Source: Homework Set 0, Problems 3–8
Tags: circle geometry, chord length, arc length, sector area, Pythagorean theorem, distance formula, parallelogram area, triangle area, shoelace formula, surveyor's formula, Gauss area formula, orientation, counterclockwise, ad minus bc
Difficulty: Introductory review | Prerequisites: Basic trigonometry, Euclidean geometry
Problems 3 through 8 of Homework 0 build from basic circle geometry up to a general formula for the area of any polygon in the plane. The progression is deliberate: first you recall how lengths and areas work on circles, then you prove the Pythagorean theorem and the distance formula, then you derive the area of a parallelogram in coordinates, and finally you reach the shoelace formula. This is the geometric reasoning toolkit that underlies surface integrals, cross products, and orientation in MATH 215. If you are shaky on trigonometry or on how area scales in coordinate geometry, this is the place to sort it out.
Circle geometry gives you the formulas for chord length, arc length, and sector area in terms of the radius R and the central angle θ. The Pythagorean theorem leads to the Euclidean distance formula, and the expression ad – bc gives the (signed) area of a parallelogram from its coordinate vertices. The shoelace formula generalises this to any polygon by summing cross-product-like terms around the vertices.
Chord
The straight line segment connecting two points on a circle. Think of it as the shortcut across the curve.
Arc length
The distance measured along the curved part of a circle between two points. In simple terms: how far you would walk if you followed the circle's edge.
Sector (wedge)
The region enclosed by two radii and the arc between them, like a slice of pie. In simple terms: the pizza-slice-shaped area bounded by two straight lines from the centre and the curve.
Pythagorean theorem
For a right triangle with legs a and b and hypotenuse c: a² + b² = c². In simple terms: the two shorter sides, squared and added, give the square of the longest side.
Distance formula
The distance between (a, b) and (c, d) in the plane is √((c – a)² + (d – b)²). A direct application of the Pythagorean theorem. Think of it as making a right triangle between the two points and measuring the hypotenuse.
Orientation (of vertices)
The direction in which you list the vertices of a polygon: counterclockwise (positive) or clockwise (negative). Orientation determines the sign of the area in the shoelace formula. In simple terms: which way you walk around the shape matters for whether the formula gives a positive or negative number.
Shoelace formula (surveyor's formula, Gauss' area formula)
A formula that computes the signed area of any simple polygon from the coordinates of its vertices. For a triangle with counterclockwise vertices (x₁, y₁), (x₂, y₂), (x₃, y₃): Area = ½(x₁y₂ + x₂y₃ + x₃y₁ – y₁x₂ – y₂x₃ – y₃x₁). In simple terms: multiply coordinates in a criss-cross pattern (like lacing a shoe), subtract the two products, and halve.
Consider a circle of radius R centred at the origin O. Let A and B be two points on the circle, and let θ = ∠AOB (in radians).
Drop a perpendicular from O to the midpoint of AB. This creates two right triangles, each with hypotenuse R and an angle of θ/2 at O.
Half the chord = R sin(θ/2)
Full chord length = 2R sin(θ/2)
This follows directly from the definition of sine in a right triangle.
The full circumference is 2πR. The arc from A to B subtends an angle θ out of the full 2π.
Arc length = Rθ
This is the definition of radian measure: one radian is the angle that subtends an arc equal in length to the radius.
The triangle has two sides of length R (the radii OA and OB) with an included angle θ.
Area = ½R² sin(θ)
This uses the standard formula: area = ½ ab sin(C), where a and b are two sides and C is the included angle.
The sector is a fraction θ/(2π) of the full disc (area πR²).
Sector area = ½R²θ
Note the similarity to the arc length formula: both are proportional to θ.
Start with a right triangle of legs a and b and hypotenuse c. Arrange four copies of this triangle inside a large square of side (a + b), with the hypotenuses forming an inner tilted square of side c.
Compute the area of the large square in two ways:
As a single square: (a + b)² = a² + 2ab + b²
As the inner square plus four triangles: c² + 4(½ab) = c² + 2ab
Set them equal: a² + 2ab + b² = c² + 2ab. Cancel 2ab from both sides: a² + b² = c².
The inner shape is a square because each of its angles is 90° (the two acute angles of the right triangle sum to 90°, and the straight angle along the side of the large square fills the remaining 90°).
Given two points (a, b) and (c, d), draw a horizontal segment from (a, b) to (c, b) and a vertical segment from (c, b) to (c, d). This forms a right triangle whose legs have lengths |c – a| and |d – b|.
By the Pythagorean theorem, the hypotenuse (the distance between the two points) is:
distance = √((c – a)² + (d – b)²)
The squaring removes the need for absolute values.
A parallelogram P with vertices at (0, 0), (a, b), (a + c, b + d), and (c, d) has area |ad – bc|.
Why ad – bc?
The geometric proof (Problem 6) works by cutting and rearranging P into a simpler parallelogram. When a, b, c, d > 0 and ad > bc, you can shear the parallelogram so that one pair of sides lies along an axis, making the base and height visible. The rearranged parallelogram has base d and height (ad – bc)/d, giving area ad – bc.
When ad < bc, the same construction gives bc – ad, so the general answer is |ad – bc|. The absolute value handles all sign combinations of a, b, c, d.
A triangle with one vertex at the origin and the other two at (a, b) and (c, d) is exactly half of the parallelogram above.
Area = ½|ad – bc|
The expression ad – bc (without absolute value) is the signed area. Its sign encodes orientation:
Positive when the vertices are listed counterclockwise.
Negative when listed clockwise.
This signed version is exactly what appears in the shoelace formula and, later in MATH 215, in cross products and determinants.
For a triangle with vertices listed counterclockwise as (x₁, y₁), (x₂, y₂), (x₃, y₃), the signed area is:
Area = ½(x₁y₂ + x₂y₃ + x₃y₁) – ½(y₁x₂ + y₂x₃ + y₃x₁)
How to prove it (Problem 7)
Translate the triangle so that one vertex is at the origin (say, subtract (x₁, y₁) from all three vertices). Then apply the parallelogram area result from Problem 6 (ad – bc) to the two translated edge vectors. The algebra, after expanding and simplifying, yields the shoelace formula.
The "shoelace" mnemonic
List the coordinates in a column and repeat the first pair at the bottom:
(x₁, y₁)
(x₂, y₂)
(x₃, y₃)
(x₁, y₁)
Multiply along the "down diagonals" (x₁y₂, x₂y₃, x₃y₁) and add them. Multiply along the "up diagonals" (y₁x₂, y₂x₃, y₃x₁) and add them. Subtract the second sum from the first, then take half.
What happens with clockwise orientation?
If the vertices are listed clockwise, the formula gives the negative of the area. Taking the absolute value recovers the correct positive area.
For a polygon with n vertices listed counterclockwise as (x₁, y₁), (x₂, y₂), ..., (xₙ, yₙ), the shoelace formula extends naturally:
Signed area = ½ ∑ from i = 1 to n of (xᵢ yᵢ₊₁ – xᵢ₊₁ yᵢ)
where indices wrap around (so xₙ₊₁ = x₁ and yₙ₊₁ = y₁).
The idea: decompose the polygon into triangles that share the origin (or any reference point) as a vertex. Each triangle contributes a signed area term of the form ½(xᵢ yᵢ₊₁ – xᵢ₊₁ yᵢ). Triangles that "overshoot" the polygon contribute negative terms that cancel exactly, leaving only the polygon's area.
\text{Chord length} = 2R\sin\!\left(\tfrac{\theta}{2}\right)\text{Arc length} = R\theta\text{Triangle area (two radii, included angle)} = \tfrac{1}{2}R^2\sin\theta\text{Sector area} = \tfrac{1}{2}R^2\theta\text{Pythagorean theorem: } a^2 + b^2 = c^2\text{Distance} = \sqrt{(c-a)^2 + (d-b)^2}\text{Parallelogram area} = |ad - bc|\text{Triangle area (one vertex at origin)} = \tfrac{1}{2}|ad - bc|\text{Shoelace: Area} = \tfrac{1}{2}\left|\sum_{i=1}^{n}(x_i y_{i+1} - x_{i+1} y_i)\right|The distance formula is the basis of GPS positioning and any algorithm that computes distances between points (route planning, nearest-neighbour search, collision detection in games).
The shoelace formula is used in surveying to compute land areas from GPS coordinates, in computer graphics to calculate polygon areas for rendering, and in geographic information systems (GIS) to measure the area of irregular regions.
Students often confuse arc length (Rθ) with chord length (2R sin(θ/2)). They are equal only when θ = 0. For small angles they are approximately equal, but for larger angles the arc is always longer than the chord.
When using the sector area formula ½R²θ, the angle must be in radians. Using degrees here is one of the most common errors.
Students sometimes write the parallelogram area as ad + bc instead of ad – bc. The subtraction is essential, and the sign encodes the geometry.
With the shoelace formula, students forget to wrap around: the last vertex must connect back to the first. Leaving out the (xₙy₁ – x₁yₙ) term gives the wrong answer.
⚠️ The expression ad – bc is the 2×2 determinant. It reappears as the cross-product magnitude in MATH 215 and is central to computing areas of parallelograms spanned by vectors in 3D.
⚠️ Orientation (counterclockwise vs clockwise) is a recurring theme. In multivariable calculus, the sign of a line integral or a surface integral depends on orientation. Building the intuition now saves confusion later.
⚠️ The circle geometry formulas (arc length = Rθ, sector area = ½R²θ) appear throughout polar coordinates, which MATH 215 uses heavily.
True or False: The arc length between two points on a circle of radius 5 with central angle π/3 is 5π/3.
Fill in the blank: The area of a sector with radius R and central angle θ is ______.
True or False: The distance between (1, 2) and (4, 6) is √(9 + 16) = 5.
Fill in the blank: A parallelogram with vertices (0,0), (3,1), (5,4), (2,3) has area |3 · __ – __ · 1| = |__ – __| = ______.
True or False: If you list the vertices of a triangle clockwise and apply the shoelace formula, you get a negative number.
Answers: 1. True (5 × π/3 = 5π/3). 2. ½R²θ. 3. True (√((4–1)² + (6–2)²) = √(9+16) = 5). 4. |3 · 3 – 2 · 1| = |9 – 2| = 7. 5. True.
Q: A circle has radius 10 and two points on it subtend a central angle of π/4. What is the chord length?
A: Chord = 2(10) sin(π/8) = 20 sin(π/8) ≈ 7.65.
Q: What is the area of the triangle with vertices (0, 0), (4, 1), and (2, 5)?
A: Using ½|ad – bc| with (a, b) = (4, 1) and (c, d) = (2, 5): ½|4 · 5 – 2 · 1| = ½|20 – 2| = 9.
Q: Apply the shoelace formula to the triangle with counterclockwise vertices (1, 0), (4, 0), (2, 3).
A: Down diagonals: 1 · 0 + 4 · 3 + 2 · 0 = 12. Up diagonals: 0 · 4 + 0 · 2 + 3 · 1 = 3. Signed area = ½(12 – 3) = 4.5.
Q: A surveyor measures a quadrilateral plot with vertices (0, 0), (6, 0), (7, 4), (1, 5), listed counterclockwise. What is the area?
A: Shoelace: down diagonals = 0·0 + 6·4 + 7·5 + 1·0 = 0 + 24 + 35 + 0 = 59. Up diagonals = 0·6 + 0·7 + 4·1 + 5·0 = 0 + 0 + 4 + 0 = 4. Area = ½|59 – 4| = 27.5.
Q: What happens to the shoelace formula result if you reverse the vertex order from counterclockwise to clockwise?
A: The result flips sign. Taking the absolute value gives the same area.
The expression ad – bc is the determinant of a 2×2 matrix. In MATH 215, this extends to 3×3 determinants and the cross product, which computes the area of a parallelogram in 3D spanned by two vectors.
Orientation becomes central when you study line integrals and surface integrals: the direction you traverse a curve or the choice of normal vector on a surface determines the sign of the integral, exactly as counterclockwise vs clockwise determines the sign of the shoelace formula here.
The circle geometry (arc length, sector area) is the foundation for polar coordinates and for computing areas and integrals in polar form, which MATH 215 covers extensively.
chord length, arc length, sector area, central angle, radian, Pythagorean theorem, distance formula, Euclidean distance, parallelogram area, triangle area, determinant, 2x2 determinant, ad minus bc, signed area, shoelace formula, surveyor's formula, Gauss area formula, polygon area, orientation, counterclockwise, clockwise, cross product, MATH 215, Calculus III, University of Michigan, plane geometry, coordinate geometry, polar coordinates