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Question Number 197680 by mr W last updated on 26/Sep/23

Commented by mr W last updated on 26/Sep/23

$$\text{You can't use 'macro parameter character \#' in math mode}$$

Answered by mr W last updated on 26/Sep/23

Commented by mr W last updated on 26/Sep/23

$$R=radiusofyellowsemicircle$$$$r=radiusofgreensemicircle$$$$p=radiusofpinksemicircle$$$$\sqrt{{(R+r)}^2-{(R-r)}^2}=\sqrt{{(R+p)}^2-R^2}+\sqrt{{(r+p)}^2-r^2}$$$$2\sqrt{Rr}=\sqrt{p(2R+p)}+\sqrt{p(2r+p)}$$$$2\sqrt{Rr}-\sqrt{p(2R+p)}=\sqrt{p(2r+p)}$$$$2Rr+(R-r)p=2\sqrt{Rrp(2R+p)}$$$$4R^2{r}^2+{(R-r)}^2{p}^2+4Rr(R-r)p=4Rrp(2R+p)$$$$\Rightarrow (6Rr-R^2-r^2)p^2+4Rr(R+r)p-4R^2{r}^2=0$$$$\Rightarrow p=\frac{-2Rr(R+r)+4Rr\sqrt{2Rr}}{6Rr-R^2-r^2}$$$$\frac{yellowarea}{greenarea}={\left(\frac{R}{r}\right)}^2=\frac{2\pi }{\pi /2}=4\Rightarrow R=2r$$$$\Rightarrow p=\frac{4r}{7}$$$$\frac{pinkarea}{greenarea}={\left(\frac{p}{r}\right)}^2={\left(\frac{4}{7}\right)}^2=\frac{16}{49}$$$$\Rightarrow pinkarea=\frac{16}{49}\times \frac{\pi }{2}=\frac{8\pi }{49}✓$$

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