GeometryArea, Volume and ModelingG-GPE.B.7

Perimeter and Area on the Coordinate Plane

Use the distance formula to find perimeters and areas.

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When a figure's vertices are given as coordinates, the distance formula from Distance and Midpoint Formulas finds every side, so you can find its perimeter. For area, a rectangle drawn around the figure turns a slanted shape into pieces you already know.

Why it works

Sides along grid lines are easy: count. Slanted sides come from the Pythagorean theorem, so they are often square roots. Only like radicals combine: $\sqrt{8} + \sqrt{2} = 2\sqrt{2} + \sqrt{2} = 3\sqrt{2}$, but $\sqrt{5} + \sqrt{13}$ stays as it is. For area, the box method works because the box is a rectangle and the leftover corner pieces are right triangles (and rectangles) whose legs lie along the grid.
Drag the vertices. The readout gives each side length, the perimeter and the area.

Key idea

Perimeter: add the side lengths from the distance formula, simplifying radicals and combining like ones. Area: enclose the figure in a grid-aligned rectangle and subtract the right triangles and rectangles in the corners.

Worked example: A perimeter with radicals

Find the perimeter of the triangle with $A(1, 1)$, $B(7, 1)$ and $C(3, 5)$.
  1. $AB = 6$ (horizontal).
  2. $BC = \sqrt{(3 - 7)^2 + (5 - 1)^2}$ $= \sqrt{32} = 4\sqrt{2}$.
  3. $CA = \sqrt{(1 - 3)^2 + (1 - 5)^2}$ $= \sqrt{20} = 2\sqrt{5}$.

Answer: $6 + 4\sqrt{2} + 2\sqrt{5} \approx 16.13$

Worked example: The box method

Find the area of the triangle with $A(1, 1)$, $B(7, 2)$ and $C(3, 6)$.
  1. The box from $x = 1$ to $7$ and $y = 1$ to $6$ is $6 \times 5 = 30$.
  2. Corner triangles: $\frac{1}{2} \cdot 6 \cdot 1 = 3$ (below $\overline{AB}$), $\frac{1}{2} \cdot 4 \cdot 4 = 8$ (right of $\overline{BC}$), $\frac{1}{2} \cdot 2 \cdot 5 = 5$ (left of $\overline{CA}$).
  3. Area $= 30 - 3 - 8 - 5 = 14$.

Answer: $14$ square units

Worked example: A right triangle on the grid

Find the area and perimeter of the triangle with $P(-2, -1)$, $Q(4, -1)$ and $R(-2, 7)$.
  1. $\overline{PQ}$ is horizontal (length 6) and $\overline{PR}$ is vertical (length 8): a right angle at $P$.
  2. Area $= \frac{1}{2} \cdot 6 \cdot 8 = 24$. $QR = \sqrt{36 + 64} = 10$, so the perimeter is $24$.

Answer: Area $24$, perimeter $24$

Common mistake

Adding unlike radicals, like $\sqrt{32} + \sqrt{20} = \sqrt{52}$. Square roots do not add that way: $\sqrt{32} + \sqrt{20} \approx 5.66 + 4.47 = 10.13$, but $\sqrt{52} \approx 7.21$. Simplify each radical, then combine only the ones with the same number under the root.

Practice

Pass with 3 right answers, the last 2 in a row. Each right answer earns 2 XP.

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Builds on

Leads to

This is one of the top rungs of Geometry.

Sources and further reading

  • OpenStax Intermediate Algebra 2e, 11.1 Distance and Midpoint Formulas; Circles
  • Common Core State Standards for Mathematics, High School Geometry, G-GPE.B.7
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