Perimeter and Area from Given Dimensions | Grade 8 Math | LevelKite

You already know how to name and classify polygons, the closed flat shapes whose sides are straight line segments. Now you will measure them, and there are two different things to measure.

How far is it around the edge? That distance is the perimeter. How much flat surface does the shape cover? That amount is the area.

The two answers use different units. A perimeter is a length, so it is measured in length units such as cm or m. An area is measured in square units such as $\text{cm}^2$ or $\text{m}^2$ . One square unit is a square whose sides are each $1$ unit long, so $1\text{ cm}^2$ is a square that is $1\text{ cm}$ on each side.

In this lesson you will find the perimeter of polygons, find the area of triangles and quadrilaterals (four-sided polygons), and decide which measurement a real situation needs.

Imagine walking along every side of a polygon until you are back where you started. The total distance you walk is the perimeter, written $P$ .

That walk gives the rule: add the lengths of all the sides, counting each side exactly once. The rule works only when every length is in the same unit, so convert first if the units differ. Treating $3\text{ m}$ plus $40\text{ cm}$ as $3 + 40 = 43$ gives a number that is not a length in either unit.

When several sides are equal, multiplying does the repeated adding:

A rectangle has two sides of length $l$ and two of width $w$ , where $l$ and $w$ are positive and in the same unit. So $P = 2l + 2w = 2(l + w)$ . Any parallelogram works the same way with its two side lengths, because its opposite sides are equal.
A regular polygon has all sides equal and all angles equal. If it has $n$ sides, each of positive length $s$ , then $P = n \cdot s$ . A regular hexagon has $n = 6$ , so with $s = 4.5\text{ cm}$ its perimeter is $P = 6 \cdot 4.5 = 27\text{ cm}$ .

Worked example

Perimeter when the units do not match

A pentagon has side lengths $1.2\text{ m}$ , $85\text{ cm}$ , $0.9\text{ m}$ , $65\text{ cm}$ and $70\text{ cm}$ . Find its perimeter in centimeters.

Check the units before adding. Three lengths are in centimeters and two are in meters, so they cannot be added yet.
Convert the two lengths given in meters. Because 1 m equals 100 cm, multiplying the number of meters by 100 renames the same length in centimeters.
$1.2 \cdot 100 = 120\text{ cm}, \quad 0.9 \cdot 100 = 90\text{ cm}$
Write the perimeter as the sum of all five side lengths, because the perimeter is the total distance around. Each side appears once, and every length is now in centimeters.
$P = 120 + 85 + 90 + 65 + 70$
Add from left to right, one length at a time, so that each partial sum can be checked.
$P = 205 + 90 + 65 + 70 = 295 + 65 + 70 = 360 + 70 = 430\text{ cm}$

Answer

The perimeter of the pentagon is $430\text{ cm}$ , which is the same length as $4.3\text{ m}$ .

Area, written $A$ , is the amount of flat surface inside a shape. You measure it by counting the unit squares, including parts of squares, that cover the shape with no gaps and no overlaps.

Start with a rectangle $5\text{ cm}$ long and $3\text{ cm}$ wide. It holds $3$ rows of $5$ squares, each $1\text{ cm}$ on a side, so its area is $5 \cdot 3 = 15\text{ cm}^2$ . Multiplying the two side lengths counts the squares, and the same product gives the area when the lengths are decimals or fractions. Both lengths must be in the same unit, just as for perimeter.

Other shapes need two more words. A base, written $b$ , is a side you choose to measure from. The height, written $h$ , is the perpendicular distance, meaning the distance measured at a right angle, from the line along the base to the opposite corner of a triangle or the opposite side of a parallelogram. Both are positive lengths in the same unit.

In a rectangle, the width meets the length at a right angle, so the width is the height. In a shape with slanted sides, the height is often a separate measurement. A side can serve as the height only when it meets the base at a right angle.

Use the height that the problem states. Do not guess it from how a shape looks.

Three more area rules grow out of the rectangle. With the rectangle rule, they cover every shape whose area you will find in this lesson: rectangles, parallelograms, triangles and trapezoids.

Parallelogram. Think of a parallelogram that leans only a little. One cut at a right angle to the base slices a right triangle off one end. Slide that triangle to the other end, and the two pieces form a rectangle with the same base $b$ and the same height $h$ . Nothing was added or thrown away, so the area is $b \cdot h$ . The same rule holds for every parallelogram, however far it leans.

Two copies of a triangle fit together to make a parallelogram, and so do two copies of a trapezoid. That is why each of their area rules starts with one half.

Triangle. Take a triangle with base $b$ and height $h$ . Turn a copy of it half a turn and join the two along a side that is not the base. Together they form a parallelogram with base $b$ and height $h$ . The triangle is half of it, so its area is $\tfrac{1}{2} b h$ .

Trapezoid. A trapezoid has a pair of parallel sides. Call their lengths $b_1$ and $b_2$ , and call the perpendicular distance between them $h$ . All three are positive lengths in the same unit. A half-turned copy joins the trapezoid along one of its other sides to form a parallelogram with base $b_1 + b_2$ and height $h$ . The trapezoid is half of it, so its area is $\tfrac{1}{2}(b_1 + b_2)h$ .

Area rules: every length is positive and in one unit; each height is perpendicular to its base.

Shape	Area rule	Dimensions you need	Example (lengths in m)
Rectangle	$A = l \cdot w$	Length and width, which meet at a right angle	$l = 9$ , $w = 4$ : $A = 9 \cdot 4 = 36\text{ m}^2$
Parallelogram	$A = b \cdot h$	A base and the height perpendicular to it	$b = 9$ , $h = 4$ : $A = 9 \cdot 4 = 36\text{ m}^2$
Triangle	$A = \tfrac{1}{2} b h$	A base and the height perpendicular to it	$b = 9$ , $h = 4$ : $A = \tfrac{1}{2} \cdot 9 \cdot 4 = 18\text{ m}^2$
Trapezoid	$A = \tfrac{1}{2}(b_1 + b_2)h$	Two parallel sides and the perpendicular distance between them	$b_1 = 9$ , $b_2 = 5$ , $h = 4$ : $A = \tfrac{1}{2}(9 + 5) \cdot 4 = 28\text{ m}^2$

Key concept

Around the edge or across the surface

Perimeter is a distance around the edge. Add every side length once, with all lengths in the same unit, and give the answer in length units.

Area is the surface inside. Build it from a base and the height perpendicular to that base, both in the same unit, and give the answer in square units.

A side that is not perpendicular to the base counts toward the perimeter, but it is never the height for that base.

Worked example

A parallelogram: which number goes where?

A parallelogram has a base of $12\text{ cm}$ and slanted sides of $7\text{ cm}$ . The perpendicular distance between the base and the opposite side is $5\text{ cm}$ . Find its perimeter and its area.

Sort the three numbers. The 12 cm base and the 7 cm slanted side are side lengths. The 5 cm distance is the height, because it is measured perpendicular to the base. All three are in centimeters, so no conversion is needed.
Find the perimeter. Opposite sides of a parallelogram are equal, so there are two sides of 12 cm and two sides of 7 cm. The height is not a side, so it is left out.
$P = 2(12 + 7) = 2 \cdot 19 = 38\text{ cm}$
Find the area. Multiply the base by the height that is perpendicular to it. The slanted side is not perpendicular to the base, so it is left out.
$A = b \cdot h = 12 \cdot 5 = 60\text{ cm}^2$
Check the units and the size. The perimeter is a length, and the area is in square units. A 12 cm by 7 cm rectangle covers 84 square centimeters, and a leaning parallelogram with the same sides covers less, so 60 is reasonable.

Answer

The perimeter of the parallelogram is $38\text{ cm}$ , and its area is $60\text{ cm}^2$ .

A slanted side is not the height. Take the parallelogram from the example: base $12\text{ cm}$ , slanted sides $7\text{ cm}$ , height $5\text{ cm}$ . A common mistake is to multiply the two side lengths: $12 \cdot 7 = 84\text{ cm}^2$ . That is the area of a $12\text{ cm}$ by $7\text{ cm}$ rectangle. Now imagine pushing that rectangle over so that it leans. Its sides are still $12\text{ cm}$ and $7\text{ cm}$ long, but the top sinks closer to the base and the shape covers less surface. So the two side lengths alone cannot give a parallelogram's area. What shrinks is the perpendicular distance from the base to the top, here $5\text{ cm}$ , and that is why the area is $12 \cdot 5 = 60\text{ cm}^2$ .

A word problem rarely uses the words perimeter or area. You decide which one fits by asking what is being measured.

If it is a length that follows the edge, you need the perimeter.
If it is a surface that gets covered, you need the area.

Once you know which measurement you need, work in four moves: list the stated dimensions with their units, make the units match, calculate with the rule that fits the shape, and answer the question that was asked in a sentence with units.

Some answers must be whole numbers of things, such as pots or rolls. When the exact result is not a whole number, round up so that you have enough.

The table below shows both kinds of question about the same object.

One object, two different measurements

Situation	What is measured	Measurement and unit
Fence around a garden	A length that follows the edge	Perimeter, in m
Grass seed for that garden	A surface that gets covered	Area, in $\text{m}^2$
Ribbon along the edge of a poster	A length that follows the edge	Perimeter, in cm
Paint to cover that poster	A surface that gets covered	Area, in $\text{cm}^2$

Worked example

Edging and mulch for a triangular flower bed

A triangular flower bed has sides of $13\text{ m}$ , $14\text{ m}$ and $15\text{ m}$ . The perpendicular distance from the $14\text{ m}$ side to the opposite corner is $12\text{ m}$ . A gardener puts edging along every side and spreads mulch over the whole bed. How many meters of edging are needed, and how many square meters must the mulch cover?

Match each question to a measurement. Edging follows the sides, so it needs the perimeter. Mulch covers the surface inside, so it needs the area. Every dimension is in meters, so the units already match.
Add the three side lengths, each once. The 12 m distance is a height, not a side, so it is not part of the perimeter.
$P = 13 + 14 + 15 = 27 + 15 = 42\text{ m}$
For the area, use the 14 m side as the base, because the 12 m distance is perpendicular to that side. A triangle is half of a parallelogram with the same base and height. Halving 14 first is allowed because factors can be multiplied in any order.
$A = \tfrac{1}{2} \cdot 14 \cdot 12 = 7 \cdot 12 = 84\text{ m}^2$
Check the units. Edging is a length, so meters fit. Mulch covers a surface, so square meters fit.

Answer

The gardener needs $42\text{ m}$ of edging, and the mulch must cover $84\text{ m}^2$ .

Worked example

Ribbon and paint for a trapezoid banner

A banner is shaped like a trapezoid. Its parallel edges are $1.4\text{ m}$ and $0.8\text{ m}$ long, and the perpendicular distance between them is $40\text{ cm}$ . The other two edges are $50\text{ cm}$ each. Ribbon is sewn along the whole edge, and one face is painted. One pot of paint covers $2000\text{ cm}^2$ . How many centimeters of ribbon are needed, and how many pots of paint must be bought?

Match each question to a measurement. Ribbon follows the edge, so it needs the perimeter. Paint covers a face, so it needs the area. The pot is labeled in square centimeters, so work in centimeters.
Convert the two lengths given in meters. Because 1 m equals 100 cm, multiplying by 100 renames each length without changing it.
$1.4 \cdot 100 = 140\text{ cm}, \quad 0.8 \cdot 100 = 80\text{ cm}$
Add the four edges. The 40 cm distance is the height, not an edge, so it stays out of the sum.
$P = 140 + 80 + 50 + 50 = 220 + 100 = 320\text{ cm}$
Use the trapezoid rule: add the two parallel edges, multiply by the perpendicular distance between them, and take half.
$A = \tfrac{1}{2}(140 + 80) \cdot 40 = \tfrac{1}{2} \cdot 220 \cdot 40 = 110 \cdot 40 = 4400\text{ cm}^2$
Divide the area by the area one pot covers. The exact result is 2.2 pots. Two pots cover only 4000 square centimeters, which is less than 4400, and pots are sold whole, so round up to 3.
$4400 \div 2000 = 2.2, \quad 2 \cdot 2000 = 4000, \quad 3 \cdot 2000 = 6000$

Answer

The banner needs $320\text{ cm}$ of ribbon, and $3$ pots of paint must be bought to cover its $4400\text{ cm}^2$ face.

Perimeter and area answer different questions about the same shape. For perimeter, add every side length once, all in one unit, and answer in length units. For area, pair a base with the height perpendicular to it: use $b \cdot h$ for a rectangle or parallelogram, half of that for a triangle, and $\tfrac{1}{2}(b_1 + b_2)h$ for a trapezoid. Answer in square units. In a word problem, ask whether the quantity follows the edge or covers the surface.

Next, you will find the area of circles and the distance around them, called the circumference, and you will find and correct errors in perimeter and area calculations.