Problem
Perimeter of a rectilinear figure
Six unit squares are glued edge to edge into one connected figure. Different gluings give different perimeters. Figures matching after a turn or flip count as one. Find every figure whose perimeter is 12.
Reasoning · Grade 5-1 Perimeter and Area
Your answer
Watch out A glued edge seems to cost 1, but both sides vanish, so it costs 2.
How to solve
Strategy Make a Systematic List — There are 35 different figures that can be made from 6 squares, far too many to draw and measure one by one. So I change focus: instead of measuring each figure, I use the fact that sliding the jagged sides of a figure outwards turns it into the smallest rectangle that surrounds it without changing the total distance around, as long as the figure has no dent bitten out of it. That says the perimeter is 12 cm exactly when the surrounding rectangle has perimeter 12 cm and the figure has no dents. Only two rectangles can hold 6 squares and have perimeter 12 cm, so the search shrinks to filling those two rectangles, which is small enough to list completely. Cutting 6 paper squares and laying one arrangement on another settles every turn-and-flip duplicate honestly, and a separate count of the glued edges checks each surviving figure.
1STEP 1
Count the perimeter by counting glued edges
A perimeter of 12 needs 6 glued edges.
6 × 4 = 24, 24 - 2 × (joins) = 12, joins = 6
Counting what is hidden is easier than tracing what is showing, and each glued edge always hides the same 2 cm, so the bookkeeping never gets complicated.
3.MD.D.8Change Focus Count The ComplementThe perimeter is found by counting the glued edges, because each glued edge hides two unit sides from the outside.
Why?
A glued edge belongs to two squares at once and is counted once from each, so every glue costs two sides of boundary.
🧱Counted twice, so halveIf every item got counted from both of its two ends, the total is exactly double the truth, so halving fixes it.
Why?
The boundary is what remains once the hidden sides are taken from all the squares' sides put together.
🧱Whole is the sum of its partsBreak a thing into pieces with no gaps or overlaps and the pieces add back to the whole.
2STEP 2
Slide the sides out to the surrounding rectangle
Then the box's width plus height is 6.
2 × (width + height) = 12, width + height = 6
This is the trick the whole unit is built on: a wiggly outline and the straight box around it are the same length, so a hard shape is measured by measuring an easy one.
3.MD.D.8Change Focus Count The Complement3STEP 3
List the rectangles that could hold the figure
Only 2 by 4 and 3 by 3 can hold six cells.
1 × 5 = 5 ＜ 6, 2 × 4 = 8 ≥ 6, 3 × 3 = 9 ≥ 6
Counting the squares a box can hold rules out the long thin box immediately, which cuts the whole search down to two boxes a child can draw on the grid.
3.MD.C.6Make A Systematic List4STEP 4
Fill the 2 by 4 box: 3 figures
The 2 by 4 box gives 3.
4 + 2 = 6, 3 + 3 = 6
Sorting the cases by how the 6 squares split between the two rows is a short, complete list, so nothing can hide between the cases.
1.G.A.2Make A Systematic List5STEP 5
Fill the 3 by 3 box: 4 figures
The 3 by 3 box gives 4.
1 + 2 + 3 = 6, 2 + 2 + 2 = 6
Inside a 3 by 3 box a figure is completely described by how long each row is and where each row sits, so listing the row lengths first makes the search finite and easy to finish.
1.G.A.2Make A Systematic List6STEP 6
Remove the turn-and-flip duplicates honestly
Removing turn-and-flip duplicates leaves 7.
3 + 4 = 7
Laying one cut-out on top of another settles a duplicate question with no argument, and the box size gives a quick reason why two figures from different boxes can never be the same.
8.G.A.2Create A Physical Representation7STEP 7
Check each of the 7 by counting its glued edges
All seven measure 12 around.
3 + 1 + 2 = 6 joins, 24 - 2 × 6 = 12 cm
The join count and the surrounding-rectangle trick are two completely different ways to get the perimeter, so agreement between them is real evidence rather than the same mistake twice.
3.MD.D.8Draw A DiagramAnswer
7 figures
24 − 2 × 6 = 12
The answer is the right size and the right shape. All 6 squares together have area 6 square cm in every figure, and 12 cm of perimeter is sensible for that: the tidiest arrangement, the 2 by 3 rectangle, has the smallest possible perimeter of 10 cm, and the most stretched-out one, a row of 6, has 14 cm, so 12 cm sits in between and should have several answers. Out of the 35 figures that 6 squares can make, exactly 1 has perimeter 10 cm, 7 have 12 cm, and the remaining 27 have 14 cm, and 1 + 7 + 27 = 35, so nothing has been lost or double-counted. Every perimeter is an even number of centimetres, as it must be, because the outline of a figure drawn on the grid always comes back to where it started. The sample figure printed in the book is figure (1) on the list, so the book and the list agree.
Takeaway
Push the jagged sides of a figure out to the box around it and the distance around stays the same, so a wiggly perimeter is really just a rectangle's perimeter in disguise.
- Count the perimeter by counting glued edges
- Slide the sides out to the surrounding rectangle
- List the rectangles that could hold the figure
- Fill the 2 by 4 box: 3 figures
- Fill the 3 by 3 box: 4 figures
- Remove the turn-and-flip duplicates honestly
- Check each of the 7 by counting its glued edges