IB MYP Year 4-5: Standard Mathematics : Unit 4: Geometry -Volume of regular polyhedra MYP Style Questions

Ans:

Given:
Radius of the melted sphere, $r = 2.80 \, \mathrm{cm}$
Radius of the cone, $R = 2.86 \, \mathrm{cm}$

We can still use the concept of equal volumes to find the height $h_{\text{cone}}$.

The volume of the cone is given by:

$V_{\text{cone}} = \frac{1}{3} \pi R^2 h_{\text{cone}}$

The volume of the sphere is given by:

$V_{\text{sphere}} = \frac{4}{3} \pi r^3$

Setting the volumes equal to each other:

$\frac{1}{3} \pi R^2 h_{\text{cone}} = \frac{4}{3} \pi r^3$

Canceling out $\pi$ and simplifying, we have:

$R^2 h_{\text{cone}} = 4r^3$

Substituting the values $r = 2.80 \, \mathrm{cm}$ and $R = 2.86 \, \mathrm{cm}$, we get:

$(2.86)^2 h_{\text{cone}} = 4(2.80)^3$

$8.1796 h_{\text{cone}} = 87.5456$

Dividing both sides by $8.1796$, we find:

$h_{\text{cone}} \approx \frac{87.5456}{8.1796} \approx 10.69$

Therefore, the height of the melted ice in the cone is approximately $10.7 \, \mathrm{cm}$.