+273 Consider rhombus ABCD with side length 6:
This is rolled to form a cylinder of volume \(\frac{27}{2}\) by taping \(\overline{AD}\) to \(\overline{BC}\):
What is \(\sin\left(\angle BAD \right)\)?
The cylinder formed has height h and radius r. Since the volume of the cylinder is 27/2, we have:
\pi r^2 h = \frac{27}{2}
Since ABCD is a rhombus, we know that AB=BC=6 and AC=BD=2*sqrt(6). The circumference of the cylinder's base is 2πr=AB+BC=12, so r=6/π.
Substituting this into the equation for the volume of the cylinder, we get:
\pi \left( \frac{6}{\pi} \right)^2 h = \frac{27}{2}
h = \frac{27/2}{9/\pi} = \frac{2\pi}{9}
The angle BAD is the angle between the bases of the cylinder. Since the height of the cylinder is h=2π/9, the sine of angle BAD is:
\sin \angle BAD = \frac{h}{2r} = \frac{2\pi/9}{6/\pi} = \boxed{\frac{\pi}{9}}
