Problem
Draw parallelograms from a given side
A pegboard holds 5 dots across and 4 dots down, evenly spaced. One segment is already drawn. Only parallelograms using that segment as a side count. Count those that are not rhombuses.
Reasoning · Grade 4-2 Counting Figures Formed by Joining Dots
Your answer
Watch out All eight look valid, but the two rhombuses have to come out.
How to solve
Strategy Make a Systematic List — A parallelogram built on a fixed side is decided by one thing only: the step taken by the next side, described as so many dots right and so many dots down. That turns a drawing question into a short list of steps. The board edges cut the list down hard, because the given segment already touches the left edge and the top row, so the candidate list is tiny and can be written out in full. Then the rhombus rule removes exactly the steps whose length matches the given side, and drawing each survivor on a board confirms it really fits.
1STEP 1
See that one step decides the whole parallelogram
One step for the second side fixes the shape.
A = (col 1, row 2), B = (col 3, row 1)
Fourth graders test whether two sides are parallel by counting the dots across and down rather than eyeballing the slant, and sliding one side to make its partner keeps them parallel automatically.
4.G.A.2Draw A DiagramOnce the given side is fixed, a single step from one of its ends decides the whole parallelogram.
Why?
The opposite side has to be the same step from the other end, so one step names one parallelogram and no more.
🧱One-to-one correspondenceIf each thing here pairs with exactly one thing there, the two groups are the same size.
Why?
Copying the same step at both ends makes the two new sides run in the same direction and never meet, which is what a parallelogram promises.
🧱Parallel lines keep a constant gapTwo parallel lines never meet and the distance between them is the same everywhere you measure it.
2STEP 2
Let the edges of the board rule out most steps
Ruling out steps off the board leaves 8.
3 × 3 - 1 = 8
Checking whether a dot still exists on the board is pure counting, and it wipes out whole families of steps in one go.
4.G.A.1Eliminate Possibilities3STEP 3
Write the 8 candidates out and draw them
Draw each of the eight.
Listing the steps in order, across then down, is the surest way for a young solver to be certain nothing has been skipped.
4.G.A.1Make A Systematic List4STEP 4
Find which of the 8 are rhombuses
Two of them are rhombuses.
A young solver can just turn the pegboard a quarter turn and watch the slanted step land on the given segment, which is the hands-on version of the rule that turns and flips preserve length.
8.G.A.1Eliminate Possibilities5STEP 5
Count and draw the survivors
Removing them leaves 6.
8 - 2 = 6
Every corner listed is a real dot on the 5 by 4 board, so each drawing can be checked simply by finding the dots.
4.G.A.2Make A Systematic ListAnswer
6 parallelograms
8 − 2 = 6
Six answers is exactly the number of blank pegboards printed with the problem, which is a strong sign the list is complete and nothing has been double counted. Each drawing can also be checked one at a time. Take the step of 1 right: the corners are (1,2), (3,1), (4,1) and (2,2). The side from (1,2) to (3,1) and the side from (2,2) to (4,1) both go 2 right and 1 up, so they are parallel and equal, while the sides from (1,2) to (2,2) and from (3,1) to (4,1) are both single steps right, so they are parallel and equal too. It is a genuine parallelogram, and since a 1-dot side is plainly shorter than the slanted side, it is not a rhombus. Every corner in all six drawings has a column between 1 and 5 and a row between 1 and 4, so all six really fit on the board.
Takeaway
Describe the next side as so many dots right and so many dots down, and the whole hunt becomes a short list you can finish.
- See that one step decides the whole parallelogram
- Let the edges of the board rule out most steps
- Write the 8 candidates out and draw them
- Find which of the 8 are rhombuses
- Count and draw the survivors