So I've tried everything! I'm stuck with this problem for at least an hour...

A sphere is divided in 2 sphere caps. One bigger and one smaller, of course. The surface area of one ($P_1$) is $16 cm^2$ and the other ($P_2$) $48 cm^2$. In these areas the area of the circle is also included. Where $A$ is surface of the whole ball, and $a$ is the surface of the circle: $16+48=64=A+2a$. The task is to find the perimeter of the circle.

Where $R$ is the radius of the sphere, and $h_1$ is the height of the smaller sphere cap, $r$ (radius of the circle I'm looking for) is: $$r=\sqrt{R^2-(R-h_1)^2}$$

Surface area of a sphere cap is $$A_1=2R\pi h_1$$surface of a circle is $$a=r^2\pi$$ and perimiter of the circle is $$o=2r\pi$$

Therefore, $$P_1=A_1+a$$ and $$P_2=A_2+a$$ Now if I enter the data: $$16=2R\pi h_1+a$$ $$48=2R\pi (R-h_1)+a$$ If $a$ is substituted: $$16=2R\pi h_1+\pi(R^2-(R-h_1)^2)$$ $$48=2R\pi (R-h_1)+\pi(R^2-(R-h_1)^2)$$ Subtracting first from the second one:$$32=2R\pi (R-h_1)-2R\pi h_1$$

Even if I enter it all in Wolfram Alpha, it doesn't give the correct result. ($\pi4\sqrt{3}cm$) I know there's a simpler solution and that I'm missing something... (Just give me a hint, it doesn't have to be full solution.)

EDIT: What I've also noticed is that I get just the surface of a part of the sphere when I subtract $48-16$, with height: $2R-2h_1$, that's without the areas of the circles. But still, I can't move from there.

  • $\begingroup$ Surface area is not $A_1=R\pi h_1$ , it's $A_1={\bf 2}R\pi h_1$ $\endgroup$ – PA6OTA Jun 8 '14 at 19:41
  • $\begingroup$ Indeed, thank you. Still, I can't get the right solution. $\endgroup$ – Guest Jun 8 '14 at 19:55

A result similar to that you provide in your question ($4\sqrt{3\pi}$) is obtained only assuming that 48 and 16 are the two areas of the spherical caps without inclusion of the common circle.

In fact, under this assumption, because a spherical cap area is proportional to its heigth and the ratio of the areas is 3, it follows that the cut was made at 3/4 of the sphere. The radius of the circle is then $\sqrt{3}/2 \,R$, its area is $3/4 \,\pi R^2$, and its circumference is $\sqrt{3}\,\pi R$.

Because the total area is 64, it is easy to obtain $R=4/\sqrt{\pi}$. Therefore, the circumference of the circle is $4\sqrt{3\pi}$.

It can also be noted that if the areas were $48\pi$ and $64\pi$, the circumference of the circle would be $4\sqrt{3} \pi$, i.e. exactly the result that you state in your question.

I also made calculations assuming that 48 and 64 include the area of the circle, but obtained more complex results, very different from the final result that you provide.


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