Question : If the side of a square is $\frac{1}{2}(x+1)$ units and its diagonal is $\frac{3-x}{\sqrt{2}}$ units, then the length of the side of the square would be:
Option 1: $\frac{4}{3}$ units
Option 2: $\frac{1}{2}$ unit
Option 3: 1 unit
Option 4: 2 units
Correct Answer: 1 unit
Solution :
Given: Side = $\frac{1}{2}(x+1)$ and diagonal = $\frac{3−x}{\sqrt2}$
We know that diagonal = $\sqrt 2$ × side
Putting the values, we get:
⇒ $\frac{3−x}{\sqrt2} = \sqrt2(\frac{1}{2}(x+1))$
⇒ $3−x =x+1$
⇒ $x = 1$
Hence, the correct answer is 1 unit.
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