What is the ratio of $A_1$ and $A_2$

Question

The sphere is formed by disrupting the cube that is made by play dough with $r = 2$. The surface area of sphere is $A_1$ and surface area of cube is $A_2$

What is the ratio of $\frac {A_1}{A_2}$? Use the approximation $\pi\approx3$.

As first, I thought that

$$\text{Volume of sphere}=\text{Volume of cube}$$

However, I'm stuck here. I mean I don't know what to do more.

Answer 1

volume of a sphere is given by $$V_1=\frac{4}{3}\pi r^3$$ and the cube $$V_2=a^3$$ and forhter $$A_1=4\pi r^2$$ (surface of a sphere) and $$A_2=6a^2$$ for a dice. to find $$a$$ solve the equation $$a^3=\frac{4}{3}\pi\cdot 2^3$$ for $a$ we find $$a=2\sqrt[3]{\frac{4\pi}{3}}$$

Answer 2

You already managed to find that $$a^3=\frac{4}{3}\pi r^3$$ So $$\left(\frac{a}{r}\right)^3=\frac{4}{3}\pi$$ We also know that $$A_1=6a^2$$ and $$A_2=4\pi r^2$$ You should be able to proceed from here using a similar trick on the area.

Answer 3

Because of The sphere is formed by disrupting the cube a=2r

48/96=1/2 That's the answer