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

• A. $$\frac{4}{3} units$$
• B. $\frac{1}{2} units$
• C. 1 unit
• D. 2 unit

Answer 1

Answer:

C. 1 unit

Diagonal of square $=\sqrt{2}\times{side}$

=>\frac{3-x}{\sqrt{2}}=\sqrt{2}\times{\frac{1}{2}(x+1)}

=>3-x=\sqrt{2}\times{\sqrt{2}}\times{\frac{1}{2}}(x+1)

$3-x=x+1$

$x+x=3-1$

$2x=2$

$x=1 unit$