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Question Number 67775 by TawaTawa last updated on 31/Aug/19

Answered by MJS last updated on 31/Aug/19

$$ABCD\mathrm{is}\mathrm{a}\mathrm{square}\mathrm{with}\mathrm{side}s=\sqrt{196}=14$$$$\Rightarrow \mathrm{the}\mathrm{yellow}{\mathrm{sector}}^{\prime }\mathrm{s}\mathrm{area}=\frac{\pi }{8}{s}^2=\frac{49\pi }{2}\mathrm{minus}$$$$\mathrm{the}\mathrm{white}\mathrm{segment}\mathrm{which}\mathrm{intersects}\mathrm{the}$$$$\mathrm{diagonal}\mathrm{in}\mathrm{the}\mathrm{center}\mathrm{of}\mathrm{the}\mathrm{square}$$$$$$$$\mathrm{the}\mathrm{area}\mathrm{of}\mathrm{the}\mathrm{white}\mathrm{segment}=\mathrm{quarter}$$$$\mathrm{circle}\mathrm{minus}\mathrm{half}\mathrm{square}=\frac{49\pi }{4}-\frac{49}{2}$$$$\Rightarrow \mathrm{the}\mathrm{yellow}\mathrm{area}=\frac{49}{4}\pi +\frac{49}{2}\approx 62.9845$$

Commented by MJS last updated on 31/Aug/19

$$\mathrm{corrected}\mathrm{my}\mathrm{mistake}$$

Commented by TawaTawa last updated on 31/Aug/19

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

Answered by mr W last updated on 31/Aug/19

$$a^2=196$$$$\Rightarrow a=\sqrt{196}=14$$$$R=a=14$$$$r=\frac{a}{2}=7$$$$A_{shaded}=\frac{\pi R^2}{8}-\frac{r^2}{2}(\frac{\pi }{2}-\sin \frac{\pi }{2})$$$$=\frac{\pi {14}^2}{8}-\frac{7^2}{2}(\frac{\pi }{2}-1)$$$$=\frac{49(\pi +2)}{4}$$$$=63{cm}^2$$

Commented by TawaTawa last updated on 31/Aug/19

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}.\mathrm{Thanks}\mathrm{for}\mathrm{your}\mathrm{time}$$

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