Perimeters, Word Problems, and Missing Vertices | Grade 6 Math | LevelKite

You already know how to find the length of a side that runs straight across or straight up and down the grid. You use the coordinates of its two ends. In this lesson you put those side lengths to work on a polygon, which is a closed shape with straight sides. Each corner of a polygon, where two sides meet, is a vertex (the plural is vertices).

You will use side lengths in three ways:

You add them to find the distance around a polygon.
You solve real-world problems about shapes drawn on a grid.
You work backward to find a vertex that is missing.

The perimeter of a polygon is the total distance around it. Picture a walk along its sides. You start at one vertex, walk every side in turn, and stop when you are back at the start. The distance you walked is the perimeter.

That is why you find a perimeter by adding the lengths of all the sides. If you leave out a side, your walk does not get back to the start.

In the coordinate plane, each vertex is a point with an $x$ -coordinate and a $y$ -coordinate. In every figure in this lesson, the horizontal axis is the $x$ -axis and the vertical axis is the $y$ -axis. You name a side by its two endpoints, the vertices at its ends. Side $AB$ joins vertex $A$ to vertex $B$ . In an equation, a side name such as $AB$ stands for the length of that side.

The coordinates tell you what kind of side you have:

A side is horizontal when its endpoints have the same $y$ -coordinate. Its length is the distance between the two $x$ -coordinates on the number line.
A side is vertical when its endpoints have the same $x$ -coordinate. Its length is the distance between the two $y$ -coordinates on the number line.
A side that is neither is slanted. You cannot get its length this way, so the problem must tell you how long it is.

You measure these lengths in grid units. In every figure in this lesson, one unit is the distance between two neighboring grid lines.

Polygon

ABCD

:

A(-3, 1)

is the lower-left corner,

B(4, 1)

the lower-right,

C(4, 5)

the upper-right and

D(-3, 5)

the upper-left.

Worked example

Perimeter from four vertices

The figure shows polygon $ABCD$ with vertices $A(-3, 1)$ , $B(4, 1)$ , $C(4, 5)$ and $D(-3, 5)$ . Its sides join the vertices in that order, and the last side joins $D$ back to $A$ . What is the perimeter of the polygon?

Sort the sides. $A$ and $B$ share the $y$ -coordinate $1$ , and $C$ and $D$ share the $y$ -coordinate $5$ , so $AB$ and $CD$ are horizontal. $B$ and $C$ share the $x$ -coordinate $4$ , and $D$ and $A$ share the $x$ -coordinate $-3$ , so $BC$ and $DA$ are vertical. Every side is horizontal or vertical, so the coordinates give every length.
Find $AB$ from its $x$ -coordinates, $-3$ and $4$ . They are on opposite sides of $0$ . It takes $3$ units to go from $-3$ to $0$ and $4$ more units to go from $0$ to $4$ , so the two distances add. In the figure, you can count the same $7$ units along the bottom side.
$AB = 3 + 4 = 7$
Find $BC$ from its $y$ -coordinates, $1$ and $5$ . Both are greater than $0$ , so the distance between them is the larger minus the smaller.
$BC = 5 - 1 = 4$
Find $CD$ and $DA$ the same way. $CD$ runs between the $x$ -coordinates $4$ and $-3$ , so it is as long as $AB$ . $DA$ runs between the $y$ -coordinates $5$ and $1$ , so it is as long as $BC$ .
$CD = 4 + 3 = 7, \quad DA = 5 - 1 = 4$
Add all four side lengths, because the perimeter is the whole distance around the polygon.
$7 + 4 + 7 + 4 = 22$

Answer

The perimeter of polygon $ABCD$ is $22$ units.

In a real-world problem, the grid is a map or a plan of something real, such as a park or a yard. Each grid unit stands for a real length, and the problem tells you what that length is. For example, one unit might stand for $10$ meters. This is the scale of the map.

You still walk around the polygon and add its sides. There are two extra jobs:

Decide what the question asks for. The distance once around a shape is its perimeter. So is the length of a fence that goes all the way around it.
Use the scale last. Find the perimeter in grid units first. Then change the units into real lengths.

If a side of the shape is slanted, look for its length in the problem.

The park:

A(-4, -2)

is the lower-left corner,

B(5, -2)

the lower-right,

C(2, 2)

the upper-right and

D(-4, 2)

the upper-left. Side

BC

is the slanted one.

Worked example

One lap around a park

The figure shows a park on a map. Each grid unit stands for $10$ meters. The corners of the park, in order, are $A(-4, -2)$ , $B(5, -2)$ , $C(2, 2)$ and $D(-4, 2)$ . The slanted side $BC$ is exactly $5$ units long. Maya jogs once around the edge of the park. How many meters does she jog?

Decide what to find. Once around the edge is the perimeter. $A$ and $B$ share the $y$ -coordinate $-2$ , and $C$ and $D$ share the $y$ -coordinate $2$ , so $AB$ and $CD$ are horizontal. $D$ and $A$ share the $x$ -coordinate $-4$ , so $DA$ is vertical. $B$ and $C$ share neither coordinate, so $BC$ is slanted and you use its given length of $5$ units.
Find $AB$ from its $x$ -coordinates, $-4$ and $5$ . They are on opposite sides of $0$ , so add their distances from $0$ . In the figure, the bottom side reaches $4$ units left of the $y$ -axis and $5$ units right of it.
$AB = 4 + 5 = 9$
Find $CD$ from its $x$ -coordinates, $2$ and $-4$ , and $DA$ from its $y$ -coordinates, $2$ and $-2$ . Each pair is on opposite sides of $0$ , so add the distances from $0$ again.
$CD = 2 + 4 = 6, \quad DA = 2 + 2 = 4$
Add all four side lengths to get the perimeter in grid units. Remember to include the given length of the slanted side $BC$ .
$9 + 5 + 6 + 4 = 24$
Use the scale last. Each unit stands for $10$ meters, so $24$ units stand for $24$ groups of $10$ meters.
$24 \times 10 = 240$

Answer

Maya jogs $240$ meters when she goes once around the park.

Key concept

Coordinates and sides work both ways

Along a horizontal or vertical side, one coordinate stays the same and the other changes. The distance between the two values of the changing coordinate is the length of the side.

From coordinates to lengths. Find the length of every side, then add them all to get the perimeter.
From sides to coordinates. If you know a side is horizontal or vertical, you know which coordinate its endpoints share. If you also know which way it goes and how long it is, you can find the vertex at its other end.

Both directions work only for horizontal and vertical sides. A slanted side needs a length that the problem states.

Now use the key idea in the second direction: from sides back to coordinates. Suppose you know what kind of polygon you have and where some of its vertices are, but one vertex is missing. The properties of the polygon are your clues.

A rectangle is a four-sided polygon with four right angles, which are square corners. Its opposite sides are parallel, which means they run in the same direction. They are also equal in length. So if one side of a rectangle is horizontal, the opposite side is horizontal too, and the other two sides are vertical. A square is a rectangle whose four sides are all equal, so this is true for a square as well.

In a rectangle with a horizontal side, that tells you what the missing vertex shares with its neighbors:

the same $y$ -coordinate as the vertex at the other end of its horizontal side,
the same $x$ -coordinate as the vertex at the other end of its vertical side.

Check from the coordinates that a known side really is horizontal or vertical. Do not judge it from how a picture looks.

Worked example

The fourth vertex of a rectangle

Rectangle $JKLM$ has vertices $J(-5, -3)$ , $K(2, -3)$ and $L(2, 1)$ . The vertices are named in order around the rectangle. What are the coordinates of the fourth vertex, $M$ ?

Name what you need. Let $M = (x, y)$ , where $x$ and $y$ are the coordinates to find. The vertices are named in order, so $JK$ and $KL$ are sides. $J$ and $K$ share the $y$ -coordinate $-3$ , so $JK$ is horizontal. $K$ and $L$ share the $x$ -coordinate $2$ , so $KL$ is vertical.
$M = (x, y)$
Use side $LM$ . It is opposite side $JK$ , and opposite sides of a rectangle are parallel, so $LM$ is horizontal too. The $y$ -coordinate does not change along a horizontal side, so $M$ has the same $y$ -coordinate as $L(2, 1)$ .
$y = 1$
Use side $MJ$ . It is opposite side $KL$ , so it is vertical. The $x$ -coordinate does not change along a vertical side, so $M$ has the same $x$ -coordinate as $J(-5, -3)$ .
$x = -5$
Put the two coordinates together: $M = (-5, 1)$ . Then check with lengths, because opposite sides of a rectangle are equal. $JK$ and $LM$ both run between the $x$ -coordinates $-5$ and $2$ . $KL$ and $MJ$ both run between the $y$ -coordinates $-3$ and $1$ . Each pair of coordinates is on opposite sides of $0$ , so add the distances from $0$ .
$JK = LM = 5 + 2 = 7, \quad KL = MJ = 3 + 1 = 4$

Answer

The fourth vertex is $M(-5, 1)$ .

Rectangle

JKLM

with all four vertices:

J(-5, -3)

,

K(2, -3)

,

L(2, 1)

and the vertex you found,

M(-5, 1)

.

The figure shows the finished rectangle $JKLM$ . The three vertices you were given are the lower-left corner $J$ , the lower-right corner $K$ and the upper-right corner $L$ . The vertex you found, $M(-5, 1)$ , is the upper-left corner. It has the same $x$ -coordinate as $J(-5, -3)$ , so it sits straight above $J$ . It has the same $y$ -coordinate as $L(2, 1)$ , so it sits straight to the left of $L$ .

In the last example, the clue is a length. The perimeter of a rectangle, together with one side you already know, tells you how long the other sides must be. Those lengths tell you where the missing vertices go.

Worked example

Placing the last two fence posts

A family plans a rectangular dog run $PQRS$ on a grid where each unit stands for $1$ meter. The corner posts are named in order around the run. Posts $P(-2, 1)$ and $Q(6, 1)$ are the ends of one side. The run lies north of that side, where the $y$ -coordinates are greater. The fence around the run must be exactly $28$ meters long. Where do posts $R$ and $S$ go?

Find the known side. $P$ and $Q$ share the $y$ -coordinate $1$ , so $PQ$ is horizontal. Its $x$ -coordinates, $-2$ and $6$ , are on opposite sides of $0$ , so add their distances from $0$ . Each unit stands for $1$ meter, so $PQ$ is $8$ meters long.
$PQ = 2 + 6 = 8$
Use the rectangle. The fence goes all the way around, so its $28$ meters is the perimeter. Side $SR$ is opposite $PQ$ , so it is also horizontal and also $8$ meters long. Add to see how much fence the two horizontal sides use.
$8 + 8 = 16$
Find what is left for the other two sides, $QR$ and $SP$ . All four sides add up to $28$ meters. Subtraction undoes addition, so taking $16$ away from $28$ leaves the sum of the other two sides. You can check that $16 + 12 = 28$ .
$28 - 16 = 12$
Split that length. $QR$ and $SP$ are opposite sides of the rectangle, so they are equal. Two equal lengths that add up to $12$ meters are $6$ meters each, because $6 + 6 = 12$ . Each unit stands for $1$ meter, so that is $6$ grid units.
$12 \div 2 = 6$
Locate the posts. $QR$ and $SP$ meet the horizontal side $PQ$ at right angles, so they are vertical. That means $R$ keeps the $x$ -coordinate of $Q$ , and $S$ keeps the $x$ -coordinate of $P$ . The run lies north, so each side goes up $6$ units from $y = 1$ .
$1 + 6 = 7, \quad R = (6, 7), \quad S = (-2, 7)$
Check the perimeter. $S$ and $R$ share the $y$ -coordinate $7$ , so $SR$ is horizontal. Its $x$ -coordinates are $-2$ and $6$ again, so its length is $2 + 6 = 8$ units. Add all four sides.
$8 + 6 + 8 + 6 = 28$

Answer

Post $R$ goes at $(6, 7)$ and post $S$ goes at $(-2, 7)$ . The run is $8$ meters by $6$ meters, so its fence is exactly $28$ meters long.

Here is the idea to keep: on a horizontal or vertical side, coordinates and side lengths give you each other.

To find a perimeter, walk around the polygon and add the length of every side. Get horizontal and vertical lengths from the coordinates. Use the given length for any slanted side.
In a real-world problem, decide what the question asks for, work in grid units, and use the scale last.
To find a missing vertex, use the properties of the polygon. In a rectangle with a horizontal side, opposite sides are parallel and equal, so the missing vertex shares one coordinate with each of its two neighbors.