← Count all the rhombuses in a figure of small triangles · Systematically Count Shapes in a Figure

Count all the rhombuses in a figure of small triangles · 5 practice problems

4.G.A.2

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Generated variants — 5

Freshly produced from the archetype’s parameters — problem, figure, and solution derived together.

Variant 1 easy

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 3 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 6 small equilateral triangles in all (3 on top, 3 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).

Your answer

Show solution

1 · Understandwhat's really being asked

An elongated hexagon is tiled by 6 identical small equilateral triangles in two rows of 3, 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 6 identical small equilateral triangles: 3 in the top row and 3 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 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.

2 · Planchoose the strategy

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

Label the 6 small triangles (top row t1..t3 left to right, bottom row b1..b3 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.

3 · Execute5 carry out the plan

1Label the small triangles

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

Number the top row left to right as t1 ... t3 and the bottom row left to right as b1 ... b3. In the top row the odd-indexed triangles point up and the even-indexed point down; the bottom row is the mirror. Every rhombus is one up-triangle glued to a neighboring down-triangle along a shared edge.

\text{top: } t_1,\dots,t_{3};\quad \text{bottom: } b_1,\dots,b_{3}

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

2Count the tall vertical rhombuses (shared edge on the centerline)

#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, making a tall vertical diamond. There are 2 such aligned pairs.

\text{vertical rhombuses} = \frac{3+1}{2} = 2

Each up-triangle whose tip touches the top edge caps a down-triangle below, giving one upright diamond apiece.

3Count the left-leaning rhombuses

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

A left-leaning rhombus is an up+down pair sharing a slanted edge tilted like a forward slash. Sweeping each row left to right, every up-triangle pairs with the down-triangle on one side; across both rows there are 2 of them.

\text{left-leaning} = 3 - 1 = 2

Each up-triangle pairs with the down-triangle on one side to make a slanted diamond; doing this in both rows gives the count.

4Count the right-leaning rhombuses

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

A right-leaning rhombus is an up+down pair sharing the other slanted edge, tilted like a backslash. Pairing each up-triangle with the down-triangle on its other side gives the mirror set, again 2.

\text{right-leaning} = 3 - 1 = 2

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

5Add the three orientations

#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.

2 + 2 + 2 = 6

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

Answer: 6

4 · Reviewdoes it hold up?

Listing every up+down pair by orientation gives 2 vertical + 2 left-leaning + 2 right-leaning = 6, with no pair counted twice and none missed.

Another way: Use Draw a Diagram (tool 1): shade each rhombus in its own copy of the figure, grouping copies by tilt, then add the three groups to confirm the total.

Standardsmin 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.

💡Takeaway. Sort the diamonds by how they lean: some stand upright and the rest lean two ways. Organized counting gives 2 + 2 + 2 = 6!

Variant 2 medium

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 5 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 10 small equilateral triangles in all (5 on top, 5 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).

Your answer

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1 · Understandwhat's really being asked

An elongated hexagon is tiled by 10 identical small equilateral triangles in two rows of 5, 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 10 identical small equilateral triangles: 5 in the top row and 5 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 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.

2 · Planchoose the strategy

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

Label the 10 small triangles (top row t1..t5 left to right, bottom row b1..b5 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.

3 · Execute5 carry out the plan

1Label the small triangles

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

Number the top row left to right as t1 ... t5 and the bottom row left to right as b1 ... b5. In the top row the odd-indexed triangles point up and the even-indexed point down; the bottom row is the mirror. Every rhombus is one up-triangle glued to a neighboring down-triangle along a shared edge.

\text{top: } t_1,\dots,t_{5};\quad \text{bottom: } b_1,\dots,b_{5}

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

2Count the tall vertical rhombuses (shared edge on the centerline)

#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, making a tall vertical diamond. There are 3 such aligned pairs.

\text{vertical rhombuses} = \frac{5+1}{2} = 3

Each up-triangle whose tip touches the top edge caps a down-triangle below, giving one upright diamond apiece.

3Count the left-leaning rhombuses

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

\text{left-leaning} = 5 - 1 = 4

Each up-triangle pairs with the down-triangle on one side to make a slanted diamond; doing this in both rows gives the count.

4Count the right-leaning rhombuses

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

A right-leaning rhombus is an up+down pair sharing the other slanted edge, tilted like a backslash. Pairing each up-triangle with the down-triangle on its other side gives the mirror set, again 4.

\text{right-leaning} = 5 - 1 = 4

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

5Add the three orientations

#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.

3 + 4 + 4 = 11

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

Answer: 11

4 · Reviewdoes it hold up?

Listing every up+down pair by orientation gives 3 vertical + 4 left-leaning + 4 right-leaning = 11, with no pair counted twice and none missed.

Another way: Use Draw a Diagram (tool 1): shade each rhombus in its own copy of the figure, grouping copies by tilt, then add the three groups to confirm the total.

Standardsmin 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.

💡Takeaway. Sort the diamonds by how they lean: some stand upright and the rest lean two ways. Organized counting gives 3 + 4 + 4 = 11!

Variant 3 medium

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).

Your answer

Show solution

1 · Understandwhat's really being asked

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 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.

2 · Planchoose the strategy

#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.

3 · Execute5 carry out the plan

1Label the small triangles

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

Number the top row left to right as t1 ... t7 and the bottom row left to right as b1 ... b7. In the top row the odd-indexed triangles point up and the even-indexed point down; the bottom row is the mirror. 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.

2Count the tall vertical rhombuses (shared edge on the centerline)

#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, making a tall vertical diamond. There are 4 such aligned pairs.

\text{vertical rhombuses} = \frac{7+1}{2} = 4

Each up-triangle whose tip touches the top edge caps a down-triangle below, giving one upright diamond apiece.

3Count the left-leaning rhombuses

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

\text{left-leaning} = 7 - 1 = 6

Each up-triangle pairs with the down-triangle on one side to make a slanted diamond; doing this in both rows gives the count.

4Count the right-leaning rhombuses

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

A right-leaning rhombus is an up+down pair sharing the other slanted edge, tilted like a backslash. Pairing each up-triangle with the down-triangle on its other side gives the mirror set, again 6.

\text{right-leaning} = 7 - 1 = 6

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

5Add the three orientations

#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

4 · Reviewdoes it hold up?

Listing every up+down pair by orientation gives 4 vertical + 6 left-leaning + 6 right-leaning = 16, with no pair counted twice and none missed.

Another way: Use Draw a Diagram (tool 1): shade each rhombus in its own copy of the figure, grouping copies by tilt, then add the three groups to confirm the total.

Standardsmin 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.

💡Takeaway. Sort the diamonds by how they lean: some stand upright and the rest lean two ways. Organized counting gives 4 + 6 + 6 = 16!

Variant 4 medium

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 9 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 18 small equilateral triangles in all (9 on top, 9 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).

Your answer

Show solution

1 · Understandwhat's really being asked

An elongated hexagon is tiled by 18 identical small equilateral triangles in two rows of 9, 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 18 identical small equilateral triangles: 9 in the top row and 9 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 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.

2 · Planchoose the strategy

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

Label the 18 small triangles (top row t1..t9 left to right, bottom row b1..b9 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.

3 · Execute5 carry out the plan

1Label the small triangles

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

Number the top row left to right as t1 ... t9 and the bottom row left to right as b1 ... b9. In the top row the odd-indexed triangles point up and the even-indexed point down; the bottom row is the mirror. Every rhombus is one up-triangle glued to a neighboring down-triangle along a shared edge.

\text{top: } t_1,\dots,t_{9};\quad \text{bottom: } b_1,\dots,b_{9}

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

2Count the tall vertical rhombuses (shared edge on the centerline)

#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, making a tall vertical diamond. There are 5 such aligned pairs.

\text{vertical rhombuses} = \frac{9+1}{2} = 5

Each up-triangle whose tip touches the top edge caps a down-triangle below, giving one upright diamond apiece.

3Count the left-leaning rhombuses

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

\text{left-leaning} = 9 - 1 = 8

Each up-triangle pairs with the down-triangle on one side to make a slanted diamond; doing this in both rows gives the count.

4Count the right-leaning rhombuses

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

A right-leaning rhombus is an up+down pair sharing the other slanted edge, tilted like a backslash. Pairing each up-triangle with the down-triangle on its other side gives the mirror set, again 8.

\text{right-leaning} = 9 - 1 = 8

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

5Add the three orientations

#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.

5 + 8 + 8 = 21

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

Answer: 21

4 · Reviewdoes it hold up?

Listing every up+down pair by orientation gives 5 vertical + 8 left-leaning + 8 right-leaning = 21, with no pair counted twice and none missed.

Another way: Use Draw a Diagram (tool 1): shade each rhombus in its own copy of the figure, grouping copies by tilt, then add the three groups to confirm the total.

Standardsmin 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.

💡Takeaway. Sort the diamonds by how they lean: some stand upright and the rest lean two ways. Organized counting gives 5 + 8 + 8 = 21!

Variant 5 hard

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 11 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 22 small equilateral triangles in all (11 on top, 11 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).

Your answer

Show solution

1 · Understandwhat's really being asked

An elongated hexagon is tiled by 22 identical small equilateral triangles in two rows of 11, 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 22 identical small equilateral triangles: 11 in the top row and 11 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 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.

2 · Planchoose the strategy

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

Label the 22 small triangles (top row t1..t11 left to right, bottom row b1..b11 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.

3 · Execute5 carry out the plan

1Label the small triangles

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

Number the top row left to right as t1 ... t11 and the bottom row left to right as b1 ... b11. In the top row the odd-indexed triangles point up and the even-indexed point down; the bottom row is the mirror. Every rhombus is one up-triangle glued to a neighboring down-triangle along a shared edge.

\text{top: } t_1,\dots,t_{11};\quad \text{bottom: } b_1,\dots,b_{11}

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

2Count the tall vertical rhombuses (shared edge on the centerline)

#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, making a tall vertical diamond. There are 6 such aligned pairs.

\text{vertical rhombuses} = \frac{11+1}{2} = 6

Each up-triangle whose tip touches the top edge caps a down-triangle below, giving one upright diamond apiece.

3Count the left-leaning rhombuses

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

\text{left-leaning} = 11 - 1 = 10

Each up-triangle pairs with the down-triangle on one side to make a slanted diamond; doing this in both rows gives the count.

4Count the right-leaning rhombuses

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

A right-leaning rhombus is an up+down pair sharing the other slanted edge, tilted like a backslash. Pairing each up-triangle with the down-triangle on its other side gives the mirror set, again 10.

\text{right-leaning} = 11 - 1 = 10

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

5Add the three orientations

#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.

6 + 10 + 10 = 26

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

Answer: 26

4 · Reviewdoes it hold up?

Listing every up+down pair by orientation gives 6 vertical + 10 left-leaning + 10 right-leaning = 26, with no pair counted twice and none missed.

Another way: Use Draw a Diagram (tool 1): shade each rhombus in its own copy of the figure, grouping copies by tilt, then add the three groups to confirm the total.

Standardsmin 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.

💡Takeaway. Sort the diamonds by how they lean: some stand upright and the rest lean two ways. Organized counting gives 6 + 10 + 10 = 26!

Where next?

← Back to the problem: Count all the rhombuses in a figure of small triangles