AREA & PERIMETER

Overview

Area and perimeter problems are among the most frequent geometry topics on the AMC 10. Most of these problems do not require exotic theorems — they reward students who can decompose a complicated figure into simpler pieces, identify cancellations, and reason carefully about what is being measured.

A composite region is built by adding or subtracting simpler shapes. For example, an L-shaped floor plan is a large rectangle minus a smaller rectangle. Always ask: can I get the answer by addition and subtraction of areas I already know? For perimeters, trace the boundary carefully — interior edges never appear on the boundary, and a circular arc replaces a straight edge when a semicircle is appended.

Shaded regions often involve a circle inside or outside a polygon. The key insight is that the shaded area equals (larger shape) $-$ (unshaded shape), so you compute two separate areas and subtract. The AMC loves setups where unexpected cancellations — like the Lune of Hippocrates — let you avoid computing $\pi$ at all.

Key facts

• Rectangle: $A = \ell w$, $P = 2(\ell + w)$.
• Triangle: $A = \tfrac{1}{2} b h$. For a right triangle with legs $a,b$ and hypotenuse $c$: $a^2+b^2=c^2$.
• Circle: $A = \pi r^2$, circumference $= 2\pi r$. A sector with central angle $\theta$ (radians) has area $\tfrac{1}{2}r^2\theta$.
• Trapezoid: $A = \tfrac{1}{2}(b_1+b_2)h$.
• Composite area: $A_{\text{composite}} = A_1 \pm A_2 \pm \cdots$ (add the parts you include, subtract holes/cutouts).
• Pick's theorem (lattice polygons): $A = I + \tfrac{B}{2} - 1$, where $I$ = number of interior lattice points and $B$ = number of boundary lattice points. Useful when a coordinate grid is given.
• Area-preserving dissection: cutting and rearranging a region without overlap preserves total area. Use this to equate areas that look different.
• Similar figures: if two similar figures have linear scale factor $k$, their areas are in ratio $k^2$.

Worked example 1

Problem. A 12 by 8 rectangle has a 3 by 4 rectangle removed from one corner. What is the area of the remaining region?

Solution. The large rectangle has area $12 \times 8 = 96$. The removed piece has area $3 \times 4 = 12$. The remaining area is $96 - 12 = \boxed{84}$.

This is a one-step composite: large region $-$ cutout. Note the perimeter of the remaining L-shaped figure includes the two new edges created by the cut (of lengths 3 and 4), replacing the single corner, so the perimeter is the same as the original rectangle's perimeter: $2(12+8) = 40$. (The 3 and 4 edges substitute for the corner, not adding new length.) Actually trace it: $12 + 8 + 3 + 4 + (12-3) + (8-4) = 12+8+3+4+9+4 = 40$ — yes, equal!

Worked example 2

Problem. A 6-by-6 square has its inscribed circle drawn (the circle just touches all four sides). Inside the square, the region outside the circle is shaded. What is the shaded area?

Solution. The inscribed circle has diameter equal to the side length 6, so radius $r = 3$.

$A_{\text{circle}} = \pi(3)^2 = 9\pi.$

$A_{\text{square}} = 36.$

$A_{\text{shaded}} = 36 - 9\pi.$

Numerically this is about $36 - 28.27 = 7.73$. The answer $36 - 9\pi$ is exact and in simplified form. This "subtract the circle" template appears repeatedly on the AMC.

Common traps

• Perimeter of a composite figure: students often add or subtract edge lengths the same way they add/subtract areas. Be careful — when a rectangle is cut from a corner, the new boundary has two new edges that replace the corner, so the total perimeter can stay the same, increase, or decrease depending on the shape.
• Diameter vs. radius: the most common arithmetic slip. Always re-read whether the problem gives the diameter or the radius before squaring.
• Pick's theorem prerequisite: the theorem applies only to simple polygons with vertices at lattice points. It does not apply to curved boundaries or non-lattice vertices.
• Area-preserving dissection confusion: moving a piece preserves area but not perimeter. Never equate perimeters of a figure and its rearrangement.
• Shaded region sign: when subtracting a circular sector from a triangle (or vice versa), decide at the start which region is larger and subtract the smaller from the larger to avoid negative areas.