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: $\frac{5}{2} a b$
Option 2: $3 a b$
Option 3: $4 a b$
Option 4: $\frac{3}{2} a b$

Question

Question : $△ P Q R$ is an isosceles triangle and $P Q = P R = 2 a$ unit, $Q R = a$ unit. Draw $P X ⊥ Q R$ , and find the length of $P X$ .
Option 1: $\sqrt{5} a$
Option 2: $\frac{\sqrt{5} a}{2}$
Option 3: $\frac{\sqrt{10} a}{2}$
Option 4: $\frac{\sqrt{15} a}{2}$

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Answer 1

Correct Answer: $\frac{3}{2} a b$

Solution :
Given that the 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.
Let's assume that $a = b$ .
The sides of the triangle become = $\sqrt{2 a^{2}}$ , $\sqrt{5 a^{2}}$ , $\sqrt{5 a^{2}} = a \sqrt{2}$ , $a \sqrt{5}$ , $a \sqrt{5}$ .
From this, we can say the triangle is an isosceles triangle with $\sqrt{5 a^{2}}$ being the two equal sides.
The area of an isosceles triangle is $\frac{1}{2} \times base \times \sqrt{(side)^{2} - (\frac{base}{2})^{2}}$ .
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 = $\frac{1}{2} \times a \sqrt{2} \times \sqrt{(a \sqrt{5})^{2} - (\frac{a \sqrt{2}}{2})^{2}}$
= $\frac{a}{2} \times \sqrt{2 [(a \sqrt{5})^{2} - (\frac{a \sqrt{2}}{2})^{2}]}$
= $\frac{a}{2} \times \sqrt{2 [(5 a^{2} - (\frac{a^{2}}{2})]}$
= $\frac{a}{2} \times \sqrt{9 a^{2}}$
= $\frac{3 a^{2}}{2}$
Since $a = b$ ⇒ $\frac{3}{2} a b = \frac{3 a^{2}}{2}$
Hence, the correct answer is $\frac{3}{2} a b$ .

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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: $\frac{5}{2} a b$
Option 2: $3 a b$
Option 3: $4 a b$
Option 4: $\frac{3}{2} a b$