18.74 Given that the diagonal of square I is $a+b$, and the area of square II is twice the area of square I. Then the perimeter of square II is
(A) $(a+b)^{2}$.
(B) $\sqrt{2}(a+b)^{2}$.
(C) $2(a+b)$.
(D) $\sqrt{8}(a+b)$.
(E) $4(a+b)$.
(11th American High School Mathematics Examination, 1960)
[Solution] The area of square I is $\left(\frac{a+b}{\sqrt{2}}\right)^{2}=\frac{1}{2}(a+b)^{2}$, so the area of square II is $(a+b)^{2}$, and its perimeter is $4(a+b)$.
Therefore, the answer is $(E)$.