Semi-circle folded into a cone with a circular base

Question

From my 7th-grade math book:

The semicircle shown is folded to form a right circular cone so that the arc PQ becomes the circumference of the base. Find the diameter of the base,

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Let $\text{circumference of cone base}=C$ and $\text{diameter}=d$.

I think the diameter should be $\frac{2C}{\pi}=\frac{2 \cdot 5cm}{\pi}\approx 3.183cm$. But my book's answer key says $d=2.5cm$, which seems to be half the circumference ($\frac{5cm}{2}$). But I don't get the reasoning behind it. The semicircle's diameter is the cone base's circumference. And if that's the circumference ($\frac{\pi d}{2}$), then the diameter should be $2C/\pi$, shouldn't it?

Why? I'm very confused. I feel the amount of information provided is insufficient somehow.

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The question is unique and original. This question seems quite similar, with an answer which is appropriate in my case too. But the problem is the answer only states (facts I've already deduced), but doesn't explain.

Answer 1

So for the first part, you must find the circumference of the semicircle that is pictured (not including the straight side), which is $$c=0.5d\pi=2.5\pi$$ When the cone is formed, this circumference of the semicircle becomes the new circumference of the base of the cone, which is $$c=D\pi=2.5\pi$$ It can be implied that the diameter must be 2.5 centimeters.

Answer 2

The flat sector is folded to form a cone. Equating arc length before and after cone folding,

$$ 2 \pi r = L \theta,\; \sin \alpha = \frac{r}{L}$$

To eliminate $L$ multiply the equations and cancel $r$

$$\sin \alpha = \frac{\theta}{2 \pi}$$

The formula is adequate, can be used by sheet metal workers to make cones.

In our case

$$\sin \alpha = \frac{\pi}{2 \pi} = \frac12 \to \alpha = 30^{\circ}$$

so that when viewed sideways triangle $OPQ$ is equilateral,

$$ PQ=OP=OQ= L=\frac{5}{2}= 2.5 \;units. $$

If in case you want to make a cone with semi vertical angle $\alpha= 60^{\circ}$ then sector angle required from above formula is $ \theta = \pi \sqrt 3 $ radians or 311.77 degrees . One half of side view would appear as the triangle (60,30,90) degrees after roll folding shown at right.