Question Number 60045 by Kunal12588 last updated on 17/May/19
Commented by Kunal12588 last updated on 17/May/19
$${Ans}\::\:\mathrm{177}\frac{\mathrm{1}}{\mathrm{4}}\pi\:\:\left({given}\:{in}\:{book}\right) \\ $$
Commented by mr W last updated on 17/May/19
$${correct}! \\ $$
Answered by mr W last updated on 17/May/19
Commented by mr W last updated on 17/May/19
$${A}_{\mathrm{1}} =\frac{\mathrm{3}}{\mathrm{4}}×\pi×\mathrm{15}^{\mathrm{2}} =\frac{\mathrm{675}\pi}{\mathrm{4}} \\ $$$${A}_{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{4}}×\pi×\mathrm{5}^{\mathrm{2}} =\frac{\mathrm{25}\pi}{\mathrm{4}} \\ $$$${A}_{\mathrm{3}} =\frac{\mathrm{1}}{\mathrm{4}}×\pi×\mathrm{3}^{\mathrm{2}} =\frac{\mathrm{9}\pi}{\mathrm{4}} \\ $$$$\Sigma{A}=\frac{\left(\mathrm{675}+\mathrm{25}+\mathrm{9}\right)\pi}{\mathrm{4}}=\frac{\mathrm{709}\pi}{\mathrm{4}} \\ $$
Commented by Kunal12588 last updated on 17/May/19
$${thanks}\:{sir} \\ $$
Commented by ajfour last updated on 17/May/19
$$\mathcal{W}{onderful}\:{Sir}! \\ $$
Commented by otchereabdullai@gmail.com last updated on 17/May/19
$$\mathrm{Fantastic}\:\mathrm{Prof}\:\mathrm{W}\:\mathrm{too}\:\mathrm{brilliant} \\ $$