Question : Three sides of a triangle are $\sqrt{a^2+b^2}, \sqrt{(2 a)^2+b^2}$, and $\sqrt{a^2+(2 b)^2}$ units. What is the area (in unit squares) of the triangle?
Option 1: $\frac{5}{2} ab $
Option 2: $3 ab$
Option 3: $4 ab$
Option 4: $\frac{3}{2} ab $
Correct Answer: $\frac{3}{2} ab $
Solution :
Given that the three sides of a triangle are $\sqrt{a^2+b^2}$, $\sqrt{(2a)^2+b^2}$, and $\sqrt{a^2+(2b)^2}$ units.
Let's assume that $a=b$.
The sides of the triangle become = $\sqrt{2a^2}$, $\sqrt{5a^2}$, $\sqrt{5a^2} =a\sqrt{2}$, $a\sqrt{5}$, $a\sqrt{5}$.
From this, we can say the triangle is an isosceles triangle with $\sqrt{5a^2}$ being the two equal sides.
The area of an isosceles triangle is $\frac{1}{2} \times \text{base} \times \sqrt{(\text{side})^2 - (\frac{\text{base}}{2})^2}$.
Here, the base of the isosceles triangle is $a\sqrt{2}$ and the equal sides are $a\sqrt{5}$.
Substituting these values into the formula,
Area = $\frac{1}{2} \times a\sqrt{2} \times \sqrt{(a\sqrt{5})^2 - (\frac{a\sqrt{2}}{2})^2}$
= $\frac{a}{2} \times \sqrt{2[(a\sqrt{5})^2 - (\frac{a\sqrt{2}}{2})^2]}$
= $\frac{a}{2} \times \sqrt{2[(5a^2 - (\frac{a^2}{2})]}$
= $\frac{a}{2} \times \sqrt{9a^2}$
= $\frac{3a^2}{2}$
Since $a=b$ ⇒ $\frac{3}{2}ab=\frac{3a^2}{2}$
Hence, the correct answer is $\frac{3}{2}ab$.
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