Concept:Count triangles formed by the grid lines and the two main diagonals of the square.
Explanation:Type I: Each of the 4 corner squares is split by a diagonal into 2 small right triangles.
⇒4×2=8 triangles.
Type II: Right triangles whose hypotenuse lies on a main diagonal.
For one diagonal: 3 hypotenuses of length 1, 2 of length 2, 1 of length 3.
Each gives 2 triangles (one above and one below). So per diagonal:
(3+2+1)×2=12.
The 4 corner triangles are already counted in Type I, so unique triangles per diagonal =
12−4=8? No, careful: In the original, they subtract the 4 corner triangles from the total for both diagonals. But simpler: Excluding the corner triangles, each diagonal has: (3+2+1) = 6 hypotenuses, each giving 2 triangles, but the two triangles at the endpoints are the corner ones? Actually, the corner unit triangles are exactly those from hypotenuse of length 1 at the ends. So per diagonal, there are 6 triangles excluding the two corner ones? The original says per diagonal unique = 6, so both diagonals give 12 unique Type II triangles.
⇒ Type II unique count = 12.
Type III: Triangles with apex at the center (intersection of diagonals) and base along a grid line.
There are 4 horizontal rows and 4 vertical columns that form such triangles.
⇒4+4=8 triangles.
Total triangles =
8+12+8=28.
Answer:28 (Option D)