Question

The diagonal of square I is $a+b$. The perimeter of square II with twice the area of I is

• A. $(a+b{)}^2$
• B. $\sqrt{2}(a+b{)}^2$
• C. $2(a+b)$
• D. $\sqrt{8}(a+b)$
• E. $4(a+b)$

Answer 1

The correct option is E $4(a+b)$

Let $s$ be a side of square I; S, a side of square II. Area $I=s^2$,
Area $II=S^2$. Since $S^2=2s^2,S=s\sqrt{2}$. But
$s=\frac{d}{\sqrt{2}}=(a+b)/\sqrt{2};\therefore S=\frac{(a+b)}{\sqrt{2}}\sqrt{2}=a+b$;
$\therefore$ the perimeter of $II=4S=4(a+b)$.