answer:We start by labeling everything first: [asy] fill((6,0)--(9,0)--(9,12)--(6,8)--cycle,gray(0.7)); draw((0,0)--(9,0)--(9,12)--cycle,linewidth(0.7)); draw((6,8)--(6,0),linewidth(0.7)); draw((5.6,0)--(5.6,0.4)--(6,0.4)); draw((8.6,0)--(8.6,0.4)--(9,0.4)); label("6",(3,0),S); label("10",(3,4),NW); label("3",(7.5,0),S); label("A",(0,0),SW); label("B",(9,0),SE); label("C",(9,12),NW); label("D",(6,0),S); label("E",(6,8),NW); [/asy] First of all, we can see that DE = 8 since we recognize a 3:4:5 triangle in ADE. Also, thanks to SAS Similarity (or AA also works), we see that triangle ABC sim triangle ADE. That means that ABC is also a 3:4:5 triangle. Since AB = 9, that means that AC = 15 and BC = 12. We find the shaded area of BCED by subtracting the areas of ABC and ADE. The area of ABC is simply dfrac{9 cdot 12}{2} = 54, and the area of ADE is dfrac{6 cdot 8}{2} = 24. Therefore, our desired area is 54 - 24 = 30} text{ cm^2.

answer:The wrapping with least rope length is the wrapping in which the four wraps are evenly spaced out. In this case, we can split the cylinder into four identical smaller cylinders, each of height 3 feet with a rope wrapped around once. The lateral area of each smaller cylinder is a rectangle with length 3 feet (the height of the cylinder) and width 2 feet (the circumference of the cylinder base). When this lateral area rectangle is rolled out with the rope atop it, the rope stretches from one corner of the rectangle to a diagonally opposite corner. Hence the rope length is the diagonal length of the rectangle, or sqrt{2^2+3^2}=sqrt{13} feet. Finally, the total length of the rope is four times this length, or 4sqrt{13} feet.

answer:Let x be the distance from the center O of the circle to the chord of length 10, and let y be the distance from O to the chord of length 14. Let r be the radius. Then, begin{align*} x^2+25&=r^2, y^2+49&=r^2, {rm so}qquad x^2+25&=y^2+49. {rm Therefore,}qquad x^2-y^2&=(x-y)(x+y)=24. end{align*}[asy] import olympiad; import geometry; size(100); defaultpen(linewidth(0.8)); draw(unitcircle); pair midpoint14 = (dir(40)+dir(140))/2; pair midpoint10 = (dir(-30)+dir(-150))/2; draw(Label("7",align=N),dir(40)--midpoint14); draw(Label("7",align=N),midpoint14--dir(140)); draw(Label("5",align=S),dir(-30)--midpoint10); draw(Label("5",align=S),midpoint10--dir(-150)); draw(Label("y",align=E),origin--midpoint14); draw(Label("x",align=E),origin--midpoint10); draw(Label("r",align=E),dir(40)--origin); draw(Label("r",align=E),dir(-30)--origin); label("O",origin,W); [/asy] If the chords are on the same side of the center of the circle, x-y=6. If they are on opposite sides, x+y=6. But x-y=6 implies that x+y=4, which is impossible. Hence x+y=6 and x-y=4. Solve these equations simultaneously to get x=5 and y=1. Thus, r^2=50, and the chord parallel to the given chords and midway between them is two units from the center. If the chord is of length 2d, then d^2+4=50, d^2=46, and a=(2d)^2=184.

answer:I know that angle bisectors in a triangle divide the opposite side in the same ratio as the adjacent sides. So, if AD is an angle bisector, then BD/DC = AB/AC. I can plug in the given values to get BD/DC = 6/10 = 3/5. This means that BD = 3k and DC = 5k for some constant k. Since BC = 8, I can add up BD and DC to get 8 = 3k + 5k. Solving for k, I get k = 1. So, BD = 3 and DC = 5. Now, I can use the Stewart's theorem, which relates the lengths of the sides and the angle bisector in a triangle. Stewart's theorem says that AB^2 cdot DC + AC^2 cdot BD = AD^2 cdot BC + BD cdot DC cdot BC. Plugging in the values, I get 6^2 cdot 5 + 10^2 cdot 3 = AD^2 cdot 8 + 3 cdot 5 cdot 8. Simplifying, I get 180 + 300 = AD^2 cdot 8 + 120. Subtracting 120 from both sides, I get 360 = AD^2 cdot 8. Dividing by 8, I get AD^2 = 45.