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

A. $$25 \,c{m^2}$$

B. $$\frac{{25}}{2}\sqrt 2 \,c{m^2}$$

C. $$25\sqrt 2 \,c{m^2}$$

D. $$25\sqrt 3 \,c{m^2}$$

Answer: Option C

Solution(By Examveda Team)

$$\eqalign{ & {\text{Area of the traingle :}} \cr & = \frac{1}{2}ab\sin \theta \cr & = \left( {\frac{1}{2} \times 10 \times 10 \times \sin {{45}^ \circ }} \right)c{m^2} \cr & = \left( {\frac{1}{2} \times 10 \times 10 \times \frac{1}{{\sqrt 2 }}} \right)c{m^2} \cr & = \left( {\frac{{50}}{{\sqrt 2 }} \times \frac{{\sqrt 2 }}{{\sqrt 2 }}} \right)c{m^2} \cr & = 25\sqrt 2 \,c{m^2} \cr} $$