Counting the Triangles

Let the area of the smallest $\triangle$ 's at the corners of the big square be $a$ .

Then, there are:

• $8\times a$ - $\triangle$ 's at the four corners of the big square
• $4\times 2a$ - $\triangle$ 's, each formed from two $a$ - $\triangle$ 's, at the four corners of the big square
• $4\times 4a$ - $\triangle$ 's whose right-angles meet at the center of the big square
• $4\times 8a$ - $\triangle$ 's whose right-angles are at the four corners of the big square
• $\boxed {20} \space \triangle$ 's in total

8 triangles on corners. 8 triangles on sides and 4 in center. So total = 20 8 triangles in four corner. 4 triangles in four side. 4 triangles in diagonally. 4 triangles in middle. Best way to solve, just counting.

8 triangles on corners. 8 triangles on sides and 4 in center.

So total = 20

Diagram In the diagram $AMB,BMC,CMD,DMA,ABC,BCD,CDA,DAB,AEF,FBG,GCH,HDE,AIF,FJB,BJG,GKC,CKH,HLD,DLE,EIA=\boxed{20}$ triangles.

8 triangles in four corner. 4 triangles in four side. 4 triangles in diagonally. 4 triangles in middle. Best way to solve, just counting.

start counting from the larger triangle nd then move to the smaller possible ones

you can count the triangles to get to 20 but I was wondering if we could use some combinatorics to solve the problem.

Just counting :D

8 △ one part area wich hypotenuses trace the perimeter of the fig.

4 △ composed of 2 △ with right angle in corners

4 △ composed of 4 △ and 2 □ with right angle in corners

4 △ composed of 2 △ and 1 □ with right angle in the middle