# If the length of the equal two sides of a isosceles triangle is 20 cm and the angle between those two sides is 45 degrees, then what is the area of the triangle?

$100 \sqrt{2}$ $u n i {t}^{2}$
Area of triangle $= \frac{1}{2} b c \sin A$ where $A = {45}^{\circ} , b = c = 20 c m$
Area of triangle = $\frac{1}{2} \cdot 20 \cdot 20 \cdot \sin {45}^{\circ} = 200 \cdot \frac{1}{\sqrt{2}} = \frac{200}{\sqrt{2}} = 100 \sqrt{2}$