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With Rhombus $ABCD$, with side length $x$, a cylinder with volume $x$ is formed by taping side $AB$ to side $CD$. Find $\sin \angle ABC$.

I drew a diagram of a rhombus, but I am unable to figure out the relation between the volume of the cylinder and the angle of the rhombus

I don't know if this useful or not, but I found the radius of the cylinder as $x/2\pi$

Other than that, I don't even know where to begin.

Any help is appreciated. Thanks in Advance!

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Start by drawing the rhombus, such that $AD$ and $BC$ are horizontal. Let's call the middle of these sides $M$ and $N$. You can keep $MN$ fixed and slowly move $AB$ over $CD$. You will form a cylinder where the axis is parallel to $MN$. But since $MN$ is also parallel to $AB$ and $CD$, the axis of the cylinder will form the same angle with the base circles as $AB$ forms with $BC$ or $AD$ (note that the angles are complementary, such that $\angle ABC=180^\circ-\angle BAD$. The height of the cylinder is then $x\sin\angle ABC$, and the volume is $$V=\pi r^2h=\pi\left(\frac{x}{2\pi}\right)^2x\sin\angle ABC$$ You should be able to finish from here

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