We define a square with side lengths of 1 unit to have an area of 1 square unit. With this definition we can easily find that the area of a rectangle will be the product of its length and width, and we can then use this to establish area formulas for a number of other polygons:

A=lw

A=bh

A=\dfrac{1}{2}bh

A=\dfrac{1}{2}\left(b_1+b_2\right)h

Worked examples

Example 1

A \left(2,1\right), B \left(7,3\right), and C \left(7,-5\right) are the vertices of a triangle.

Determine the area of the triangle using A=\dfrac{1}{2}bh.

Approach

-1

-5

-4

-3

-2

-1

It can help to do a rough sketch of the triangle to determine which side is the base and how to calculate the height.

Solution

Using a rough sketch or looking at the coordinates, we can see that we can use \overline{BC} as the base and the line segment between A and \overline{BC} as the height.

The base is a vertical line segment, so its length can be calculated using the difference in the y-coordinates of B and C.

\displaystyle b	\displaystyle =	\displaystyle \left\vert 3--5\right\vert
	\displaystyle =	\displaystyle 8

The height is a horizontal line segment, so its length can be calculated using the difference in the x-coordinates of A and \overline{BC}.

\displaystyle h	\displaystyle =	\displaystyle \left\vert 2-7\right\vert
	\displaystyle =	\displaystyle 5

\displaystyle A_{\triangle ABC}	\displaystyle =	\displaystyle \dfrac{1}{2}bh	Formula for area of a triangle
	\displaystyle =	\displaystyle \dfrac{1}{2}\left(8\right)\left(5\right)	Substitute in the values for b and h
	\displaystyle =	\displaystyle 20	Simplify

The area of \triangle ABC is 20 units^2.

Example 2

Consider a quadrilateral with vertices A \left(-4,3\right), B \left(-2,-4\right), C \left(4,-4\right), and D \left(5,3\right).

Determine the perimeter of the quadrilateral, rounding your answer to two decimal places.

Approach

-4

-3

-2

-1

-5

-4

-3

-2

-1

It can be very helpful to make a sketch to see what type of quadrilateral we have.

From this sketch we have a trapezoid.

Solution

To find the perimeter, we want to find the length of each side and then find their sum by adding them all together.

From the sketch, the two parallel sides are horizontal line segments. This means that we can find their lengths by calculating the difference in the x-coordinates of their end points.

\displaystyle AD	\displaystyle =	\displaystyle \left\vert -4-5\right\vert
	\displaystyle =	\displaystyle 9

\displaystyle BC	\displaystyle =	\displaystyle \left\vert -2-4\right\vert
	\displaystyle =	\displaystyle 6

For other two sides we will need to use the distance formula or the Pythagorean theorem, substituting in the coordinates of the end points for each side.

\displaystyle AB	\displaystyle =	\displaystyle \sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}	State distance formula
	\displaystyle =	\displaystyle \sqrt{\left(-4--2\right)^2+\left(3--4\right)^2}	Substitute known values
	\displaystyle =	\displaystyle \sqrt{\left(2\right)^2+\left(7\right)^2}	Simplify terms in parentheses
	\displaystyle =	\displaystyle \sqrt{4+49}	Square terms
	\displaystyle =	\displaystyle \sqrt{53}	Simplify the sum

\displaystyle CD	\displaystyle =	\displaystyle \sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}	State distance formula
	\displaystyle =	\displaystyle \sqrt{\left(1\right)^2+\left(7\right)^2}	Substitute and simplify
	\displaystyle =	\displaystyle \sqrt{1+49}	Square terms
	\displaystyle =	\displaystyle \sqrt{50}	Simplify the sum

Now that we have all of the side lengths we can find their sum on the calculator and round to two decimal places.

\displaystyle P_{ABCD}	\displaystyle =	\displaystyle 9+6+\sqrt{53}+\sqrt{50}
	\displaystyle =	\displaystyle 29.35

The perimeter of ABCD is 29.35 units.

Reflection

It is generally best to wait until the end to round for the most precise answer.

Determine the area of the quadrilateral, rounding your answer to two decimal places.

Approach

Much of the work has been done in the previous part.

Solution

This is a trapezoid, so we will use the formula:

A_{ABCD}=\dfrac{1}{2}\left(a + b\right)h

From the previous part, we have:

In the calculation of AB and CD, we found the vertical distance as a part of the calculation (the difference between the y-coordinates).

\displaystyle h	\displaystyle =	\displaystyle \left\vert 3--4 \right\vert
	\displaystyle =	\displaystyle 7

Now we can substitute these values into the formula for the are of a trapezoid.

\displaystyle A_{ABCD}	\displaystyle =	\displaystyle \dfrac{1}{2}\left(a+b\right)h
	\displaystyle =	\displaystyle \dfrac{1}{2}\left(6+9\right)\left(7\right)
	\displaystyle =	\displaystyle \dfrac{1}{2}\left(15\right)\left(7\right)
	\displaystyle =	\displaystyle 52.5

The area of the quadrilateral is 52.5 units^2.

Reflection

Since the exact solution is already to one decimal place, we do not need to do any rounding.

Outcomes

M1.N.Q.A.1.A

Choose and interpret the scale and the origin in graphs and data displays.*

M1.G.GPE.A.1

Use coordinates to justify geometric relationships algebraically and to solve problems.

M1.MP1

Make sense of problems and persevere in solving them.

M1.MP2

Reason abstractly and quantitatively.

M1.MP3

Construct viable arguments and critique the reasoning of others.

M1.MP4

Model with mathematics.

M1.MP5

Use appropriate tools strategically.

M1.MP6

Attend to precision.

M1.MP7

Look for and make use of structure.

M1.MP8

Look for and express regularity in repeated reasoning.

10.03 Perimeter and area in the coordinate plane

Concept summary

Worked examples

Example 1

Approach

Solution

Example 2

Approach

Solution

Reflection

Approach

Solution

Reflection

Outcomes

M1.N.Q.A.1.A

M1.G.GPE.A.1

M1.MP1

M1.MP2

M1.MP3

M1.MP4

M1.MP5

M1.MP6

M1.MP7

M1.MP8

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