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Question 66
Length of each equal side of an isosceles triangle is 10 cm and the included angle between those two sides is $$45^\circ$$. Find the area of the triangle.
Solution
Area of isosceles triangle = $$\frac{1}{2}\times(a)^2\times sin(\theta)$$, where $$a$$ is one of the equal sides and $$\theta$$ is the angle between them.
=> Area =Â $$\frac{1}{2}\times(10)^2\times sin(45^\circ)$$
= $$50\times\frac{1}{\sqrt2}$$
= $$25\sqrt2$$ $$cm^2$$
=> Ans - (A)
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