Triangles $OAB,OBC,OCD$ are isoceles triangles with $\angle OAB=\angle OBC=\angle OCD=\angle 90^o$. Find the area of the triangle $OAB$ if the area of the triangle $OCD$ is 12.
1. Given that triangles \(OAB\), \(OBC\), and \(OCD\) are isosceles right triangles with \(\angle OAB = \angle OBC = \angle OCD = 90^\circ\), we can infer that these triangles are 45-45-90 triangles. In a 45-45-90 triangle, the legs are equal, and the hypotenuse is \(\sqrt{2}\) times the length of each leg.

2. Let \(OC = CD = x\). Since \(\triangle OCD\) is a 45-45-90 triangle, the area of \(\triangle OCD\) is given by:
   \[
   \text{Area of } \triangle OCD = \frac{1}{2} \times OC \times CD = \frac{1}{2} \times x \times x = \frac{x^2}{2}
   \]
   We are given that the area of \(\triangle OCD\) is 12, so:
   \[
   \frac{x^2}{2} = 12 \implies x^2 = 24 \implies x = 2\sqrt{6}
   \]

3. Next, we need to find the area of \(\triangle OAB\). Since \(\triangle OAB\) is also a 45-45-90 triangle, let \(OA = AB = y\). The hypotenuse \(OB\) of \(\triangle OAB\) is:
   \[
   OB = y\sqrt{2}
   \]

4. Since \(\triangle OBC\) is also a 45-45-90 triangle, and \(OB = BC\), we have:
   \[
   OB = BC = \frac{x}{\sqrt{2}} = \frac{2\sqrt{6}}{\sqrt{2}} = \sqrt{12} = 2\sqrt{3}
   \]

5. Therefore, \(y\sqrt{2} = 2\sqrt{3}\), solving for \(y\):
   \[
   y = \frac{2\sqrt{3}}{\sqrt{2}} = \frac{2\sqrt{3} \cdot \sqrt{2}}{2} = \sqrt{6}
   \]

6. The area of \(\triangle OAB\) is:
   \[
   \text{Area of } \triangle OAB = \frac{1}{2} \times OA \times AB = \frac{1}{2} \times y \times y = \frac{1}{2} \times (\sqrt{6})^2 = \frac{1}{2} \times 6 = 3
   \]

The final answer is \(\boxed{3}\).