5) The answer is (B). Let $I$ and $L$ be the midpoints of sides $DC$ and $AD$ respectively. Applying the Pythagorean theorem to the right triangle $DIG$ gives $GI=\sqrt{13^2-12^2}=5$, and applying the same theorem to the right triangle $DHL$ gives $HL=\sqrt{15^2-12^2}=9$. The diagonals of quadrilateral $EFGH$ are perpendicular, and we have $EG=5+24+5=34$, $HF=9+24+9=42$. The required area is therefore $34 \cdot \frac{42}{2}=714$.

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5) Consider the square $A B C D$ with side length 24. Outside the square, construct the isosceles triangles $A E B, C G D$ with side length 13 and bases $A B$ and $C D$, and the isosceles triangles $B F C, D H A$ with side length 15 and bases $B C, D A$. What is the area of the quadrilateral EFGH?
(A) 357
(B) 714
(C) 912
(D) 952
(E) 1428.