2.5 Area and perimeter

Achievable AMC

2. Geometry

Area and perimeter

5 min read

Font

Discuss

Feedback

This chapter applies to all AMC 8/10/12 test takers.

This chapter focuses on formulas for the area and perimeter of common polygons. Circles and triangles have their own dedicated chapters, so the emphasis here is on quadrilaterals and other polygons with four or more sides. For each shape, you’ll see the standard area and perimeter formulas, along with a few useful extensions for less common cases.

Quadrilaterals

Here is a table of area and perimeter formulas for common quadrilaterals. Notice a common pattern: most area formulas (except for the kite) are essentially “base (or average of bases) times height.” Also notice that the second area formula for a rhombus matches the kite area formula. That’s because every rhombus is a type of kite, but not every kite is a rhombus.

Shape	Area	Perimeter
Square	$s^{2}$	$4 s$
Rectangle	$lw$	$2 (l + w)$
Parallelogram	$bh$	$2 (a + b)$
Rhombus	$bh$ or $\frac{1}{2} (d_{1} d_{2})$	$4 s$
Trapezoid	$\frac{1}{2} (b_{1} + b_{2}) h$	Sum of all sides
Kite	$\frac{1}{2} (d_{1} d_{2})$	$2 a + 2 b$

In harder AMC problems, you may run into irregular quadrilaterals. Brahmagupta’s formula finds the area of any cyclic quadrilateral (a quadrilateral whose vertices lie on a circle). In this formula, $a, b, c, d$ are the side lengths, and $s$ is the semiperimeter (half the sum of the side lengths).

$A = (s - a) (s - b) (s - c) (s - d)$

$where s = \frac{a + b + c + d}{2}$

Example: The question below is from 2022 AMC 10A

Quadrilateral $A BC D$ with side lengths $A B = 7, BC = 24, C D = 20, D A = 15$ is inscribed in a circle. The area interior to the circle but exterior to the quadrilateral can be written in the form $\frac{aπ - b}{c},$ where $a, b,$ and $c$ are positive integers such that $a$ and $c$ have no common prime factor. What is $a + b + c ?$

A. $260$
B. $855$
C. $1235$
D. $1565$
E. $1997$

(spoiler)

Answer: D. $1565$

The area of the circle is $156.25 π$ , or $625 π /4$ . If you’re not sure where this comes from, use the fact that $A C$ is the hypotenuse of right triangle $A C D$ , so $A C$ is the diameter of the circle.

Next, apply Brahmagupta’s formula to the cyclic quadrilateral to find that its area is $234$ .

The desired region is the area inside the circle but outside the quadrilateral, so subtract:

circle area minus quadrilateral area

This simplifies to $(625 π - 936) /4$ .

Now match this to $\frac{aπ - b}{c}$ , so $a = 625$ , $b = 936$ , and $c = 4$ . Therefore,

$a + b + c = 625 + 936 + 4 = 1565$ .

Regular polygons

Below are a few key area formulas for commonly used regular polygons. A regular hexagon shows up often in AMC problems (after the equilateral triangle and the square). Its area is $(\frac{3 3}{2}) s^{2}$ . One way to remember why this works is to split a regular hexagon into 6 equilateral triangles. Each triangle has area $(\frac{3}{4}) s^{2}$ , so the total area is $6 \cdot (\frac{3}{4}) s^{2} = (\frac{3 3}{2}) s^{2}$ .

The regular octagon also has a useful area formula:

$A = 2 (1 + 2) s^{2}$ Because all sides are equal, the perimeter of any regular polygon is the side length multiplied by the number of sides.

As the number of sides in a regular polygon increases, the polygon (which is cyclic) more closely approximates the circle it is inscribed in.

Example: The question below is from 2022 AMC 10A

A bowl is formed by attaching four regular hexagons of side $1$ to a square of side $1$ . The edges of the adjacent hexagons coincide, as shown in the figure. What is the area of the octagon obtained by joining the top eight vertices of the four hexagons, situated on the rim of the bowl?

A. $6$
B. $7$
C. $5 + 22$
D. $8$
E. $9$

(spoiler)

Answer: B. $7$

The key observation is that this octagon is not regular.

The sides formed by the hexagons have length $1$ .
The sides formed by the corner triangles have length $2$ .

Now place the octagon inside a $3$ by $3$ square. The difference between the square and the octagon is made up of four congruent isosceles right triangles, one in each corner.

Each triangle has area $1/2$ , so the total area removed is $4 \cdot (1/2) = 2$ . Therefore, the octagon’s area is:

$9 - 2 = 7$

Common themes

Look for hidden, simpler shapes inside more complex shapes.
Do not forget that the height of a shape is the perpendicular distance to the top from the base, which is not always equivalent to a side length.
Never assume a shape is regular or drawn to scale unless it is directly stated or it is provable with information other than just its appearance.

Quadrilaterals

Standard area formulas:
- Square: $s^{2}$
- Rectangle: $lw$
- Parallelogram: $bh$
- Rhombus: $bh$ or $\frac{1}{2} (d_{1} d_{2})$
- Trapezoid: $\frac{1}{2} (b_{1} + b_{2}) h$
- Kite: $\frac{1}{2} (d_{1} d_{2})$
Perimeter formulas:
- Square: $4 s$
- Rectangle: $2 (l + w)$
- Parallelogram: $2 (a + b)$
- Rhombus: $4 s$
- Trapezoid: sum of all sides
- Kite: $2 a + 2 b$
Brahmagupta’s formula for cyclic quadrilaterals:
- $A = (s - a) (s - b) (s - c) (s - d)$
- $s = \frac{a + b + c + d}{2}$

Regular polygons

Area formulas:
- Regular hexagon: $A = (\frac{3 3}{2}) s^{2}$
- Regular octagon: $A = 2 (1 + 2) s^{2}$
Perimeter: number of sides $\times$ side length
As sides increase, regular polygon approaches a circle

Common themes

Identify hidden or simpler shapes within complex figures
Height is always the perpendicular distance from base to top, not necessarily a side
Do not assume regularity or scale from appearance alone; rely on given or provable information

Next | 2.6 Volume and surface area

Area and perimeter

This chapter applies to all AMC 8/10/12 test takers.

Quadrilaterals

Shape	Area	Perimeter
Square	$s^{2}$	$4 s$
Rectangle	$lw$	$2 (l + w)$
Parallelogram	$bh$	$2 (a + b)$
Rhombus	$bh$ or $\frac{1}{2} (d_{1} d_{2})$	$4 s$
Trapezoid	$\frac{1}{2} (b_{1} + b_{2}) h$	Sum of all sides
Kite	$\frac{1}{2} (d_{1} d_{2})$	$2 a + 2 b$

$A = (s - a) (s - b) (s - c) (s - d)$

$where s = \frac{a + b + c + d}{2}$

Example: The question below is from 2022 AMC 10A

Quadrilateral $A BC D$ with side lengths $A B = 7, BC = 24, C D = 20, D A = 15$ is inscribed in a circle. The area interior to the circle but exterior to the quadrilateral can be written in the form $\frac{aπ - b}{c},$ where $a, b,$ and $c$ are positive integers such that $a$ and $c$ have no common prime factor. What is $a + b + c ?$

A. $260$
B. $855$
C. $1235$
D. $1565$
E. $1997$

(spoiler)

Answer: D. $1565$

Next, apply Brahmagupta’s formula to the cyclic quadrilateral to find that its area is $234$ .

The desired region is the area inside the circle but outside the quadrilateral, so subtract:

circle area minus quadrilateral area

This simplifies to $(625 π - 936) /4$ .

Now match this to $\frac{aπ - b}{c}$ , so $a = 625$ , $b = 936$ , and $c = 4$ . Therefore,

$a + b + c = 625 + 936 + 4 = 1565$ .

Regular polygons

The regular octagon also has a useful area formula:

$A = 2 (1 + 2) s^{2}$ Because all sides are equal, the perimeter of any regular polygon is the side length multiplied by the number of sides.

As the number of sides in a regular polygon increases, the polygon (which is cyclic) more closely approximates the circle it is inscribed in.

Example: The question below is from 2022 AMC 10A

A bowl is formed by attaching four regular hexagons of side $1$ to a square of side $1$ . The edges of the adjacent hexagons coincide, as shown in the figure. What is the area of the octagon obtained by joining the top eight vertices of the four hexagons, situated on the rim of the bowl?

A. $6$
B. $7$
C. $5 + 22$
D. $8$
E. $9$

(spoiler)

Answer: B. $7$

The key observation is that this octagon is not regular.

The sides formed by the hexagons have length $1$ .
The sides formed by the corner triangles have length $2$ .

Now place the octagon inside a $3$ by $3$ square. The difference between the square and the octagon is made up of four congruent isosceles right triangles, one in each corner.

Each triangle has area $1/2$ , so the total area removed is $4 \cdot (1/2) = 2$ . Therefore, the octagon’s area is:

$9 - 2 = 7$

Common themes

Look for hidden, simpler shapes inside more complex shapes.
Do not forget that the height of a shape is the perpendicular distance to the top from the base, which is not always equivalent to a side length.
Never assume a shape is regular or drawn to scale unless it is directly stated or it is provable with information other than just its appearance.

Achievable AMC

2. Geometry