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Question Number 206396 by cortano21 last updated on 13/Apr/24

Answered by mr W last updated on 13/Apr/24

Commented by mr W last updated on 13/Apr/24

$$a=sidelengthofhexagon$$$$C=2\times \frac{\sqrt{3}{a}^2}{4}$$$$A+B=\frac{\sqrt{3}{a}^2}{4}$$$$11+10=6\times \frac{\sqrt{3}{a}^2}{4}-A-B-C=3\times \frac{\sqrt{3}{a}^2}{4}$$$$\Rightarrow \sqrt{3}{a}^2=28\Rightarrow a=\sqrt{\frac{28}{\sqrt{3}}}$$$$\frac{A+11-\frac{\sqrt{3}{a}^2}{4}}{B+10-\frac{\sqrt{3}{a}^2}{4}}=\frac{a-x}{x}=\frac{A}{B}$$$$\frac{A+11-7}{B+10-7}=\frac{A}{B}$$$$\Rightarrow \frac{A}{B}=\frac{4}{3}=\frac{a-x}{x}$$$$\Rightarrow \frac{x}{a}=\frac{3}{7}$$$$\Rightarrow x=\frac{3}{7}\times \sqrt{\frac{28}{\sqrt{3}}}=\frac{2\sqrt{21\sqrt{3}}}{7}\approx 1.723✓$$

Commented by cortano21 last updated on 13/Apr/24

$$⋗\cancel{\underset{―}{}}$$

Commented by mr W last updated on 13/Apr/24

$$A=\frac{(a-x)h}{2}$$$$B=\frac{xh}{2}$$$$\Rightarrow \frac{A}{B}=\frac{a-x}{x}$$$$\frac{A+11-7}{B+10-7}=\frac{a-x}{a}$$$$\Rightarrow \frac{A+11-7}{B+10-7}=\frac{A}{B}$$$$\Rightarrow \frac{A}{B}=\frac{4}{3}$$$$\Rightarrow \frac{a-x}{x}=\frac{4}{3}$$

Commented by cortano21 last updated on 13/Apr/24

$$\cancel{\underset{―}{\underbrace{\mathfrak{f}}}}$$

Commented by mr W last updated on 13/Apr/24

$$\frac{A+4}{B+3}=\frac{A}{B}$$$$\Rightarrow AB+4B=AB+3A$$$$\Rightarrow 4B=3A$$$$\Rightarrow \frac{A}{B}=\frac{4}{3}$$

Commented by cortano21 last updated on 13/Apr/24

$$\underset{―}{\underbrace{}}$$

Commented by cortano21 last updated on 17/Apr/24

$$\Rightarrow \frac{11}{10}=\frac{2x+1}{2a-2x+1}$$$$\Rightarrow 22a-22x+11=20x+10$$$$\Rightarrow 42x=22a+1$$$$\Rightarrow x=\frac{22a+1}{42}$$

Commented by mr W last updated on 17/Apr/24

$$seemsnottrue.$$

Answered by A5T last updated on 13/Apr/24

Commented by A5T last updated on 13/Apr/24

$$sx\sqrt{3}+\frac{s^2\sqrt{3}}{4}=10+\frac{sx\sqrt{3}}{4}+\frac{sx\sqrt{3}}{2}\Rightarrow \frac{s^2\sqrt{3}+sx\sqrt{3}}{4}=10$$$$\Rightarrow s^2+sx=\frac{40\sqrt{3}}{3}...\left(i\right)$$$$(s-x)s\sqrt{3}+\frac{s^2\sqrt{3}}{4}=11+\frac{s(s-x)\sqrt{3}}{4}+\frac{s(s-x)\sqrt{3}}{2}$$$$\Rightarrow \frac{s(s-x)\sqrt{3}+s^2\sqrt{3}}{4}=11\Rightarrow 2s^2-sx=\frac{44\sqrt{3}}{3}...\left(ii\right)$$$$\left(i\right)+\left(ii\right)\Rightarrow s^2=\frac{84\sqrt{3}}{9}\Rightarrow s=\sqrt{\frac{28\sqrt{3}}{3}}$$$$2\times \left(i\right)-\left(ii\right)\Rightarrow sx=4\sqrt{3}\Rightarrow x=4\sqrt{3}\times \frac{\sqrt{3}}{\sqrt{28\sqrt{3}}}\approx 1.723$$

Answered by A5T last updated on 13/Apr/24

Commented by A5T last updated on 13/Apr/24

$$\frac{s^2\sqrt{3}}{4}+\frac{s(s-x)\sqrt{3}}{2}=11+s(s-x)\times \frac{\sqrt{3}}{4}$$$$\Rightarrow \frac{s^2\sqrt{3}}{4}+\frac{s(s-x)\sqrt{3}}{4}=\frac{s^2\sqrt{3}}{2}-\frac{sx\sqrt{3}}{4}=11$$$$\Rightarrow 2s^2-sx=\frac{44\sqrt{3}}{3}...\left(i\right)$$$$\frac{s^2\sqrt{3}}{4}+\frac{sx\sqrt{3}}{2}=10+\frac{sx\sqrt{3}}{4}\Rightarrow s^2\sqrt{3}+sx\sqrt{3}=40$$$$\Rightarrow s^2+sx=\frac{40\sqrt{3}}{3}...\left(ii\right)$$$$\left(i\right)\mathrm{\&}\left(ii\right)\Rightarrow x\approx 1.723$$

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