A hip roof is shown in the diagram. The roof front and back are in the shape of congruent isosceles trapezoids. The two sides are in the shape of congruent isosceles triangles. The shingles used on the roof cover 5.50 square meters per bundle. How many bundles are required to cover the roof?
Let $B$ be the base of one of the trapezoids, let $b \ne B$ be one of the bases of the trapezoids, and let $h$ be the height of one of the isosceles trapezoids. Then the area of one of the isosceles trapezoids is
\[\frac{B + b}{2} \cdot h = \frac{B + b}{2} \cdot \frac{B \sqrt{B^2 - B^2/4}}{2B} = \frac{\sqrt{3}}{4} B^2.\]Therefore, the area of both isosceles trapezoids is $\sqrt{3} B^2.$
Also, each isosceles triangle has base length $B$ and height length $h,$ so the area of each triangle is
\[\frac{B \cdot h}{2} = \frac{\sqrt{3}}{4} B^2.\]Therefore, the four triangles have a total area of $2 \sqrt{3} B^2.$
[asy]
unitsize(0.5 cm);
pair A, B, C, D, E, F, G, H;
B = (0,0);
C = (6,0);
G = (-1.5,-5);
H = (7.5,-5);
A = intersectionpoint(arc(G,3.75,0,180),arc(H,2.75,0,180));
draw(B--C--A--cycle);
draw(B--A);
draw(C--A);
draw(A--(A + (0,-11)));
draw(B--(B + (0,-11)));
draw(C--(C + (0,-11)));
draw(A--H--G);
draw(H--F--G,dashed);
label("$A$", A, NW);
label("$B$", B, NW);
label("$C$", C, NE);
label("$D$", H, S);
label("$E$", G, S);
label("$F$", (C + H)/2, N);
[/asy]
Let $D$ and $E$ be the bases of the triangles. Let $F$ be the midpoint of $\overline{DC},$ so triangles $BEF$ and $CDH$ are congruent. Let the midpoint of $\overline{BC}$ be $G.$ Since triangles $ABC$ and $DGA$ are similar, $AG = \frac{2}{3} AD$ and $BG = \frac{2}{3} BC.$ Therefore, $AG/GF = AD/DF = 2/1,$ so $DF = 2GF.$ By the Pythagorean Theorem on right triangle $ADF,$
\[AF = \sqrt{3} GF.\](Note that $GF = DC.$)
Then the area of $ABCD$ is
\[[ABCD]=[AHD]+[HDC]=\frac{1}{2}(D C)(HG) = DC \cdot DF = 2 DC \cdot GF = 2 AF \cdot GF = 2(\sqrt{3} GF)(GF) = 2 \sqrt{3} GF^2.\]Hence, the area of the four triangles is equal to the area of trapezoid $ABCD.$ Solving for $GF$, we find that
\[GF = \frac{\sqrt{3}}{2 B}.\]We want to interpret $GF$ as the horizontal length of the bases of the triangles. This length is
\[DF = GF \cdot \frac{AB}{AF} = GF \cdot \frac{AB}{\sqrt{3} GF} = \frac{AB}{\sqrt{3}}.\]Since this length is congruent to both $AB$ and $BC,$
\[\frac{AB}{\sqrt{3}} = AB + BC,\]so $AB/BC = \sqrt{3} - 1.$ Therefore, $b = BC = AB/(\sqrt{3} - 1) = B (\sqrt{3} + 1).$
Finally, the area of the roof is $2 \sqrt{3} B^2,$ so the number of bundles required to cover the roof is
\[\frac{2 \sqrt{3} B^2}{5.50}\]bundles.