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
A. $$25{\text{ c}}{{\text{m}}^2}$$
B. $$\frac{{25}}{2}\sqrt 2 {\text{ c}}{{\text{m}}^2}$$
C. $$25\sqrt 2 {\text{ c}}{{\text{m}}^2}$$
D. $$2\sqrt 3 {\text{ c}}{{\text{m}}^2}$$
Answer: Option C
Solution (By Examveda Team)
Remember: area of isosceles triangle$$ = \frac{1}{2}{a^2}\sin \theta $$
(θ is angle between equal sides)
$$\eqalign{ & = \frac{1}{2}{\left( {10} \right)^2}\sin {45^ \circ } \cr & = \frac{{100}}{2} \times \frac{1}{{\sqrt 2 }} \cr & = \frac{{50}}{{\sqrt 2 }} \times \frac{{\sqrt 2 }}{{\sqrt 2 }} \cr & = 25\sqrt 2 {\text{ c}}{{\text{m}}^2} \cr} $$
This Question Belongs to Arithmetic Ability >> Mensuration 2D
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