perpendicular.

Traps that recur

• Building the frame side from 3/7 alone (e.g. 1 + 6/7 or similar), which produces the decoy answers with 49 in the denominator. (2023 AMC 10A #24)
• Assuming R must be parallel to the 1x3 rectangle or to the unit square, which finds only one of the two configurations. (2021 AMC Fall 10B #25)
• Trying to find the trapezoid's dimensions or P's location; only the ratio is determined, and the height h cancels immediately in the differences of squares. (2022 AMC 10A #23)
• Assuming the pentagon is regular, or adding four equilateral triangles plus a 'middle region' without noticing that segments FC and GD cross. (2021 AMC 10B #20)