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Question Number 61186 by Tawa1 last updated on 30/May/19

Commented by Prithwish sen last updated on 30/May/19

$$\mathrm{Area}\mathrm{of}\mathrm{the}\mathrm{shadded}\mathrm{portion}=\mathrm{Area}\mathrm{of}\mathrm{the}\mathrm{incircle}$$$$-\mathrm{Area}\mathrm{of}\mathrm{the}\mathrm{quarter}\mathrm{circle}+\frac{3}{4}(\mathrm{Area}\mathrm{of}\mathrm{the}$$$$\mathrm{square}-\mathrm{Area}\mathrm{of}\mathrm{the}\mathrm{incircle})$$$$=\pi 5^2-\frac{1}{4}\pi {\left(10\right)}^2+\frac{3}{4}[{\left(10\right)}^2-\pi 5^2]$$$$=\frac{1}{4}\pi 5^2-\frac{1}{4}\pi {10}^2+\frac{3}{4}{10}^2$$$$=\frac{3}{4}\times 100-\frac{\pi }{4}\times 75\backsim 16.09$$$$\mathrm{please}\mathrm{check}\mathrm{the}\mathrm{answer}$$

Commented by mr W last updated on 30/May/19

$$\text{You can't use 'macro parameter character \#' in math mode}$$$$answershouldbe\approx 0.146a^2=14.6{cm}^2$$

Commented by Tawa1 last updated on 30/May/19

$$\mathrm{I}\mathrm{will}\mathrm{draw}\mathrm{the}\mathrm{same}\mathrm{angle}\mathrm{you}\mathrm{drew}???.\mathrm{In}\mathrm{the}\mathrm{previous}\mathrm{one}?$$$$\mathrm{to}\mathrm{solve}\mathrm{this}?$$

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