8. The answer is (D). Let $H$ be the foot of the altitude from $A$ to $B C$, and let $K$ be the foot of the altitude from $D$ to $A A^{\prime}$. The product of the areas of the indicated quadrilaterals is $(A H \cdot B C)\left(D K \cdot A A^{\prime}\right)=16$. We have the following equalities: $A A^{\prime}=2 A H$ by construction and $A A^{\prime} \cdot A H=B C \cdot D K$ by similarity between $A B C$ and $D A A^{\prime}$. Substituting the second into the product gives $A H \cdot A A^{\prime} \cdot A H \cdot A A^{\prime}=16$, which, using the first equality, becomes $A A^{\prime 4}=4 \cdot 16=64$, from which $A A^{\prime}=2 \sqrt{2}$.

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8. Let $A B C$ be a triangle and let $A^{\prime}$ be the symmetric point of $A$ with respect to $B C$; let $D A A^{\prime}$ be similar to $A B C$ and let $D^{\prime}$ be the symmetric point of $D$ with respect to $A A^{\prime}$. Given that the product of the areas of the quadrilaterals $A B A^{\prime} C$ and $A D A^{\prime} D^{\prime}$ is 16, we can say that $A A^{\prime} \ldots$
(A) is 1
(B) is $2 \sqrt[4]{2}$
(C) is 2
(D) is $2 \sqrt{2}$
(E) is not uniquely determined by the data. (Note: the similarity between $D A A^{\prime}$ and $A B C$ should be understood in an ordered manner: $D A / A B = A A^{\prime} / B C = A^{\prime} D / C A$)