How to find the area of the following triangle I am stuck on the following problem: 

Let ABC be an isosceles triangle having two equal sides of length $20$ cm. and the angle between the 
  two equal sides is $45^{\circ}$. Then I have to find the area of $\triangle ABC$ . But I can not use any trigonometric formulae using $\sin$, $\cos$ etc. It is a ninth grade problem .

Can someone please help? Thanks in advance for your time. 
ADDENDUM : Though my problem has been solved , but I have got a similar problem which is identical with the one posted above apart from the fact that the angle between the two equal sides( each of length $20$ cm.) is given by $30^{\circ}$. In that case how can I solve the problem ? 
 A: Here's how I accomplished this problem...
Suppose we have $\triangle ABC$. I've drawn the triangle such that $A$ is on the right, $C$ is at the top, and $B$ is on the left. $AC = 20$ cm and $AB = 20$ cm. Drop a straight line down from $C$ to $AB$. Call the point where they meet $D$. We know that $AC = 20$ cm and that $A = 45^\circ$. 
We know that $\triangle ADC$ is a $45^\circ-45^\circ-90^\circ$ triangle. The hypotenuse $AC = 20$ cm. Thus, $AD = DC = \frac{20}{\sqrt2}$ cm. Area of a triangle is $A_\triangle = \frac12bh$, where we now know $DC = h$. So the area of $\triangle ABC$ is 
$A_{\triangle ABC} = \frac{1}{2}(20)\left(\frac{20}{\sqrt2}\right)$cm$^2 = 100\sqrt2$ cm$^2$.
A: Hint:
If the angle between the equal sides of length $l$ is $\pi/4$ than the height with respect one of this sides is $l \sin (\pi/4)$.
If you can not use $\sin$, note that the height with respect to one side is the side of the square that has $l$ as diagonal.
A: Hint: What happens if you reflect the triangle along the hypotenuse?
