Answer:
B: x√2 + 3√2
Step-by-step explanation:
[tex]\boxed{\begin{minipage}{9 cm}\underline{Pythagoras Theorem} \\\\$a^2+b^2=c^2$\\\\where:\\ \phantom{ww}$\bullet$ $a$ and $b$ are the legs of the right triangle. \\ \phantom{ww}$\bullet$ $c$ is the hypotenuse (longest side) of the right triangle.\\\end{minipage}}[/tex]
The diagonal of a square is the hypotenuse of a right triangle with legs equal to the side length of the square.
Given:
Substitute the given side length into the formula and solve for c:
[tex]\begin{aligned}\implies c^2&=(x+3)^2+(x+3)^2\\c^2&=2(x+3)^2\\c&=\sqrt{2(x+3)^2}\\c&=\sqrt{2}\sqrt{(x+3)^2}\\c&=\sqrt{2}(x+3)\\c&=x\sqrt{2}+3\sqrt{2}\end{aligned}[/tex]
Therefore, the diagonal of the square with side length (x + 3) is:
