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Question Number 69609 by TawaTawa last updated on 25/Sep/19

Commented by Prithwish sen last updated on 25/Sep/19

$$\mathrm{I}\mathrm{think}$$$$\mathrm{The}\mathrm{area}\mathrm{of}\mathrm{inner}\mathrm{section}=\pi {\left(75\right)}^2\mathrm{sq}.\mathrm{cm}$$$$\mathrm{The}\mathrm{area}\mathrm{of}\mathrm{outer}\mathrm{section}=\pi (100-75)(100+75)$$$$=\pi \times 25\times 175\mathrm{sq}.\mathrm{cm}$$$$\therefore \mathbf{Total}\mathbf{cost}=\mathit{\pi }\times 75\times 75\times 2500+\mathit{\pi }\times 25\times 175\times 1500$$$$\mathbf{now}\mathbf{do}\mathbf{the}\mathbf{calculation}$$

Commented by TawaTawa last updated on 25/Sep/19

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}$$

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