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Question Number 95707 by john santu last updated on 27/May/20

Commented by john santu last updated on 27/May/20

$$\mathrm{find}\mathrm{the}\mathrm{shaded}\mathrm{area}$$

Commented by john santu last updated on 27/May/20

Commented by john santu last updated on 27/May/20

$$\mathrm{RE}=\sqrt{{(\mathrm{a}-\mathrm{r})}^2-{\mathrm{r}}^2}=\sqrt{{\mathrm{a}}^2-2\mathrm{ar}}$$$${\mathrm{AQ}}^2={\mathrm{r}}^2+{\mathrm{AR}}^2$$$${\mathrm{a}}^2+2\mathrm{ar}+{\mathrm{r}}^2={\mathrm{r}}^2+{(\mathrm{a}+\sqrt{{\mathrm{a}}^2-2\mathrm{ar}})}^2$$$${\mathrm{a}}^2+2\mathrm{ar}={\mathrm{a}}^2+2\mathrm{a}\sqrt{{\mathrm{a}}^2-2\mathrm{ar}}+{\mathrm{a}}^2-2\mathrm{ar}$$$$4\mathrm{ar}={\mathrm{a}}^2+2\mathrm{a}\sqrt{{\mathrm{a}}^2-2\mathrm{ar}}$$$$4\mathrm{r}-\mathrm{a}=2\sqrt{{\mathrm{a}}^2-2\mathrm{ar}}$$$${16\mathrm{r}}^2-8\mathrm{ar}+{\mathrm{a}}^2={4\mathrm{a}}^2-8\mathrm{ar}$$$${\mathrm{r}}^2=\frac{3}{16}{\mathrm{a}}^2.\mathrm{area}2\mathrm{small}\mathrm{circle}$$$$=2\pi {\mathrm{r}}^2=2\pi \left(\frac{3}{16}{\mathrm{a}}^2\right)=\frac{3\pi }{8}{\mathrm{a}}^2$$$$\mathrm{shaded}\mathrm{area}=2\mathrm{a}\times \mathrm{a}-\frac{3\pi {\mathrm{a}}^2}{8}$$$$={2\mathrm{a}}^2-\frac{3\pi {\mathrm{a}}^2}{8}=\left(\frac{16-3\pi }{8}\right){\mathrm{a}}^2$$

Answered by mr W last updated on 27/May/20

$$saysidelengthofsquareis2a.$$$$radiusofsmallcirclesisr.$$$${(a+\sqrt{{(a-r)}^2-r^2})}^2+r^2={(a+r)}^2$$$$2a\sqrt{a(a-2r)}=4ar-a^2$$$$3a^2=16r^2$$$$r=\frac{\sqrt{3}a}{4}$$$$shadedareas=2a\times a-2\times \pi r^2$$$$=2a^2-2\pi \times \frac{3a^2}{16}$$$$=\frac{(16-3\pi )a^2}{8}$$

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