Question

area of triangle =?

Explanation:

Step1: Identify the figure type

The figure shows a square (ABCD) and a triangle (ACO). Since ABCD is a square, AB = BC = 6 cm (assuming side length of square is 6 cm as AB is 6 cm).

Step2: Recall the formula for the area of a triangle

The area of a triangle is given by $\frac{1}{2} \times base \times height$. For triangle ACD or triangle ACB (but here the shaded triangle ACO, we can notice that the base and height can be related to the square. Wait, actually, the area of triangle ABC (if we consider ABC as a triangle) or more accurately, the area of triangle ACO: Wait, no, looking at the figure, the shaded triangle's base and height: Wait, actually, in the square ABCD, the area of triangle ACD is $\frac{1}{2} \times AB \times BC$? Wait, no, ABCD is a square, so AB = BC = CD = DA = 6 cm. The triangle ACD: base AD, height CD? No, wait, the shaded triangle: Wait, maybe the key is that the area of triangle ACO is half of the square? Wait, no, let's think again. Wait, the square has side 6 cm. The area of the square is $6 \times 6 = 36$ cm². The area of triangle ACD (which is half of the square) is $\frac{1}{2} \times 6 \times 6 = 18$ cm²? Wait, no, maybe the shaded triangle is equal to half of the square. Wait, let's check: In a square, the diagonal divides it into two triangles of equal area, each with area $\frac{1}{2} \times side^2$. But here, the shaded triangle: Wait, maybe the base is AB = 6 cm and height is BC = 6 cm? Wait, no, the formula for the area of triangle ACO: Wait, maybe the figure is such that the triangle's base and height are both 6 cm. Wait, let's assume that the triangle has base 6 cm and height 6 cm. Then the area is $\frac{1}{2} \times 6 \times 6 = 18$ cm²? Wait, no, maybe I made a mistake. Wait, the square's side is 6 cm, so the area of the square is 36. The triangle ACD (if D is top left, C top right, B bottom right, A bottom left) then AC is the diagonal? No, wait, ABCD is a square, so AB is horizontal, BC vertical. Then triangle ABC: base AB = 6, height BC = 6, area $\frac{1}{2} \times 6 \times 6 = 18$ cm². And the shaded triangle ACO: Wait, maybe O is the center? No, but the key is that the area of the triangle is half of the square's area? Wait, no, let's re-express. Wait, the problem is to find the area of the shaded triangle. Let's see: The square has side length 6 cm (since AB is 6 cm). The area of triangle ABC (which is the same as triangle ACD? No, ABC is a right triangle with legs 6 and 6, area $\frac{1}{2} \times 6 \times 6 = 18$ cm². Wait, maybe the shaded triangle is equal to triangle ABC? So the area is 18 cm²? Wait, maybe I'm overcomplicating. Let's use the formula: Area = $\frac{1}{2} \times base \times height$. If base is 6 cm and height is 6 cm, then Area = $\frac{1}{2} \times 6 \times 6 = 18$ cm².

Answer:

18 cm²