Twelve polygons are drawn on dot boards with dots 1 cm apart. Five have no dot inside and seven have dots inside. For each, count the boundary dots and the interior dots into a table. Find the formula giving area from those counts.
Your answer
Watch outAll dots seem equal, but a boundary dot is worth only half an interior one.
How to solve
StrategyLook for a Pattern — The two tables are the whole point: once every column is filled in honestly, the rule jumps out of the numbers. So I work in a fixed order for each figure - count boundary dots, count interior dots, then find the area by cutting the figure into rectangles and right triangles (or by cutting corner triangles off a surrounding rectangle). Board (1) has all of its interior counts equal to zero, which is a deliberately easier version of the same question; I read the rule off there first, then look at what each interior dot adds when I move to board (2).
1STEP 1
Set the unit: one little square is 1 cm²
Take one little square as 1 cm².
1 cm × 1 cm = 1 cm², 1 ÷ 2 = 0.5 cm²
Measuring area by counting unit squares is the Grade 3 idea, and the dot board hands you the unit squares ready-made, so no length ever has to be measured with a ruler.
3.MD.C.6Draw A Diagram
2STEP 2
Board (1): count the dots on each boundary
The five without interior dots have 3, 4, 5, 6, 7 on the boundary.
A:3, B:4, C:5, D:6, E:7
Going around the outline in one direction, and looking at every side rather than only at the corners, is what stops you from missing the dots that sit in the middle of a long side.
4.G.A.1Make A Systematic List
3STEP 3
Board (1): find each area by cutting the figure up
Their areas rise by 0.5 each time.
C = 0.5 + 1 = 1.5, D = 1 × 2 = 2, E = 5 × 1 ÷ 2 = 2.5
Any polygon on a dot board can be chopped into rectangles and right triangles whose legs run along the grid, and each of those pieces has an area you get by multiplying and halving.
6.G.A.1Identify Subproblems
4STEP 4
Read the rule off table (1)
So the area is boundary ÷ 2 − 1.
area = (boundary dots) ÷ 2 - 1
When a table climbs by the same step (here 0.5) every time the input climbs by 1, the rule is halve-then-shift; one row is enough to find the shift and the other rows check it.
4.OA.C.5Look For A Pattern
5STEP 5
Board (2): count boundary dots and interior dots
Count the dots for the seven with interior dots too.
F 3|1, G 4|1, H 5|1, I 6|1, J 4|2, K 6|2, L 4|4
Keeping the two counts in two separate rows, and deciding for every doubtful dot whether it is ON a side or strictly inside, is the whole difficulty here - a dot sitting on a slanted side is easy to mistake for an inside dot.
4.G.A.1Make A Systematic List
6STEP 6
Board (2): find each area by cutting the figure up
Cut each up to measure its area.
F = 4 - 2.5 = 1.5, G = 2 × 2 ÷ 2 = 2, H = 4 - 1.5 = 2.5, I = 3, J = 3, K = 4, L = 9 - 4 × 1 = 5
The box-minus-corners trick works for every tilted figure on a dot board, because the corners you cut away are always right triangles whose legs run along the grid lines.
6.G.A.1Identify Subproblems
7STEP 7
See what one interior dot is worth
One interior dot adds 1 to the area.
1, 2, 3, 5 for 0, 1, 2, 4 interior dots → +1 cm² each
Holding one column of the table still and letting only the other column change is the cleanest way to see what that second quantity is doing on its own.
4.OA.C.5Look For A Pattern
Holding the boundary count still and adding one interior dot raises the area by exactly one square each time.
Why?
Because the rise is the same every time, the area is the boundary part plus that fixed step repeated once per interior dot.
🧱A constant step reaches any termWhen a list climbs by the same amount every time, any term is the starting value plus that step repeated, so you never have to write out the ones between.
Why?
The rule has to hold for every figure on the board, so what one interior dot is worth cannot depend on which figure is drawn.
🧱Proportional quantities scale togetherWhen two quantities keep a fixed ratio, doubling one doubles the other, so a ratio you know can stand in for one you do not.
8STEP 8
Write the formula and test it on all twelve figures
The formula is boundary ÷ 2 + interior − 1.
area = (boundary dots) ÷ 2 + (interior dots) - 1
A rule guessed from two or three rows is only a guess; testing it against every row of both tables is what turns it into an answer you can trust.
6.EE.A.2Look For A Pattern
Answer
boundary ÷ 2 + interior − 1 cm²
The units come out right: dots are counted (plain numbers) and the answer lands in cm² because the square between four neighbouring dots is 1 cm². The sizes are sensible - the biggest figure, the tilted square L, sits inside a 3 cm by 3 cm box and 5 cm² is comfortably under 9 cm², while the smallest, half a unit square, is 0.5 cm². Every area came out as a whole number of half-squares, which has to happen because chopping a dot-board polygon leaves only rectangles and half-rectangles, and that is exactly why the formula divides by 2. The formula also never returns a silly value: the smallest possible dot-board polygon has 3 boundary dots and no interior dot, giving 3 div 2 - 1 = 0.5 cm², the smallest triangle you can draw.
Takeaway
Count the dots on the edge, halve them, add the dots inside, take away 1 - and the area of any dot-board shape falls out without measuring a single length.
Set the unit: one little square is 1 cm²
Board (1): count the dots on each boundary
Board (1): find each area by cutting the figure up
Read the rule off table (1)
Board (2): count boundary dots and interior dots
Board (2): find each area by cutting the figure up
See what one interior dot is worth
Write the formula and test it on all twelve figures