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Coordinate Area Formulas: Core practice

10 practice problems for this lesson. Work on paper, use hints when you need them, and check the answer or the full solution when you are ready.

Difficulty: Advanced (beyond the core course) Advanced. This problem set goes beyond core Geometry. You can skip it.

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Problem 1 of 10

Problem 1 Two numbers for one edge

A directed boundary edge goes from $P(-2,4)$ to $Q(4,7)$ . Find the signed trapezoid contribution relative to $y=1$ and the shoelace product difference in the original coordinates.
1. Hint 1
  The signed width is the starting horizontal coordinate minus the ending horizontal coordinate.
2. Hint 2
  Measure the trapezoid heights above the reference line. Use the original coordinates in the diagonal products.
Answer

The trapezoid contribution is $-27$ square units. The product difference is $-30$ .

Full solution

The edge moves right by 6 units, so its signed width is $-6$ .
The heights above the reference line are 3 and 6.

Multiply the signed width by the average height.
$\begin{aligned}E&=-6\cdot\tfrac{3+6}{2}\\E&=-27.\end{aligned}$

Use the original coordinates for the product difference.
$\begin{aligned}d_{PQ}&=(-2)\cdot7-4\cdot4\\d_{PQ}&=-30.\end{aligned}$

Answer

The trapezoid contribution is $-27$ square units. The product difference is $-30$ .

Key idea

The signed trapezoid and the product difference describe different quantities on one edge.
Problem 2 Read the corners from the grid

Each square on the grid has side length one unit. The positive horizontal direction is to the right and the positive vertical direction is up. The arrows mark the order $P,Q,R,S$ .

Read the four coordinate pairs. Then find the signed area and state the direction of the boundary.

Read each point from the axes, then follow the arrows.

Text description of this figure

P is three squares left and one square up from the origin. Q is two squares right on the horizontal axis. R is four squares right and four up. S is five squares up on the vertical axis. The arrows run P to Q to R to S to P.
1. Hint 1
  Use the intersection of the labeled axes as the origin.
2. Hint 2
  Include the edge from $S$ back to $P$ .
Answer

$P(-3,1)$ , $Q(2,0)$ , $R(4,4)$ , and $S(0,5)$ . The signed area is $41/2$ square units, and the boundary is counterclockwise.

Full solution

Reading from the grid gives $P(-3,1)$ , $Q(2,0)$ , $R(4,4)$ , and $S(0,5)$ .

The product differences in the indicated order are $-2$ , $8$ , $20$ , and $15$ .

Add the four differences and halve the result.
$\begin{aligned}2S&=-2+8+20+15\\S&=41/2.\end{aligned}$

The signed area is positive, so the boundary is counterclockwise.

Answer

$P(-3,1)$ , $Q(2,0)$ , $R(4,4)$ , and $S(0,5)$ . The signed area is $41/2$ square units, and the boundary is counterclockwise.

Key idea

A unit grid gives the coordinates, while the arrows give their order.
Problem 3 Corners between grid points

The boundary of a simple quadrilateral follows $P(-1/2,1)$ , $S(0,5/2)$ , $R(2,7/2)$ , and $Q(3,1)$ before closing.

Find its exact ordinary area and explain whether the lattice condition is needed for your calculation.

The same coordinate scale applies between grid points.

Text description of this figure

A horizontal lower edge connects P at negative one half comma one to Q at three comma one. The upper corners are S at zero comma five halves and R at two comma seven halves.
1. Hint 1
  Use the fractions directly in the diagonal products.
2. Hint 2
  Simplicity and boundary order are separate from whether the points have integer coordinates.
Answer

$45/8$ square units. The shoelace formula works for these noninteger coordinates.

Full solution

The product differences in the given order are $-5/4$ , $-5$ , $-17/2$ , and $7/2$ .

$\begin{aligned}2S&=-5/4-5-17/2+7/2\\2S&=-45/4\\S&=-45/8.\end{aligned}$

The ordinary area is the magnitude, $45/8$ square units.
The derivation used distances and products for a simple boundary and never required integer coordinates.

Answer

$45/8$ square units. The shoelace formula works for these noninteger coordinates.

Key idea

Ordered fractional coordinates work in the same formula as lattice coordinates.
Problem 4 One outline, two methods

A pentagon has vertices $A(-2,0)$ , $B(5,0)$ , $C(5,3)$ , $D(1,6)$ , and $E(-2,3)$ in boundary order.

Find the ordinary area by decomposition and by shoelace. Explain which method needs less computation here.

Look for a horizontal level shared by two vertices.

Text description of this figure

A and B lie on the horizontal axis. C and E lie three units above it on opposite sides of the shape. D is above both.
1. Hint 1
  For decomposition, draw the horizontal segment $CE$ .
2. Hint 2
  The two methods must measure the same region. Retain the closing edge in the shoelace sum.
Answer

Both methods give $63/2$ square units. The rectangle and triangle make decomposition shorter here.

Full solution

Decomposition.
The rectangle below $CE$ has width 7 and height 3.
The triangle above $CE$ has base 7 and perpendicular height 3.

$\begin{aligned}K&=7\cdot3+\tfrac12(7)(3)\\K&=63/2.\end{aligned}$

Shoelace.
The five product differences are $0$ , $15$ , $27$ , $15$ , and $6$ .

$\begin{aligned}2S&=0+15+27+15+6\\S&=63/2.\end{aligned}$

Taking the magnitude gives $63/2$ square units.
Here the decomposition uses just two familiar areas, while shoelace needs a difference for each of five edges.

Answer

Both methods give $63/2$ square units. The rectangle and triangle make decomposition shorter here.

Key idea

Choose decomposition when one cut produces pieces with easy bases and heights.
Problem 5 Pair both parts of a strip

A simple polygon follows $A(0,0)$ , $B(7,0)$ , $C(7,2)$ , $D(3,2)$ , $E(3,5)$ , $F(7,5)$ , $G(7,8)$ , and $H(0,8)$ . Close at $A$ .

Use the reference line $y=-2$ . Find the net signed trapezoid contributions within the left strip between $x=0$ and $x=3$ and the right strip between $x=3$ and $x=7$ . Find the signed area.

The inward corners share one horizontal coordinate.

Text description of this figure

A and B lie along the bottom. C is directly above B, D is left of C, E is directly above D, and F is right of E. G is above F, and H lies left of G. The eight sides form a rectangular notch from the right.
1. Hint 1
  The right strip contains a lower piece and an upper piece. Pair the top and bottom of each piece separately.
2. Hint 2
  Every horizontal boundary edge moving left adds a rectangle down to the reference line. Every horizontal edge moving right subtracts one.
Answer

The left strip contributes $24$ square units. The right strip contributes $20$ square units. The signed area is $44$ square units.

Full solution

Left strip.
The leftward upper edge has height 10 above the reference line.
The rightward lower edge has height 2.
The net contribution is $3(10-2)$ , or 24.

Right strip, lower piece.
The upper and lower heights are 4 and 2.
The net contribution is $4(4-2)$ , or 8.

Right strip, upper piece.
The upper and lower heights are 10 and 7.
The net contribution is $4(10-7)$ , or 12.

Right strip total.
Add 8 and 12 to get 20.

Signed area.
The left and right strips account for all the directed contributions.
The vertical edges have zero width.
$\begin{aligned}S&=24+20\\S&=44.\end{aligned}$

Answer

The left strip contributes $24$ square units. The right strip contributes $20$ square units. The signed area is $44$ square units.

Key idea

Separate the inside pieces of a concave strip before pairing their upper and lower edges.
Problem 6 Lower the reference line

A simple quadrilateral follows $A(1,1)$ , $B(5,3)$ , $C(4,6)$ , and $D(-2,4)$ . The reference line drops from $y=0$ to $y=-3$ .

Find the change in the signed trapezoid contribution for each directed edge. Explain why dropping a reference line below any simple polygon leaves the sum unchanged.
1. Hint 1
  The average height for every edge increases by 3.
2. Hint 2
  Add the signed widths around the walk and track the horizontal position where it closes.
Answer

The changes for $AB$ , $BC$ , $CD$ , and $DA$ in that direction are $-12$ , $3$ , $18$ , and $-9$ square units. The total change is zero for any closed boundary.

Full solution

The signed widths from $A$ around the boundary are $-4$ , $1$ , $6$ , and $-3$ .

Each average height increases by 3, so multiplying each signed width by 3 gives $-12$ , $3$ , $18$ , and $-9$ .
These changes sum to zero.

For any simple polygon, a downward change in the reference line increases every average height by the same amount.
The change in each contribution is that amount times its signed width.

The signed widths add to zero because each horizontal coordinate appears once with a plus sign and once with a minus sign.
The total contribution therefore stays the same.

Answer

The changes for $AB$ , $BC$ , $CD$ , and $DA$ in that direction are $-12$ , $3$ , $18$ , and $-9$ square units. The total change is zero for any closed boundary.

Key idea

Moving the reference line adds a multiple of the total signed width, which is zero.
Problem 7 Orient both pieces

The order $A,B,C,D$ follows a simple quadrilateral counterclockwise. Its interior diagonal $AC$ divides the region into two triangles.

Triangle $ABC$ has signed area $15$ in that order. Triangle $ADC$ has signed area $-10$ in that order.

Find the signed area of $ABCD$ . Name the directed copies of the diagonal that cancel and explain why.

Each triangle uses the shared diagonal.

Text description of this figure

A is left of C. B lies below the segment AC and D lies above it. A dashed diagonal joins A to C within the quadrilateral.
1. Hint 1
  Choose the same orientation for both triangle boundaries.
2. Hint 2
  List the closing edge of the reoriented triangle, along with its first edge.
Answer

The signed area is $25$ square units. The directed diagonals $C\to A$ and $A\to C$ cancel.

Full solution

Keep the order $A,B,C$ for the first triangle.
Reverse $A,D,C$ to $A,C,D$ for the second, so its signed area becomes positive 10.

The first triangle closes with $C\to A$ .
The second opens with $A\to C$ .
Their shoelace products swap order, so the two differences add to zero.

The remaining edges follow $A,B,C,D$ and return to $A$ .
Its signed area is $15+10$ , or 25 square units.

Answer

The signed area is $25$ square units. The directed diagonals $C\to A$ and $A\to C$ cancel.

Key idea

When neighboring pieces share a diagonal, give both the same orientation to cancel the opposite copies of that edge.
Problem 8 Build a whole family of lattice triangles

Fix $P(-2,2)$ and $Q(11,2)$ . Choose a lattice point $R$ so that the triangle in the order $P,Q,R$ has signed area $-13/2$ .

Describe every possible location of $R$ . Give one example.
1. Hint 1
  Write $R=(u,v)$ and look for terms that cancel in the sum.
2. Hint 2
  Use the sign to determine whether $R$ is above or below the horizontal base.
Answer

Every point $(u,1)$ with integer $u$ works. For example, $R(0,1)$ works.

Full solution

Let $R=(u,v)$ .
The three product differences are $-26$ , $11v-2u$ , and $2u+2v$ .

Add the differences.
The terms involving $u$ cancel.
$\begin{aligned}2S&=-26+11v+2v\\2S&=13v-26\\-13&=13v-26\\v&=1.\end{aligned}$

The horizontal coordinate disappeared from the sum.
Every integer $u$ at height 1 gives a lattice point off the base and gives the required signed area.
The point $R(0,1)$ is one such choice.

Answer

Every point $(u,1)$ with integer $u$ works. For example, $R(0,1)$ works.

Key idea

With a fixed horizontal base, its directed height controls signed area.
Problem 9 Insert a vertex on a vertical edge

A simple polygon has a vertical boundary edge from $P(a,b)$ to $R(a,d)$ with $b<d$ . Insert $Q(a,c)$ between them, where $b<c<d$ .

Show that the original product difference is the sum of the two new ones. Explain why the same ordinary area follows even if some of the coordinates are negative.
1. Hint 1
  The two endpoints of each vertical edge have the same horizontal coordinate.
2. Hint 2
  Keep the middle height $c$ visible until the two new terms are added.
Answer

The new differences add to $a(d-b)$ , the original difference. Signed and ordinary area are unchanged.

Full solution

The original product difference is $ad-ab$ , or $a(d-b)$ .

The difference for $P\to Q$ is $ac-ab$ .
The difference for $Q\to R$ is $ad-ac$ .

When added, the terms $ac$ and $-ac$ cancel.
The remaining expression is $ad-ab$ , the original difference.

The other edges stay the same, and the split keeps the same simple boundary.
The signed sum is unchanged, so its magnitude is unchanged.
Cancellation works for all real values of $a$ , $b$ , $c$ , and $d$ satisfying the stated order.

Answer

The new differences add to $a(d-b)$ , the original difference. Signed and ordinary area are unchanged.

Key idea

A vertex inserted on a straight vertical edge replaces one height change with two consecutive changes.
Problem 10 Which lists describe the same boundary?

The simple pentagon with vertices $A(0,0)$ , $B(7,0)$ , $C(7,3)$ , $D(3,7)$ , and $E(0,3)$ has signed area $35$ in order $A,B,C,D,E$ . Close each list below at its first vertex.

List I. $D,E,A,B,C$ .

List II. $E,D,C,B,A$ .

List III. $A,C,D,E,B$ .

For each list that traces the same boundary, give its signed and ordinary areas. For any other list, explain how its segments differ and check for a crossing.

Explain why every list that uses each original boundary edge once must be a cyclic shift of the original order or of its reversal.

The solid sides show the original boundary.

Text description of this figure

The original boundary joins A, B, C, D, and E. A and B lie on the bottom, C and E share a horizontal level, and D is above them.
1. Hint 1
  Check which edge comes immediately after the new starting vertex. A changed start keeps every directed edge.
2. Hint 2
  The four vertices $A,B,C,E$ are the corners of a rectangle.
Answer

List I has signed and ordinary area $35$ . List II has signed area $-35$ and ordinary area $35$ . List III crosses at $(7/2,3/2)$ . Along the original boundary, the starting vertex and first direction force the rest of the list.

Full solution

List I.
Starting at $D$ keeps the same directed edges.
The signed and ordinary areas are both 35 square units.

List II.
Starting at $E$ and walking backward reverses every edge.
The signed area is $-35$ square units, and the ordinary area is 35 square units.

List III.
The new edges $AC$ and $EB$ are diagonals of rectangle $ABCE$ .
They cross at their common midpoint $(7/2,3/2)$ .
This walk cannot serve as the simple boundary of the pentagon.

Any list using each original boundary edge once has a choice of starting vertex and of which of its two neighbors comes first.
At every later vertex, the arrival edge has already been used, so the other boundary edge must come next.

This forces a complete walk around the cycle in the chosen direction.
Those two choices give exactly the cyclic shifts of the original order or its reversal.

The two slanted segments cross inside rectangle ABCE.

Answer

List I has signed and ordinary area $35$ . List II has signed area $-35$ and ordinary area $35$ . List III crosses at $(7/2,3/2)$ . Along the original boundary, the starting vertex and first direction force the rest of the list.

Key idea

A new starting point and a reversal preserve the boundary, while an arbitrary rearrangement can change its segments.

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