Count all the rhombuses in a figure of small triangles

4.G.A.2 · take · grade 4

Archetype: Systematically Count Shapes in a Figure · step in a 5-type progression

How many rhombuses, large and small, can be found in the figure in all?

Figure description: The figure is an elongated hexagon made by joining identical small equilateral triangles edge to edge with no gaps. There are two rows of 7 small triangles, one above the other, separated by a horizontal centerline that runs from the pointed left tip to the pointed right tip. In each row the triangles alternate point-up and point-down, so the figure holds 14 small equilateral triangles in all (7 on top, 7 on the bottom). Count every rhombus you can find, where each rhombus is made of exactly 2 small triangles (one point-up triangle joined to a neighboring point-down triangle).

Show solution

Understand

An elongated hexagon is tiled by 14 identical small equilateral triangles in two rows of 7, one above the other, split by a horizontal centerline from the left tip to the right tip. In each row the triangles alternate point-up and point-down. I must count every rhombus, where each rhombus is exactly one point-up triangle joined to a neighboring point-down triangle (2 small triangles), in all of its possible orientations.

Givens

There are 14 identical small equilateral triangles: 7 in the top row and 7 in the bottom row.
In each row the triangles alternate point-up and point-down, sharing edges with no gaps.
A horizontal centerline runs from the pointed left tip to the pointed right tip.
A rhombus here is the union of one point-up triangle and one neighboring point-down triangle that share an edge, forming a 4-equal-sided diamond.

Unknowns

The total number of rhombuses (each made of 2 small triangles) in the figure.

Constraints

Each rhombus is made of exactly 2 small triangles (one up, one down) sharing one edge.
The strip is only one rhombus tall, so no rhombus made of 4 or more small triangles can fit; only 2-triangle rhombuses are possible.
A 2-triangle rhombus can lean in three directions, set by which of the three edges the up-triangle and down-triangle share.

Plan

#2 Make a Systematic List · also uses: #1 Draw a Diagram#7 Identify Subproblems

Label the 14 small triangles (top row t1..t7 left to right, bottom row b1..b7 left to right) and list rhombuses by orientation so none is missed or double-counted. Each rhombus is a fixed up+down pair sharing an edge; the shared edge can be horizontal (a tall vertical diamond) or one of two slanted edges (a left-leaning or a right-leaning diamond). Count each orientation as a subproblem, then add.

Execute

#1 Draw a Diagram 4.G.A.2

Number the top row from left to right as t1, t2, t3, t4, t5, t6, t7 and the bottom row from left to right as b1, b2, b3, b4, b5, b6, b7. In the top row the point-up triangles are t1, t3, t5, t7 and the point-down triangles are t2, t4, t6. In the bottom row the point-down triangles are b1, b3, b5, b7 and the point-up triangles are b2, b4, b6. Every rhombus is one up-triangle glued to a neighboring down-triangle along a shared edge.

\text{top: } t_1,\dots,t_7;\quad \text{bottom: } b_1,\dots,b_7

Naming every triangle first makes it possible to list each rhombus by the exact two triangles it covers, so the final count is checkable.

#2 Make a Systematic List 4.G.A.2

A point-up triangle in the top row sitting directly above a point-down triangle in the bottom row share their base on the centerline, and together they make a tall vertical diamond. These pairs are t1+b1, t3+b3, t5+b5, and t7+b7. That is 4 vertical rhombuses.

t_1{+}b_1,\ t_3{+}b_3,\ t_5{+}b_5,\ t_7{+}b_7 \;\Rightarrow\; 4

The four up-triangles whose tips touch the top edge each cap a down-triangle below, giving one upright diamond apiece.

#2 Make a Systematic List 4.G.A.2

A rhombus that leans one way is an up+down pair sharing a slanted edge tilted like a forward slash. In the top row these are t1+t2, t3+t4, t5+t6 (each up-triangle joined to the down-triangle on its right). In the bottom row these are b2+b3, b4+b5, b6+b7. That is 3 + 3 = 6 left-leaning rhombuses.

t_1{+}t_2,\ t_3{+}t_4,\ t_5{+}t_6,\ b_2{+}b_3,\ b_4{+}b_5,\ b_6{+}b_7 \;\Rightarrow\; 6

Sweeping left to right, each up-triangle pairs with the down-triangle on one side to make a slanted diamond; doing this in both rows gives six.

#2 Make a Systematic List 4.G.A.2

A rhombus that leans the other way is an up+down pair sharing a slanted edge tilted like a backslash. In the top row these are t2+t3, t4+t5, t6+t7. In the bottom row these are b1+b2, b3+b4, b5+b6. That is 3 + 3 = 6 right-leaning rhombuses.

t_2{+}t_3,\ t_4{+}t_5,\ t_6{+}t_7,\ b_1{+}b_2,\ b_3{+}b_4,\ b_5{+}b_6 \;\Rightarrow\; 6

Pairing each up-triangle with the down-triangle on its other side gives the mirror set of slanted diamonds, again six.

#7 Identify Subproblems 4.OA.A.3

The three lists do not overlap, because each rhombus has exactly one orientation. Add the vertical, left-leaning, and right-leaning counts to get the total.

4 + 6 + 6 = 16

Counting by orientation splits the job into three clean piles with no double-counting, so the sum is the grand total.

Answer: 16

Review

Each of the 7 up-triangles touches at most three neighboring down-triangles. Cross-check by up-triangle: the four up-triangles that reach the top edge (t1,t3,t5,t7) each make a vertical diamond plus two slanted ones, but the slanted ones at the very ends are shared with their row only once, and the three bottom up-triangles (b2,b4,b6) each make two slanted diamonds. Listing every pair explicitly gives 4 vertical + 6 left-leaning + 6 right-leaning = 16, with no pair counted twice and none missed.

Use Draw a Diagram (tool 1): shade each rhombus in its own copy of the figure, grouping copies by tilt. You will fill exactly 4 upright diamonds, 6 forward-slash diamonds, and 6 backslash diamonds, confirming 4 + 6 + 6 = 16.

Standards · min grade 4

4.G.A.2 Classify two-dimensional figures based on presence of parallel or perpendicular lines — Recognizing which up-and-down triangle pairs form a rhombus (four equal sides, opposite sides parallel) and listing them by orientation.
4.OA.A.3 Solve multi-step word problems using four operations with whole numbers — Adding the counts of the three rhombus orientations to get the total.

💡 Sort the diamonds by how they lean: 4 stand upright, 6 lean one way, 6 lean the other. Organized counting gives 4 + 6 + 6 = 16!