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Area of a polygon from coordinates

The corners of a shape are given as coordinates, and some of its sides run slantwise across the grid. How much space does it cover?

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When you reach for it

A zoo map, a garden plan or a video game level drawn on a coordinate grid often gives a region only as a list of its corners. To find its area you need lengths, and coordinates hand you lengths in just two directions: straight across, from the first numbers, and straight up, from the second numbers. A slanted length needs the distance formula, which comes much later.

So the skill is to find an area using only lengths that run along grid lines. That is always possible, whatever the polygon and in whichever of the four quadrants its corners sit.

This lesson runs over two days. Day 1 handles a polygon with a side along a grid line, where a formula from Area of triangles or Area of parallelograms and trapezoids applies once its base and height are read off the coordinates. Day 2 handles a polygon whose sides slant, by boxing it inside a rectangle and taking away the corner pieces.

The procedure

Day 1: lengths along grid lines

Find the sides that run along grid lines. Two corners (vertices) with the same second number are joined by a horizontal side; two with the same first number, by a vertical side. Its length is how far apart the other two numbers are, found as in Horizontal and vertical distance: add the two absolute values when the ends lie on opposite sides of an axis, subtract the smaller absolute value from the larger when they lie on the same side.

A height runs along a grid line too. Take a triangle with a horizontal side as its base. Its height is the distance from the third vertex straight down (or up) to the line the base lies on, and that line is a grid line: every point on it has the base’s second number. So the height is the gap between the third vertex’s second number and the base’s second number. No first number enters, so the height is the same wherever the third vertex sits across, even beyond the end of the base. Then

A = \frac{1}{2}bh

and neither slanted side was needed. A vertical base works the same way turned on its side: the height is the gap between first numbers.

Parallel sides along grid lines. Two horizontal sides are parallel, so a quadrilateral with two of them is a trapezoid (or a parallelogram). Its height is the gap between the two second numbers, and the trapezoid formula \displaystyle A = \frac{1}{2}(b_1 + b_2)h takes over.

Every side along grid lines: cut along a grid line. Continue one side straight on until it meets the far side of the polygon. The cut runs along a grid line, so the pieces are rectangles with every side read off the coordinates. Adding their areas is allowed because the pieces do not overlap and together make the whole polygon; cutting and adding is the idea of Area of any polygon by cutting it into triangles and rectangles, and the coordinates only supply the lengths.

Day 2: slanted sides

When every side slants, no side can serve as a base with an easy height. Instead, surround the polygon with a rectangle whose sides run along grid lines.

Draw the box. Its left side goes through the smallest first number among the vertices, its right side through the largest, its bottom through the smallest second number and its top through the largest. Its width is the gap between the smallest and largest first numbers, its height the gap between the smallest and largest second numbers. The polygon sits inside and touches all four sides.

Find the pieces. The space between the polygon and the box falls into pieces. A slanted side that runs from one side of the box to the next cuts off a right triangle: the slanted side is its long side, and its two short sides lie along the box. The short side across is the gap between the first numbers at the two ends of the slanted side; the short side up is the gap between their second numbers. So each corner piece has area \displaystyle \frac{1}{2} \times \text{across} \times \text{up}, and still no slanted length is needed. A side lying along the box cuts off nothing.

Take the pieces away. The box is cut exactly into the polygon and the pieces, with no overlap and nothing left over, so

\text{area of polygon} = \text{area of box} - \text{area of the pieces}

When a vertex sits inside the box rather than on it, a piece may not be a corner triangle. It is still a shape whose lengths run along grid lines, often a triangle with one side on the box, and Day 1 finds its area (Example 7).

Or cut along a grid line. When two vertices share a coordinate, the segment joining them is horizontal or vertical. Cutting the polygon there leaves triangles with a base on that line, which is Day 1 again. Either method gives the same area, so one can check the other.

Worked examples

Examples 1 to 4 go with Day 1. Examples 5 to 7 go with Day 2.

Example 1

Day 1: a triangle with a horizontal base

Find the area of the triangle with vertices A(-4, -2), B(5, -2) and C(1, 4).

Solution

A and B share the second number -2, so AB is horizontal and makes a good base. A is 4 units left of the vertical axis and B is 5 units right of it, on opposite sides, so AB = 4 + 5 = 9 units.

The height is the vertical distance from C to the grid line through A and B, where every second number is -2. C is 4 units above the horizontal axis and that line is 2 units below it, so the height is 4 + 2 = 6 units.

−5 −4 −3 −2 −1 0 1 2 3 4 5 6 −3 −2 −1 0 1 2 3 4 5 A B C
Figure 1Example 1. The base runs along a grid line, and so does the height, so both come from the coordinates.
\text{area} = \frac{1}{2} \times 9 \times 6 = 27

The area is 27 square units. The sides AC and BC slant, and their lengths were never needed.

Example 2

Day 1: a paddock on the zoo map

On a zoo map each grid unit stands for 10 meters. The giraffe paddock is the hexagon with corners (-6, 5), (2, 5), (2, 1), (-1, 1), (-1, -3) and (-6, -3), joined in that order. Birgit needs its area in square meters to order grass seed.

Solution

Each corner shares a number with the next one, so every side runs along a grid line. Continue the side from (2, 1) to (-1, 1) straight on to (-6, 1). That cut lies on a grid line and leaves two rectangles.

−7 −6 −5 −4 −3 −2 −1 0 1 2 3 −4 −3 −2 −1 0 1 2 3 4 5 6
Figure 2Example 2. A cut along a grid line leaves two rectangles whose sides come from the coordinates.

The top rectangle runs across from -6 to 2, which is 6 + 2 = 8 units, and up from 1 to 5, which is 5 - 1 = 4 units, both ends above the axis. Its area is 8 \times 4 = 32 square units.

The bottom rectangle runs across from -6 to -1, both left of the axis, so 6 - 1 = 5 units, and up from -3 to 1, which is 3 + 1 = 4 units. Its area is 5 \times 4 = 20 square units.

The paddock covers 32 + 20 = 52 square units. Check by boxing instead: the box is 8 by 8, area 64, and the missing corner runs from -1 to 2 across (3 units) and from -3 to 1 up (4 units), area 12. And 64 - 12 = 52.

Each square unit is a square 10 meters by 10 meters, which is 100 square meters, so the paddock covers 52 \times 100 = 5{,}200 square meters.

Example 3

Day 1: the third vertex beyond the end of the base

The meerkat lawn on the same zoo map is the triangle P(-3, 4), Q(-3, -2), R(5, 7), measured in grid units. Find its area.

Solution

P and Q share the first number -3, so PQ is vertical: from 4 down to -2 crosses the horizontal axis, so PQ = 4 + 2 = 6 units. Take PQ as the base.

The height is the horizontal distance from R to the vertical grid line through P and Q, where every first number is -3. R is 5 units right of the vertical axis and that line is 3 units left of it, so the height is 5 + 3 = 8 units.

−4 −3 −2 −1 0 1 2 3 4 5 6 −3 −2 −1 0 1 2 3 4 5 6 7 8 P Q R
Figure 3Example 3. The height from R meets the line through P and Q beyond the end of the side.

R is higher than P, so the height meets the line through P and Q at (-3, 7), beyond the end of the base. That does not matter: as in Area of triangles, the height only has to be measured at right angles to the line the base lies on.

A = \frac{1}{2} \times 6 \times 8 = 24

The lawn covers 24 square units.

Example 4

Day 1: a trapezoid with decimal coordinates

Find the area of the quadrilateral with vertices (-4.5, 3), (2.5, 3), (4, -1) and (-5, -1).

Solution

The first two vertices share the second number 3, and the last two share -1, so the quadrilateral has two horizontal sides. Horizontal sides are parallel, so it is a trapezoid.

The top side runs from -4.5 to 2.5, across the axis: 4.5 + 2.5 = 7 units. The bottom side runs from -5 to 4: 5 + 4 = 9 units. The height is the gap between the two grid lines, from 3 down to -1: 3 + 1 = 4 units.

A = \frac{1}{2}(7 + 9) \times 4 = 32

The area is 32 square units. Decimals change nothing about how the gaps are found.

Example 5

Day 2: a lava pool in a game level

Mehmet is building a video game level on a grid where each square is one floor tile. A lava pool is the triangle A(-5, 1), B(3, 5), C(1, -4). How many tiles of area does it cover?

Solution

No two vertices share a first number or a second number, so every side slants. Box the triangle.

The first numbers are -5, 3 and 1, so the box runs across from -5 to 3: 5 + 3 = 8 units. The second numbers are 1, 5 and -4, so it runs up from -4 to 5: 4 + 5 = 9 units. The box has area 8 \times 9 = 72.

−6 −5 −4 −3 −2 −1 0 1 2 3 4 −5 −4 −3 −2 −1 0 1 2 3 4 5 6 A B C
Figure 4Example 5. The box, and the three corner pieces to take away from it.

B is the top right corner of the box, A is on its left side and C is on its bottom. Each side of the triangle cuts off one right triangle:

sideacrossuppiece
AB, from (-5, 1) to (3, 5)5 + 3 = 85 - 1 = 4\displaystyle \frac{1}{2} \times 8 \times 4 = 16
BC, from (3, 5) to (1, -4)3 - 1 = 25 + 4 = 9\displaystyle \frac{1}{2} \times 2 \times 9 = 9
CA, from (1, -4) to (-5, 1)1 + 5 = 64 + 1 = 5\displaystyle \frac{1}{2} \times 6 \times 5 = 15

The pieces add to 16 + 9 + 15 = 40, so the pool covers 72 - 40 = 32 square units: 32 tiles of area, though tiles along the slanted edges are only partly covered.

Example 6

Day 2: four slanted sides

Dorota designs a shield for the same game. It is the quadrilateral P(-4, 2), Q(1, 6), R(5, 0), S(-2, -3). Find its area.

Solution

The box runs across from -4 to 5, which is 4 + 5 = 9 units, and up from -3 to 6, which is 3 + 6 = 9 units, so its area is 81. Each vertex touches a different side of the box, so each side of the shield cuts off a corner triangle.

−5 −4 −3 −2 −1 0 1 2 3 4 5 6 −4 −3 −2 −1 0 1 2 3 4 5 6 7 P Q R S
Figure 5Example 6. Four slanted sides cut off four right triangles.
sideacrossuppiece
PQ4 + 1 = 56 - 2 = 410
QR5 - 1 = 46 - 0 = 612
RS5 + 2 = 70 + 3 = 310.5
SP4 - 2 = 22 + 3 = 55

The pieces add to 10 + 12 + 10.5 + 5 = 37.5, so the shield’s area is 81 - 37.5 = 43.5 square units. A half is no surprise: each piece is half a rectangle, and half of 7 \times 3 is not a whole number.

Example 7

Day 2: a dent in the hedge maze

Tavish is laying turf in the middle of a hedge maze. On his plan each grid unit stands for 2 meters, and the lawn is the pentagon A(-4, -3), B(-1, 4), C(5, 4), D(2, 0), E(5, -3). Find its area in square meters.

Solution

The box runs across from -4 to 5, which is 9 units, and up from -3 to 4, which is 7 units: area 63.

−5 −4 −3 −2 −1 0 1 2 3 4 5 6 −4 −3 −2 −1 0 1 2 3 4 5 A B C D E
Figure 6Example 7. The piece on the right is a triangle with a vertical side, not a corner triangle.

BC lies along the top of the box and EA along the bottom, so they cut off nothing. The side AB cuts off a corner triangle 4 - 1 = 3 units across and 3 + 4 = 7 units up, with area \displaystyle \frac{1}{2} \times 3 \times 7 = 10.5.

On the right, D sits inside the box, so the piece between the lawn and the box is the triangle CDE, not a corner triangle. Its side CE lies along the right of the box, vertical, from 4 down to -3: 4 + 3 = 7 units. Taking it as the base, the height is the horizontal distance from D to that side, from 2 across to 5: 5 - 2 = 3 units. This is Day 1’s move, and the piece has area \displaystyle \frac{1}{2} \times 7 \times 3 = 10.5.

The two pieces add to 10.5 + 10.5 = 21, so the lawn’s area is 63 - 21 = 42 square units.

Check by another cut. A, B, C and E make a trapezoid with horizontal sides BC = 1 + 5 = 6 and AE = 4 + 5 = 9, and height 7, so its area is \displaystyle \frac{1}{2}(6 + 9) \times 7 = 52.5. The lawn is that trapezoid with the triangle CDE taken out: 52.5 - 10.5 = 42.

Each square unit is 2 meters by 2 meters, or 4 square meters, so the lawn covers 42 \times 4 = 168 square meters.

Boundaries

The order of the vertices matters. A list of corners describes a polygon only when they are joined in the order given and no two sides cross. The same four points joined in another order can draw a different shape, or a figure whose sides cross, which is not a polygon at all.

Lengths and areas are never negative. A coordinate can be negative, because it says where a point is. A length is a gap between two numbers and is never negative, and neither is an area.

Only lengths along grid lines come off the coordinates. That is enough for the area, but not for the perimeter: a slanted side’s length needs the distance formula of Perimeter and area from coordinates.

A scale multiplies area twice. If each grid unit stands for k meters, each square unit is k by k, so multiply the area in square units by k \times k, never by k alone.

Where this goes next

In Geometry, Perimeter and area from coordinates finds slanted lengths with the distance formula, so a polygon’s perimeter comes from its coordinates too, and Classifying figures from coordinates uses those lengths to say which quadrilateral four points make. Later still, one formula, the shoelace formula, gives any polygon’s area straight from its coordinates.

Exercises

Day 1: a side along a grid line

Find the area of each triangle.

  1. (-6, -1), (2, -1), (-3, 5)
  2. (1, 2), (1, -6), (-4, -3)
  3. (-5, -4), (2, -4), (2, 3)
  4. (-2, 3), (4, 3), (7, -2)

Day 1: sides along grid lines

Find the area of each polygon, with its vertices joined in the order given.

  1. (-5, -4), (4, -4), (4, 0), (0, 0), (0, 3), (-5, 3)
  2. (-3, 4), (3, 4), (5, -1), (-6, -1)
  3. (-4, -3), (-4, 2), (3, 5), (3, 0)

Day 1: on the zoo map, and working backwards

  1. On the zoo map, where each grid unit is 10 meters, Gwenda marks the elephant yard as the triangle (-2.5, 1), (5.5, 1), (2, -4). Find its area in square meters.
  2. A triangle has vertices (-3, -2) and (5, -2), and its third vertex lies on the vertical grid line through 1, above the other two. Its area is 28 square units. Find the third vertex.

Day 2: box and subtract

Find the area of each polygon by boxing it.

  1. (-4, -1), (2, 4), (5, -3)
  2. (-3, 2), (4, 5), (1, -3)
  3. (-5, 1), (-1, 4), (6, 2), (0, -4)

Day 2: choosing a method

  1. Find the area of the quadrilateral (-5, 0), (1, 6), (4, 0), (0, -4) twice: once by cutting along the horizontal axis and once by boxing it.
  2. A platform in a game level is the parallelogram (-4, -2), (2, -4), (5, 1), (-1, 3). No side runs along a grid line. Find its area.
  3. An arrowhead has vertices (-4, -3), (0, 5), (4, -3), (0, 0), joined in that order. Find its area.

Day 2: on the map and in the game

  1. On a zoo map with 5 meters to a grid unit, the picnic lawn is the triangle (-3, 4), (5, 1), (0, -4). Find its area in square meters.
  2. A tilted square in a game level has vertices (-2, -3), (3, -1), (1, 4), (-4, 2). Find its area.

Reasoning

  1. Explain why a triangle with one horizontal side has an area you can find from its coordinates, even though its other two sides slant.
  2. A slanted side runs from the left side of a box to its top. Explain why the piece it cuts off is a right triangle, and why its two short sides can be read from the coordinates of the slanted side’s ends.
  3. Every vertex of a polygon is moved 4 units right and 3 units down. Explain why its area does not change.
  4. Joined in the order (-2, -2), (3, -2), (3, 2), (-2, 2), four points make a rectangle. Describe what is drawn when the same points are joined in the order (-2, -2), (3, 2), (3, -2), (-2, 2), find the area it encloses, and explain why a list of vertices needs its order.
Answers
  1. 24 square units. Base from -6 to 2: 6 + 2 = 8. Height from -1 up to 5: 1 + 5 = 6. \displaystyle \frac{1}{2} \times 8 \times 6 = 24.
  2. 20 square units. The vertical side from 2 down to -6 is 2 + 6 = 8. The height, from the line through 1 across to -4, is 1 + 4 = 5. \displaystyle \frac{1}{2} \times 8 \times 5 = 20.
  3. 24.5 square units. A right triangle: base from -5 to 2 is 7, height from -4 to 3 is 7. \displaystyle \frac{1}{2} \times 7 \times 7 = 24.5.
  4. 15 square units. Base from -2 to 4 is 6. Height from 3 down to -2 is 5; the third vertex sits beyond the end of the base, which changes nothing. \displaystyle \frac{1}{2} \times 6 \times 5 = 15.
  5. 51 square units. Cut along the horizontal axis: the bottom rectangle is 9 by 4, area 36, and the top one is 5 by 3, area 15. 36 + 15 = 51.
  6. 42.5 square units. A trapezoid with horizontal sides 3 + 3 = 6 and 6 + 5 = 11, and height 4 + 1 = 5. \displaystyle \frac{1}{2}(6 + 11) \times 5 = 42.5.
  7. 35 square units. A parallelogram with vertical sides of length 3 + 2 = 5, and the gap between the grid lines through -4 and 3 is 4 + 3 = 7. 5 \times 7 = 35.
  8. 2{,}000 square meters. Base from -2.5 to 5.5 is 8, height from 1 down to -4 is 5, so the yard is \displaystyle \frac{1}{2} \times 8 \times 5 = 20 square units, each 10 \times 10 = 100 square meters.
  9. (1, 5). The base is 3 + 5 = 8, so \displaystyle \frac{1}{2} \times 8 \times h = 28 gives h = 7. From the base’s line, where the second number is -2, 7 units up uses 2 to reach 0 and ends 5 above it.
  10. 28.5 square units. Box 9 by 7, area 63. Pieces: \displaystyle \frac{1}{2} \times 6 \times 5 = 15, \displaystyle \frac{1}{2} \times 3 \times 7 = 10.5, \displaystyle \frac{1}{2} \times 9 \times 2 = 9. 63 - 34.5 = 28.5.
  11. 23.5 square units. Box 7 by 8, area 56. Pieces: \displaystyle \frac{1}{2} \times 7 \times 3 = 10.5, \displaystyle \frac{1}{2} \times 3 \times 8 = 12, \displaystyle \frac{1}{2} \times 4 \times 5 = 10. 56 - 32.5 = 23.5.
  12. 44.5 square units. Box 11 by 8, area 88. Pieces: \displaystyle \frac{1}{2} \times 4 \times 3 = 6, \displaystyle \frac{1}{2} \times 7 \times 2 = 7, \displaystyle \frac{1}{2} \times 6 \times 6 = 18, \displaystyle \frac{1}{2} \times 5 \times 5 = 12.5. 88 - 43.5 = 44.5.
  13. 45 square units both ways. Cut: the segment from (-5, 0) to (4, 0) is 9 long, with triangles of height 6 and 4 on it: 27 + 18 = 45. Box: 9 by 10, area 90, pieces 18 + 9 + 8 + 10 = 45, and 90 - 45 = 45.
  14. 36 square units. Box 9 by 7, area 63. Pieces: \displaystyle \frac{1}{2} \times 6 \times 2 = 6, \displaystyle \frac{1}{2} \times 3 \times 5 = 7.5, 6 and 7.5 again. 63 - 27 = 36.
  15. 20 square units. Box 8 by 8, area 64. The left and right corner triangles are each \displaystyle \frac{1}{2} \times 4 \times 8 = 16. The piece at the bottom is a triangle with base 8 on the box and height 3, area 12. 64 - 44 = 20. Or cut along the vertical axis: two triangles with base 5 and height 4, 10 + 10 = 20.
  16. 687.5 square meters. Box 8 by 8, area 64. Pieces: \displaystyle \frac{1}{2} \times 8 \times 3 = 12, \displaystyle \frac{1}{2} \times 5 \times 5 = 12.5, \displaystyle \frac{1}{2} \times 3 \times 8 = 12. The lawn is 64 - 36.5 = 27.5 square units, each 5 \times 5 = 25 square meters.
  17. 29 square units. Box 7 by 7, area 49. The four corner triangles are each \displaystyle \frac{1}{2} \times 5 \times 2 = 5. 49 - 20 = 29.
  18. The horizontal side is a base whose length is a gap between first numbers. The height is the vertical distance from the third vertex to the grid line the base lies on, which is a gap between second numbers. Base and height are all \displaystyle \frac{1}{2}bh needs; the slanted sides are never used.
  19. The box’s left side is vertical and its top is horizontal, so they meet at a right angle at the box’s corner. The piece has that corner and the two ends of the slanted side as its vertices, so it is a right triangle. Its short sides lie along the box: the horizontal one runs between the first numbers of the two ends and the vertical one between their second numbers.
  20. Every first number goes up by 4 and every second number goes down by 3, so every gap between two first numbers, and every gap between two second numbers, stays the same. The box and every piece keep their sizes, so the area found from them is unchanged. The polygon has only slid.
  21. 10 square units. The new order draws the rectangle’s two diagonals and its two vertical sides: a bow tie, two triangles meeting where the diagonals cross, at the center (0.5, 0), halfway from -2 to 3. Each triangle has a vertical side of length 2 + 2 = 4 and reaches 2.5 units across to the center, so each is \displaystyle \frac{1}{2} \times 4 \times 2.5 = 5, and 5 + 5 = 10. The rectangle was 5 \times 4 = 20. Same corners, different order, different figure and area, so the order is part of the polygon.
kg1128 · updated 2026-10-13 · CC-BY-SA-4.0