7. $A B C D$ is an isosceles trapezoid with side lengths $A B=15$, $B C=D A=5$, $C D=7$. On the sides of $A B C D$, four squares are constructed externally to $A B C D$. Referring to the figure, what is the area of the polygon $A_{1} A_{2} B_{1} B_{2} C_{1} C_{2} D_{1} D_{2}$?
(A) $297+35 \sqrt{5}$
(E) $200 \sqrt{5}$
(B) 396
(C) 423
(D) $105+144 \sqrt{5}$

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7. The answer is $\mathbf{( C )}$. First, observe that the height of the trapezoid $A B C D$ is 3: indeed, if $H$ is the foot of the perpendicular from $C$ to $A B$, the triangle $H B C$ is right-angled at $H$ and the equalities $B C=5$ and $H B=\frac{1}{2}(A B-C D)=4$ hold, so (by the Pythagorean theorem) it follows that $H C=3$ as stated. The given polygon can be decomposed into the union of four squares (having a total area of $5^{2}+5^{2}+15^{2}+7^{2}=324$), the original trapezoid (having an area of $\left.\frac{(15+7) \cdot 3}{2}=33\right)$, and four triangles, whose total area is exactly twice the area of the trapezoid (i.e., 66). The reason for this last statement is that the triangle determined by two segments of lengths $\ell_{1}, \ell_{2}$ forming an angle $\theta$ has the same area as a triangle determined by two segments of lengths $\ell_{1}, \ell_{2}$ forming an angle $180^{\circ}-\theta$: applying this fact repeatedly, we find that the following pairs of triangles have the same area: $A A_{1} A_{2}$ and $A B D, B B_{1} B_{2}$ and $A B C, C C_{1} C_{2}$ and $B C D, D D_{1} D_{2}$ and $A D C$. Since the sum of the areas of triangles $A B D$ and $B C D$ is the area of the trapezoid, and the same holds for the areas of $A D C$ and $A B C$, it follows that the areas of triangles $A A_{1} A_{2}, B B_{1} B_{2}, C C_{1} C_{2}, D D_{1} D_{2}$ sum to twice the area of the trapezoid. The correct answer is therefore $324+33+66=423$.