Problem
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.
Naming every triangle first makes it possible to list each rhombus by the exact two triangles it covers, so the final count is checkable.
4.G.A.2Draw A DiagramA 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.
The four up-triangles whose tips touch the top edge each cap a down-triangle below, giving one upright diamond apiece.
4.G.A.2Make A Systematic ListA rhombus that leans one way is an up+down pair sharing a slanted edge tilted like a backslash. 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.
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.
4.G.A.2Make A Systematic ListA rhombus that leans the other way is an up+down pair sharing a slanted edge tilted like a forward slash. 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.
Pairing each up-triangle with the down-triangle on its other side gives the mirror set of slanted diamonds, again six.
4.G.A.2Make A Systematic ListThe 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.
Counting by orientation splits the job into three clean piles with no double-counting, so the sum is the grand total.
4.OA.A.3Identify SubproblemsThe total number of rhombuses in the figure equals the vertical count plus the left-leaning count plus the right-leaning count.
The three orientation lists cover every rhombus once, cutting the whole set into parts with no gaps and no overlaps, and parts that cover a whole with no gap and no overlap add back to that whole.
The lists really do overlap nowhere and leave no rhombus out, so sorting by lean is a clean split rather than a lossy one.
Each rhombus is one point-up triangle joined to one point-down triangle along a single shared edge, and that one edge tilts just one way, so each rhombus matches exactly one orientation group and never two.
Turning the three separate group counts into one grand total means adding three piles, which reach the same total no matter which two you join first.
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!