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1. 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. $25 \sqrt{3} c m^{2}$

Solution:

\begin{aligned} Area of the traingle : \\ = \frac{1}{2} a b \sin θ \\ = (\frac{1}{2} \times 10 \times 10 \times \sin 45^{\circ}) c m^{2} \\ = (\frac{1}{2} \times 10 \times 10 \times \frac{1}{\sqrt{2}}) c m^{2} \\ = (\frac{50}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}}) c m^{2} \\ = 25 \sqrt{2} c m^{2} \end{aligned}

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