10. An equilateral triangle has the same perimeter as a rectangle of dimensions $b$ and $h$ (with $b>h$). The area of the triangle is $\sqrt{3}$ times the area of the rectangle. What is $\frac{b}{h}$?
(A) $\sqrt{3}$
(B) 2
(C) $\frac{3+\sqrt{3}}{2}$
(D) $\frac{3+\sqrt{5}}{2}$
(E) $\frac{7+3 \sqrt{5}}{2}$.
10. The answer is $\mathbf{( E )}$. Let $a$ be the side of the equilateral triangle, then the relations are $3 a=2(b+h)$, $\frac{a^{2} \sqrt{3}}{4}=\sqrt{3} b h$, from which $\left\{\begin{array}{l}b+h=\frac{3 a}{2} \\ b h=\frac{a^{2}}{4}\end{array}\right.$. The solving equation of the symmetric system in the unknowns $b$ and $h$ is $t^{2}-\frac{3 a}{2} t+\frac{a^{2}}{4}$, where $t$ is any of the unknowns. Therefore, $4 t^{2}-6 a t+a^{2}=0$, from which $t=\frac{3 \pm \sqrt{9-4}}{4} a$, hence $\frac{b}{h}=\frac{3+\sqrt{5}}{3-\sqrt{5}}$. Rationalizing, we get $\frac{b}{h}=\frac{7+3 \sqrt{5}}{2}$.