The problem states that the area of triangle OFC is $33 \text{ cm}^2$. We need to find the area of triangle OBG. From the figure, we can see that the segments AH, HB, BC, CD, DE, EF, and FG are equal in length.

1. Problem Description

The problem states that the area of triangle OFC is $33 \text{ cm}^2$. We need to find the area of triangle OBG. From the figure, we can see that the segments AH, HB, BC, CD, DE, EF, and FG are equal in length.

2. Solution Steps

Let the length of each segment AH, HB, BC, CD, DE, EF, and FG be $x$.

Thus, AH = HB = BC = CD = DE = EF = FG = $x$.

The height of triangle OFC is $3x$.

The height of triangle OBG is $6x$.

Since triangles OFC and OBG share the same vertex O and the base of both triangles lies on the line AG, the ratio of their areas is equal to the ratio of their heights.

Let the area of triangle OFC be $A_{OFC}$ and the area of triangle OBG be $A_{OBG}$. We are given that $A_{OFC} = 33$.

We have

$\frac{A_{OBG}}{A_{OFC}} = \frac{6x}{3x} = 2$.

Therefore,

$A_{OBG} = 2 \times A_{OFC} = 2 \times 33 = 66$.

The area of triangle OBG is $66 \text{ cm}^2$.

3. Final Answer

66 cm$^2$