The figure contains 12 triangles — option 4. Counting in an ordered way, rather than at random, is what makes such a question safe.

Step 1 — the rectangle with its two diagonals alone. The diagonals cross at the centre O and cut the rectangle into four small triangles, and each pair of adjacent small triangles combines into a larger one :

Step 2 — the horizontal segment. The segment from the left side to O cuts the left-hand small triangle SOP into two smaller triangles. Both are new, and SOP itself has already been counted, so this adds 2.

Step 3 — the vertical segment. In exactly the same way, the segment from O down to the base cuts the bottom small triangle ROS into two, adding another 2.

Total :

\(8+2+2=12\)

Why the two extra lines add so little. Each of them starts at the centre and ends on a side, so each lies wholly inside one of the four quarters and can only subdivide that one quarter. Neither crosses the interior of the rectangle as a whole, so neither creates any new large triangle — which is what makes 16 and 18 wrong. Option 2, ten, is what a candidate gets by counting the four small and four large triangles and then noticing only one of the two added lines.

The general method. Count triangles by the number of smallest regions they contain: first those made of one region, then of two, then of three, and so on, until no larger combination is possible. Working in that order guarantees that nothing is counted twice and nothing is missed.

Hence, the answer is 12.