Question : If the length of each of two equal sides of an isosceles triangle is 10 cm and the adjacent angle is $45^{\circ}$, then the area of the triangle is:
Option 1: $20\sqrt{2}$ sq. cm
Option 2: $12\sqrt{2}$ sq. cm
Option 3: $25\sqrt{2}$ sq. cm
Option 4: $15\sqrt{2}$ sq. cm
Correct Answer: $25\sqrt{2}$ sq. cm
Solution :
$AD = AB \sin 45^\circ$
$= 10\times\frac{1}{\sqrt2}$
$= 5\sqrt2$ cm
$\therefore$ Area of $\triangle ABC = \frac{1}{2}\times BC \times AD$
$=\frac{1}{2}\times 10 \times 5\sqrt2$
$= 25\sqrt2$ sq. cm
Hence, the correct answer is $25\sqrt2$ sq. cm.
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