Pythagorean Theorem: Perimeter, Area and Word Problems | Grade 8 Math | LevelKite

Perimeter is the total distance around a figure, and area is the number of square units that cover it. Both depend on lengths, and sometimes the length you need is missing. It might be a slanted side you can't count on a grid, or a height nobody measured.

You already know how to find a missing side of a right triangle. Now you'll use that skill as a tool. You'll spot a right triangle hiding inside a figure or a real situation and use it to get the missing length. Then you'll finish the perimeter, the area or the question being asked.

Key concept

Find the right triangle, then the length

A right triangle has one right angle. Its legs $a$ and $b$ are the two sides that form the right angle. Its hypotenuse $c$ is the side opposite the right angle. The Pythagorean theorem says every right triangle labeled this way has $a^2 + b^2 = c^2$ . So if you know two of its sides, you can find the third.

When a perimeter or area needs a length you don't have, find or draw a right triangle with that length as one of its sides. Solve for the length, then use it in the perimeter or area formula. A length is positive, so when you undo a square, keep only the nonnegative (zero or positive) square root.

a^2 + b^2 = c^2

Here is the strategy in a figure. A quadrilateral is a four-sided figure. Quadrilateral $ABCD$ sits on a coordinate grid where 1 unit stands for 1 cm. Its vertices are $A(1, 1)$ , $B(10, 1)$ , $C(7, 5)$ and $D(1, 5)$ .

The coordinates tell you which sides are horizontal or vertical. $A$ and $B$ share the y-coordinate 1, and $D$ and $C$ share the y-coordinate 5, so $AB$ and $DC$ are horizontal. $A$ and $D$ share the x-coordinate 1, so $AD$ is vertical. $B$ and $C$ share neither coordinate, so side $BC$ is slanted, and you need a right triangle to find its length. Point $E(7, 1)$ takes its x-coordinate from $C$ and its y-coordinate from $B$ , so $CE$ is vertical and $EB$ is horizontal. A vertical segment meets a horizontal one at a right angle, so triangle $CEB$ has its right angle at $E$ .

Quadrilateral

ABCD

with

A(1, 1)

,

B(10, 1)

,

C(7, 5)

and

D(1, 5)

; triangle

CEB

has its right angle at

E(7, 1)

, and

BC

is its hypotenuse.

Worked example

Perimeter with a slanted side

Find the perimeter of quadrilateral $ABCD$ with $A(1, 1)$ , $B(10, 1)$ , $C(7, 5)$ and $D(1, 5)$ , where 1 grid unit is 1 cm.

Start with the sides you can read directly. For the horizontal sides $AB$ and $DC$ , subtract the smaller x-coordinate from the larger. For the vertical side $AD$ , subtract the y-coordinates the same way.
$AB = 10 - 1 = 9, \quad DC = 7 - 1 = 6, \quad AD = 5 - 1 = 4$
Find the legs of right triangle $CEB$ the same way. $CE$ is vertical and $EB$ is horizontal.
$CE = 5 - 1 = 4, \quad EB = 10 - 7 = 3$
The right angle of triangle $CEB$ is at $E$ , and $BC$ is opposite it, so $BC$ is the hypotenuse. Substitute the legs into $a^2 + b^2 = c^2$ .
$BC^2 = 4^2 + 3^2 = 16 + 9 = 25$
$BC$ is a length, so it is the nonnegative square root of 25. As a check, 5 is longer than both legs, as the hypotenuse of a right triangle must be.
$BC = \sqrt{25} = 5$
Add all four sides to get the distance around the figure.
$AB + BC + DC + AD = 9 + 5 + 6 + 4 = 24$

Answer

The perimeter of quadrilateral $ABCD$ is $24$ cm.

Area brings a different missing length. For a triangle, $\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}$ . The base is the side you choose to measure from. The height is the length of a segment from the opposite vertex that meets the line through the base at a right angle.

That right angle is the key. The height is a leg of a right triangle, and the side of the original triangle across from that right angle is the hypotenuse. When you know enough lengths of that right triangle, you can solve for the height like any missing leg.

Worked example

Area from three side lengths

A triangle has two sides of length $10$ cm and a third side of length $12$ cm. Find its area.

Use the 12 cm side as the base. Draw the height $h$ , in centimeters, from the opposite vertex to the base. It meets the base at a right angle, so it splits the triangle into two right triangles. In each one, a 10 cm side is opposite the right angle, so that side is the hypotenuse.
Call the two parts of the base $x$ and $y$ . Each right triangle has legs $h$ and one part, and hypotenuse 10, so $x^2 + h^2 = 10^2$ and $y^2 + h^2 = 10^2$ . Subtracting $h^2$ shows that $x^2$ and $y^2$ both equal $10^2 - h^2$ . Two positive lengths with equal squares are equal, so the 12 cm base splits evenly.
$x = y = 12 \div 2 = 6$
Solve one right triangle for $h$ . Subtracting $6^2$ from both sides keeps the equation true.
$6^2 + h^2 = 10^2 \;\Rightarrow\; h^2 = 10^2 - 6^2 = 100 - 36 = 64$
$h$ is a length, so take the nonnegative square root.
$h = \sqrt{64} = 8$
Put the base and height into the area formula.
$\text{Area} = \tfrac{1}{2} \times 12 \times 8 = 6 \times 8 = 48$

Answer

The area of the triangle is $48$ square centimeters.

The triangle from the example above, with 1 unit = 1 cm: base

AB

runs from

(1, 1)

to

(13, 1)

, and height

CD

runs from

(7, 9)

down to

(7, 1)

, so

AD = DB = 6

and

CD = 8

.

In a word problem, the right triangle sits inside a real situation, so your first job is to find it. Look for a right angle that the facts guarantee, such as two fences that meet at a right angle or a wall that stands straight up from level ground. The side opposite that angle is the hypotenuse. Then decide whether you are missing a leg or the hypotenuse, and solve.

Many answers are not whole numbers. A square root such as $\sqrt{117}$ is an exact value, while a rounded decimal such as $10.8$ is only an approximation of it. Keep the exact value while you work. Round only when the situation asks for a number you can measure or buy.

The garden bed with 1 unit = 1 m: the sides along the fences run from

A(1, 1)

to

B(10, 1)

and from

A(1, 1)

to

C(1, 7)

, so they are 9 m and 6 m long.

Worked example

Edging for a garden bed

A garden bed is a right triangle in the corner of a yard, where two fences meet at a right angle. Its sides along the fences are $9$ m and $6$ m, as shown in the figure above. Edging is sold only in whole meters. What is the least number of whole meters of edging that will go all the way around the bed?

The fences meet at a right angle at point $A$ . Side $BC$ , across the yard, is opposite that angle, so it is the hypotenuse. Let $s$ be its length in meters, with $s > 0$ .
Use $a^2 + b^2 = c^2$ with the fence sides as the legs.
$s^2 = 9^2 + 6^2 = 81 + 36 = 117$
Because 117 is not the square of a whole number, keep the exact length $\sqrt{117}$ m. A larger positive number has a larger square root, and 117 lies between $10^2 = 100$ and $11^2 = 121$ .
$s = \sqrt{117}, \qquad 10 < \sqrt{117} < 11$
Add the three sides. Adding 15 to every part of the inequality keeps it true, so 25 m of edging falls short and 26 m is enough.
$9 + 6 + \sqrt{117} = 15 + \sqrt{117}, \qquad 25 < 15 + \sqrt{117} < 26$
For a decimal estimate, compare 117 with squares of tenths. It lies between $10.8^2$ and $10.9^2$ , so $\sqrt{117}$ is between 10.8 and 10.9, and the perimeter is between 25.8 m and 25.9 m.
$10.8^2 = 116.64 < 117 < 118.81 = 10.9^2$

Answer

The bed needs exactly $15 + \sqrt{117}$ m of edging, which is between $25.8$ and $25.9$ m, so the least number of whole meters to buy is $26$ .

Takeaway: when a perimeter, an area or a real situation needs a length you don't have, find or draw a right triangle that contains it. Make sure the right angle comes from coordinates or stated facts, not from how a picture looks. Use the side opposite that angle as the hypotenuse $c$ in $a^2 + b^2 = c^2$ . Keep the nonnegative square root, and keep exact values until the situation calls for rounding.

Next, you'll apply the Pythagorean theorem in three dimensions.