Finding a missing side from the perimeter
The fence is bought and its length is known, and so are all but one side of the shape it has to go around. How long must the last side be, and when is there no such side at all?
Where you are 15 in Area and perimeter
When you reach for it
In Perimeter every side was known, and the sides were added to get the distance all the way around. Often the question comes the other way. Jurgis has 36 meters of fence for a rectangular horse corral and wants it 11 meters long: how wide can it be? Kotryna has a fixed length of border for a square herb garden: how long is each side? The perimeter is known, and so are some of the sides. One side, or one kind of side, is missing.
The idea that settles every such question is the same: the perimeter is the sum of the sides. If the total is known and some of the parts are known, the missing part is the total minus the known parts, exactly as with any sum where one addend is unknown. Subtraction undoes addition.
The lesson takes two days. Part 1 works with any shape with straight sides, a polygon, where one side is missing: subtract the known sides from the perimeter. It also writes the question as an equation with a letter for the missing side. Part 2 does rectangles and squares, where equal sides give shortcuts: half the perimeter of a rectangle is one length plus one width, and a square’s perimeter divided by 4 is its side. Part 2 ends with the cases that have no answer: no side in whole units, or no shape at all.
The procedure
Part 1: one side missing
Going all the way around a polygon walks each side once, so the perimeter is the sum of the side lengths. Suppose a shape has five sides, four of them known, and the perimeter is known too. Then
This is a sum with one addend missing, and the missing addend is found by subtraction:
There are two ways to carry out the subtraction, and both give the same answer. You can add the known sides first and take that one sum away from the perimeter. Or you can take the known sides away from the perimeter one at a time. The first way is usually shorter; the second shows the fence being used up, side by side, until only the last side’s share is left.
An equation with a letter. The question can be written down before it is solved, using a letter for the missing side, as in Using a letter for the unknown number. For a five-sided fence of 40 meters with sides 9, 6, 8 and 7 meters, call the missing side s:
The letter stands for one number, the one that makes the sentence true. Adding the known sides gives 30 + s = 40, and the related subtraction fact gives s = 40 - 30 = 10. A box in place of the letter, 30 + \square = 40, says the same thing.
Check by adding. Put the answer back with the other sides and add all the way around. If the sum is the perimeter, the side is right. The check costs one addition, and it catches a slip in the subtraction.
Two equal missing sides. Sometimes two sides are missing but are known to be equal, as in a triangle with two equal sides. Subtracting the known side leaves the length the two equal sides share. They share it in two equal parts, so divide by 2.
Part 2: rectangles and squares
A rectangle’s opposite sides are equal (see Rectangles, rhombuses and squares), so going around it walks length, width, length, width. The first length and width take you to the opposite corner, halfway around, and the second pair brings you back the same distance. So
That gives the quickest route to a missing side of a rectangle: halve the perimeter, then subtract the side you know. A second route says the same thing in a different order. Two lengths use 2 \times \text{length} of the fence; what is left covers the two widths, and half of that is one width. Both routes are worked in Example 4.
A square has four equal sides, so its perimeter is 4 \times \text{side}. Going backward, the side is the perimeter divided by 4, because division undoes multiplication (Multiplication and division undo each other): 4 \times s = 28 asks which number times 4 is 28, and 28 \div 4 is that number.
When there is no answer. A subtraction or a division can give a result that no real shape could have.
- If the perimeter does not divide evenly, there is no side in whole units. A square with 30 feet of border would need sides between 7 feet (which uses 28) and 8 feet (which uses 32).
- If the known sides already add up to the perimeter or more, nothing is left for the last side, and there is no shape. The same goes for a rectangle whose given length is half the perimeter or more.
- Even a missing side that comes out positive can be too long: the other sides together must stretch farther than it does, or they cannot reach around from one end of it to the other.
Checking by adding catches a wrong subtraction, but not these. They are caught by asking whether the shape could be built.
Worked examples
Examples 1 to 3 go with Part 1, and Examples 4 to 7 with Part 2.
The duck pond fence
Tadas has 40 meters of fence for a five-sided pen at the duck pond. Going around, four of the sides are 9 meters, 6 meters, 8 meters and 7 meters. How long is the fifth side?
Call the fifth side s meters. The fence goes once along every side, so the five sides add up to the whole fence:
Add the known sides first, grouping them in pairs that are easy to add (the associative property lets the grouping change): 9 + 6 = 15 and 8 + 7 = 15, so the known sides use 15 + 15 = 30 meters. That leaves 30 + s = 40, and subtraction undoes the addition:
The fifth side is 10 meters. Check by adding all the way around: 9 + 6 + 8 + 7 + 10 = 40, the whole fence.
A long fence, taken away side by side
A four-sided field beside the duck pond has a perimeter of 315 meters. Three of its sides are 84 meters, 96 meters and 78 meters. Find the fourth side.
SolutionThe known sides add up to
and the fourth side is what the perimeter has left:
This subtraction regroups twice (Multi-digit subtraction): 5 ones is too few to take 8, so one ten becomes ten ones and 15 - 8 = 7; then 0 tens is too few to take 5, so one hundred becomes ten tens and 10 - 5 = 5; and 2 - 2 = 0 hundreds.
The other way takes the sides away one at a time, as if walking the fence and using it up:
Both ways give 57 meters, as they must: taking away 84, then 96, then 78 takes away 258 in all. Check: 258 + 57 = 315.
A triangle with two equal sides
Kotryna edges a triangle-shaped herb bed with 25 feet of border. One side is 7 feet, and the other two sides are equal to each other. How long is each of the equal sides?
SolutionTake away the side that is known. The two equal sides share what is left:
Two equal sides make 18 feet, so each one is half of 18, which is 18 \div 2 = 9 feet, because 2 \times 9 = 18.
Each equal side is 9 feet. Check by adding: 7 + 9 + 9 = 25.
The horse corral
Jurgis has 36 meters of fence for a rectangular horse corral. He wants it 11 meters long. How wide is it?
Call the width w meters. Going around, the sides are 11, w, 11 and w.
Halfway around. One length and one width, the thick part of the figure, take half of the fence, and 36 \div 2 = 18 meters. So
and w = 18 - 11 = 7.
Lengths first. The two long sides use 2 \times 11 = 22 meters. The rest, 36 - 22 = 14 meters, goes on the two widths, and they are equal, so each is 14 \div 2 = 7 meters.
The corral is 7 meters wide. Check: 11 + 7 + 11 + 7 = 36.
The square herb garden
Kotryna has 28 feet of border for a square herb garden and uses all of it. How long is each side?
Call the side s feet. A square’s four sides are equal, so the border is s + s + s + s, four groups of s:
The side is the number that, taken 4 times, makes 28, and that number is 28 \div 4. The multiplication fact 4 \times 7 = 28 gives 28 \div 4 = 7.
Each side is 7 feet. Check: 7 + 7 + 7 + 7 = 28.
A border that does not share out
Tadas has 30 feet of edging and wants a square bed that uses all of it, with each side a whole number of feet. Can he make one? If not, what rectangle with whole-number sides could he make instead?
SolutionEach side would be 30 \div 4 feet. But 4 \times 7 = 28 and 4 \times 8 = 32, and 30 lies between them: 30 \div 4 is 7 with 2 left over. Sides of 7 feet leave 2 feet of edging unused, and sides of 8 feet need 2 feet more than he has. No square with whole-number sides uses exactly 30 feet. (A side of 7 and a half feet would, which needs fractions.)
A rectangle is easier to fit. Half of 30 is 15, so any length and width with \text{length} + \text{width} = 15 uses all the edging. One choice is 8 feet by 7 feet: 8 + 7 + 8 + 7 = 30. There are others, such as 10 by 5; how differently such rectangles fill the ground inside is the subject of Perimeter versus area.
When there is no shape at all
(a) A four-sided fence at the duck pond is to have a perimeter of 20 meters. Three of its sides are 9 meters, 6 meters and 7 meters. Find the fourth side.
(b) A triangle is to have a perimeter of 20 meters, with two sides of 4 meters and 5 meters. Find the third side.
(a) The known sides add up to 9 + 6 + 7 = 22 meters. That is already more than the 20 meters of fence. The fourth side would have to be 20 - 22, less than nothing, and no side has a length below zero. There is no such fence: these three sides alone need 22 meters.
(b) The subtraction goes through: 20 - 4 - 5 = 11 meters. But try to build it. Lay the 11-meter side down. The other two sides start at its two ends and have to meet at a corner. Together they are 4 + 5 = 9 meters long, and the gap between their starting points is 11 meters. However they are tipped, they cannot reach each other, as the figure shows. There is no triangle.
In both parts the arithmetic produced a number, or tried to, and only asking whether the shape could exist showed there is none. Adding all the way around would not have caught (b): 4 + 5 + 11 = 20.
Boundaries
Only one unknown length. Subtraction finds one missing side, or several sides known to be equal. If two sides are missing and nothing says how they compare, the perimeter alone cannot decide them: a four-sided fence of 30 meters with sides 8 and 9 leaves 13 meters for the other two, which could be 6 and 7, 5 and 8, or many other pairs.
A rectangle needs one side besides the perimeter. The perimeter gives only length plus width. A perimeter of 16 meters fits 7 by 1, 5 by 3, 4 by 4 and more.
The halving shortcut needs a rectangle. Length plus width is half the perimeter only because opposite sides are equal. For a four-sided shape whose sides are not two equal pairs, subtract all the known sides instead.
Dividing by 4 needs a square. The four sides must be equal. Likewise dividing by 3 needs a triangle with three equal sides, and in general by the number of sides when every side is equal.
All lengths in one unit. A perimeter in meters and a side in centimeters must be put into the same unit before subtracting (Metric length units, Customary length units).
The answer must make a shape. The missing side has to be more than zero, and it has to be shorter than all the other sides put together, or they cannot reach around from one of its ends to the other. If a whole number of units is required, the division for a square or for equal sides must come out even.
Where this goes next
Perimeter versus area keeps the perimeter fixed, as in Example 6, and asks how much ground rectangles with the same fence enclose. The answer is that it changes a lot. A different backward question starts from the area instead of the fence: Finding a missing side from the area knows how much ground a rectangle covers and one side, and finds the other by division rather than by halving and subtracting. Later, an equation with a letter for an unknown length becomes the ordinary way to write any such question, with the perimeter formula P = 2l + 2w solved for the length or the width.
Exercises
Part 1: one side missing
- A triangle-shaped flag has a perimeter of 24 inches. Two of its sides are 7 inches and 9 inches. How long is the third side?
- A four-sided herb bed has 30 feet of border. Three of its sides are 6 feet, 9 feet and 8 feet. Find the fourth side.
- A six-sided fence at the duck pond is 52 meters long. Five of its sides are 8, 7, 9, 6 and 10 meters. How long is the sixth side?
- A four-sided pen beside the corral has a perimeter of 64 meters, with three sides of 15, 18 and 12 meters. Write an equation with a letter for the fourth side, then solve it.
- A four-sided field has a perimeter of 400 meters. Three of its sides are 125, 98 and 107 meters. Find the fourth side.
- A five-sided pasture next to the corral has a perimeter of 612 meters. Four of its sides are 145, 118, 96 and 137 meters. Find the fifth side, and check by adding.
- A triangle-shaped herb bed has 32 feet of border. One side is 10 feet and the other two are equal. How long is each equal side?
- A triangle with three equal sides has a perimeter of 27 centimeters. How long is each side?
- A four-sided fence has a perimeter of 45 meters, and three of its sides are 12, 10 and 14 meters. Is the fourth side 9 meters? Decide by adding all the way around.
- A six-sided pen has all its sides equal and a perimeter of 42 meters. How long is each side?
Part 2: rectangles and squares
- A rectangular corral has a perimeter of 26 meters and a length of 8 meters. Find its width.
- A rectangular herb garden has 50 feet of border and is 9 feet wide. How long is it?
- A square herb bed has 36 feet of border. How long is each side?
- A square pen at the duck pond takes 32 meters of fence. How long is each side?
- A rectangular paddock has a perimeter of 90 meters and a length of 30 meters. Find its width, both by halving first and by taking away the two lengths first.
- A rectangular corral takes 140 meters of fence and is 25 meters wide. How long is it?
- Can a square with whole-number sides, in meters, have a perimeter of 22 meters? Explain.
- A four-sided fence is to have a perimeter of 25 meters, with three sides of 9, 8 and 10 meters. Find the fourth side, or explain why there is none.
- A triangle is to have a perimeter of 30 meters, with two sides of 6 meters and 8 meters. Find the third side, or explain why there is none.
- Jurgis wants a rectangular corral with 24 meters of fence, every side a whole number of meters. List every length and width he could choose. (Count 11 by 1 and 1 by 11 as the same corral.)
- A rectangle has a perimeter of 40 meters, and its length is 4 meters more than its width. Find the length and the width.
Explain your thinking
- Explain why half of a rectangle’s perimeter is always one length plus one width.
- A rectangle is said to have a perimeter of 36 meters and a length of 20 meters. Explain why no such rectangle exists.
- Can you find the length and width of a rectangle when you know only its perimeter? Explain, using an example.
- Explain why the perimeter of a square with whole-number sides is always a number you say when counting by 4s, and use this to say why no such square has a perimeter of 30.
Answers
- 8 inches. The known sides use 7 + 9 = 16, and 24 - 16 = 8. Check: 7 + 9 + 8 = 24.
- 7 feet. 6 + 9 + 8 = 23, and 30 - 23 = 7. Check: 6 + 9 + 8 + 7 = 30.
- 12 meters. The five sides add to 8 + 7 + 9 + 6 + 10 = 40, and 52 - 40 = 12.
- 15 + 18 + 12 + s = 64. The known sides add to 45, so 45 + s = 64 and s = 64 - 45 = 19 meters.
- 70 meters. 125 + 98 + 107 = 330, and 400 - 330 = 70. Check: 330 + 70 = 400.
- 116 meters. 145 + 118 = 263, 263 + 96 = 359, 359 + 137 = 496, and 612 - 496 = 116. Check: 496 + 116 = 612.
- 11 feet. 32 - 10 = 22 is shared by the two equal sides, and 22 \div 2 = 11. Check: 10 + 11 + 11 = 32.
- 9 centimeters. Three equal sides: 27 \div 3 = 9, since 3 \times 9 = 27.
- Yes. 12 + 10 + 14 + 9 = 45, which is the perimeter.
- 7 meters. 42 \div 6 = 7, since 6 \times 7 = 42.
- 5 meters. Half the perimeter is 26 \div 2 = 13, and 13 - 8 = 5. Check: 8 + 5 + 8 + 5 = 26.
- 16 feet. Half the border is 50 \div 2 = 25, and 25 - 9 = 16. Check: 16 + 9 + 16 + 9 = 50.
- 9 feet. 36 \div 4 = 9, since 4 \times 9 = 36.
- 8 meters. 32 \div 4 = 8, since 4 \times 8 = 32.
- 15 meters. Halving first: 90 \div 2 = 45 and 45 - 30 = 15. Lengths first: 2 \times 30 = 60, 90 - 60 = 30 for the two widths, and 30 \div 2 = 15.
- 45 meters. Half the fence is 140 \div 2 = 70, and 70 - 25 = 45. Check: 45 + 25 + 45 + 25 = 140.
- No. The side would be 22 \div 4, but 4 \times 5 = 20 and 4 \times 6 = 24, so 22 is not 4 times any whole number: 22 \div 4 is 5 with 2 left over.
- There is none. The three sides already add to 9 + 8 + 10 = 27 meters, more than the 25 meters of fence, so nothing is left for a fourth side.
- There is none. The subtraction gives 30 - 6 - 8 = 16 meters, but the other two sides together are only 6 + 8 = 14 meters, shorter than 16, so they cannot reach from one end of the 16-meter side to the other.
- Six corrals. Length plus width is 24 \div 2 = 12, so the pairs are 11 by 1, 10 by 2, 9 by 3, 8 by 4, 7 by 5 and 6 by 6 (a square).
- Length 12 meters, width 8 meters. Length plus width is 40 \div 2 = 20. Taking away the extra 4 meters leaves 20 - 4 = 16 for two equal widths, so the width is 16 \div 2 = 8 and the length is 8 + 4 = 12. Check: 12 + 8 + 12 + 8 = 40.
- Going around a rectangle, the sides come as length, width, length, width. Opposite sides are equal, so the second length and width are the same as the first. The perimeter is the same pair walked twice, and half of it is one length plus one width.
- Half of 36 is 18, which must be the length plus the width. A length of 20 is already more than 18, so the width would have to be less than nothing. The two lengths alone would need 40 meters of fence.
- No. The perimeter gives only the length plus the width, and many pairs share the same sum. A perimeter of 20 meters fits a rectangle 9 by 1, 6 by 4, or 5 by 5. One more fact, such as the length, is needed.
- The perimeter of a square is 4 \times \text{side}, which is 4 groups of the side, so with a whole-number side it lands on the count by 4s: 4, 8, 12, \ldots. Counting by 4s goes 28, 32 and skips 30, so no square with whole-number sides has a perimeter of 30.