Respuesta :
The volume of a basketball can be given as volume of a shpere below,
[tex]\text{Volume, V, of a Sphere =}\frac{4}{3}\pi r^3[/tex]Let the volumes of the original and inflated basketball be given below respectively as,
[tex]\begin{gathered} \text{Vol of original = V}_1 \\ \text{Vol of inflated basketball =V}_2=225\operatorname{cm}^3 \end{gathered}[/tex]To find the radius of the inflated basketball below,
[tex]\begin{gathered} V_2=\frac{4}{3}\pi r^3=225\operatorname{cm}^3 \\ \text{Crossmultiply} \\ \frac{4}{3}\pi r^3=3\times225=675 \\ 4\pi r^3=675 \\ r^3=\frac{675}{4\pi} \\ \text{Where }\pi=3.14 \\ r^3=\frac{675}{4\times3.14}=53.742 \\ r=\sqrt[3]{53.742} \\ r=3.77\operatorname{cm} \end{gathered}[/tex]Radius r₂ of the inflated basketball is 3.77cm
Given the radius ratio of the original to the inflated basketball below as,
[tex]\begin{gathered} Ratio\text{ of original to inflated baskteball=}\frac{4}{9} \\ \text{Where the radius of the inflated baskteball is 3.77} \\ To\text{ find the radius of the original basketball} \\ \frac{4}{9}=\frac{r_1}{r_2}=\frac{r_1}{3.77} \\ \text{Crossmultiply} \\ 4\times3.77=r_1\times9 \\ r_1=\frac{4\times3.77}{9}=1.68\operatorname{cm} \\ r_1=1.68\operatorname{cm} \end{gathered}[/tex]To find the volume of the original basketball using the formula for the volume of a sphere,
[tex]\begin{gathered} V_1=\frac{4}{3}\pi r^3=\frac{4}{3}\times3.14\times(1.68)^3=19.85\operatorname{cm}^3 \\ V_1=20\operatorname{cm}^3(nearest\text{ whole number)} \end{gathered}[/tex]The volume V₁ of the original basketball is 20cm³(nearest whole no)
