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Question Number 207482 by cherokeesay last updated on 17/May/24

Answered by mr W last updated on 17/May/24

$$x(a-x)=yb=zc$$$$\Rightarrow y=\frac{x(a-x)}{b}$$$$\Rightarrow z=\frac{x(a-x)}{c}$$$$d^2=y^2+z^2-2yz\times \frac{b^2+c^2-a^2}{2bc}$$$$d^2=\frac{x^2{(a-x)}^2}{b^2}+\frac{x^2{(a-x)}^2}{c^2}-\frac{x^2{(a-x)}^2(b^2+c^2-a^2)}{b^2{c}^2}$$$${\left(\frac{\sqrt{11}a}{12}\right)}^2=\frac{x^2{(a-x)}^2{a}^2}{b^2{c}^2}$$$$\Rightarrow \frac{x^2{(a-x)}^2}{b^2{c}^2}=\frac{11}{144}$$$$\frac{yellow+blue}{yellow}=\frac{bc}{yz}=\frac{b^2{c}^2}{x^2{(a-x)}^2}=\frac{144}{11}$$$$\Rightarrow \frac{blue}{yellow}=\frac{144}{11}-1=\frac{133}{11}✓$$

Commented by cherokeesay last updated on 17/May/24

$$Great!thankyoumaster!$$

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