In an isosceles triangle, the measure of each of equal sides is 10 cm and the angle between them is 45°, then area of the triangle is

Answer: Option C

Remember: area of isosceles triangle
$$=\frac{1}{2}{a}^2\sin \theta$$
(θ is angle between equal sides)
$$\begin{array}{rl}& =\frac{1}{2}{\left(10\right)}^2\sin 45^{\circ }\\ & =\frac{100}{2}\times \frac{1}{\sqrt{2}}\\ & =\frac{50}{\sqrt{2}}\times \frac{\sqrt{2}}{\sqrt{2}}\\ & =25\sqrt{2}\text{ c}{\text{m}}^2\end{array}$$