Bethany ordered satin balloons for the celebration of her brother?s birthday party. A packet
of satin balloons had conical and spherical shaped balloons, all of which were different in
colours. The packet also mentioned that the volume of air that is required to be filled in the
conical and spherical balloon is the same. Bethany measured the height of the conical
balloon, after it is filled with air, as 16 inches and its radius as 4 inches.
Another packet of satin balloons have hemispherical balloons with 4 times the radius of the
spherical balloons of the first packet.
A.What volume of air is to be filled in the spherical balloon
B.What is the radius of the spherical balloon after it is filled with air completely?
C.What will be the ratio of the total surface areas of the hemispherical balloons of the
second packet and spherical balloons of the first packet?

Solution:
A. It is given that the volume of the conical and spherical balloons are the same.
$\therefore$ The volume of the conical balloon = $\frac{1}{3}{\mathrm{\pi r}}^2\mathrm{h}$
                                                              = $\frac{1}{3}\times \frac{22}{7}\times 16\times 16$                  $\left(\because \mathrm{radius}=4, \mathrm{height}=16\right)$
                                                              = 268.16 sq. inches
So, the volume of the spherical balloon  = 268.16 sq. inches

B.  It is given that the volume of the conical and spherical balloons are the same.
Let the radius of the spherical balloon be R
$\therefore$ The volume of the Conical balloon = The volume of a spherical balloon
        $\frac{1}{3}{\mathrm{\pi r}}^2\mathrm{h}$ = $\frac{4}{3}\pi {\mathrm{R}}^3$
     $$\begin{gathered} \Rightarrow R^3=\frac{r^2\times h}{4} \\ =\frac{4\times 4\times 16}{4} \\ =64 \\ R=4 \end{gathered}$$

Thus, the radius of the spherical balloon after it is filled with air completely is 4 inches.

C.  Let the radius of the hemispherical balloon be ${\mathrm{r}}_1$.
Now, the radius of the hemispherical balloon is 4 times the radius of the spherical balloon.
$\Rightarrow {\mathrm{r}}_1=4\times 4=16$

We know,
$\begin{array}{rcl}\frac{\mathrm{Total} \mathrm{surface} \mathrm{area} \mathrm{of} \mathrm{hemispherical} \mathrm{ballons}}{\mathrm{Total} \mathrm{surface} \mathrm{area} \mathrm{of} \mathrm{spherical} \mathrm{balloons}}& =& \frac{3{{\mathrm{\pi r}}_1}^2}{4{\mathrm{\pi R}}^2}\\ & =& \frac{{{\mathrm{r}}_1}^2}{{\mathrm{R}}^2}\\ & =& \frac{16\times 16}{4\times 4}\\ & =& 16\end{array}$

Thus, the ratio of the total surface areas of the hemispherical balloons of the second packet and the spherical balloons of the first packet is 16:1.