Question

The cost of an 8-inch pizza is $4. The cost of a 16-inch pizza is $13. Use 3.14 as an approximation for π.
(a) How much greater is the area of the 16-inch pizza than the area of the 8-inch pizza?
(b) Which is a better deal? Explain your reasoning.

Answer 1

Dear student,

$$\begin{gathered} Areaof16inchpizza={\mathrm{\pi R}}^2=3.14\times 8\times 8=200.96 \\ \mathrm{Area}\mathrm{of}8\mathrm{inch}\mathrm{pizza}={\mathrm{\pi r}}^2=3.14\times 4\times 4=50.24 \\ \mathrm{a})\mathrm{Area}\mathrm{of}16\mathrm{inch}\mathrm{pizza}>\mathrm{Area}\mathrm{of}8\mathrm{inch}\mathrm{pizza}\mathrm{by}\mathrm{area} \\ 200.96-50.24=150.717 \\ \mathrm{b})\mathrm{cost}\mathrm{of}16-\mathrm{inch}\mathrm{pizza}=\&13 \\ \mathrm{cost}\mathrm{of}1-\mathrm{inch}\mathrm{pizza}=\frac{\&13}{16}=\&0.8 \\ \mathrm{cost}\mathrm{of}8-\mathrm{inch}\mathrm{pizza}=\&4 \\ \mathrm{cost}\mathrm{of}1-\mathrm{inch}\mathrm{pizza}=\frac{\&4}{8}=\&0.5 \\ \mathrm{Therefore},\mathrm{it}\mathrm{is}\mathrm{bettere}\mathrm{to}\mathrm{buy}8-\mathrm{inch}\mathrm{pizza}\mathrm{as}\mathrm{in}1\mathrm{inch}\mathrm{it}\mathrm{cost}\&0.5\mathrm{while}, \\ \mathrm{the}\mathrm{cost}\mathrm{of}16-\mathrm{inch}\mathrm{pizza}\mathrm{per}\mathrm{inch}\mathrm{is}\&0.8 \end{gathered}$$
Regards