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Question Number 125337 by ajfour last updated on 10/Dec/20

Commented by ajfour last updated on 10/Dec/20

$$Ifallcolouredareasare10units$$$$each,findcoordinatesofG.$$$$(FGEandCGDarestraight$$$$segments)$$

Answered by ajfour last updated on 10/Dec/20

$$letG(p,q),C(0,c),F(0,f)$$$$eq.ofAB:x+y=10$$$$E(2p,2q-f)$$$$\Rightarrow 2p+2q-f=10...\left(i\right)$$$$D(d,0)$$$$\frac{q}{d-p}=\frac{c}{d}....\left(ii\right)$$$$p(c-f)=20...\left(iii\right)$$$$(10-c)\left(2p\right)=20....\left(iv\right)$$$$cd=40....\left(v\right)$$$$\Rightarrow q=\frac{c^2}{40}(\frac{40}{c}-p)....\left(I\right)$$$$f=c-\frac{20}{p}$$$$2p+\frac{c^2}{20}(\frac{40}{c}-p)=10+c-\frac{20}{p}..\left(II\right)$$$$from\left(iv\right)$$$$c=10-\frac{10}{p}$$$$substitutingforcin\left(II\right)$$$$2p+10-\frac{10}{p}-\frac{p}{20}{(10-\frac{10}{p})}^2=10-\frac{20}{p}$$$$\Rightarrow 2p+\frac{10}{p}-5p(1-\frac{2}{p}+\frac{1}{p^2})=0$$$$\Rightarrow 2p+\frac{10}{p}-5p+10-\frac{5}{p}=0$$$$\Rightarrow 3p-\frac{5}{p}=10$$$$\Rightarrow 3p^2-10p-5=0$$$$p=\frac{5}{3}+\sqrt{\frac{25}{9}+\frac{15}{9}}$$$$p=\frac{5+2\sqrt{10}}{3}$$$$c=10(1-\frac{3}{5+2\sqrt{10}})=10\left(\frac{2+2\sqrt{10}}{5+2\sqrt{10}}\right)$$$$=4(1+\sqrt{10})(2\sqrt{10}-5)/3$$$$=\frac{4}{3}(2\sqrt{10}-5+20-5\sqrt{10})$$$$c=4(5-\sqrt{10})$$$$Andfrom\left(I\right):q=\frac{c^2}{40}(\frac{40}{c}-p)$$$$q=c-\frac{c^{2}p}{40}$$$$=20-4\sqrt{10}-\frac{16{(5-\sqrt{10})}^2(5+2\sqrt{10})}{40\times 3}$$$$q=20-4\sqrt{10}-\frac{2(7-2\sqrt{10})(5+2\sqrt{10})}{3}$$$$=\frac{60-12\sqrt{10}-2(7-2\sqrt{10})(5+2\sqrt{10})}{3}$$$$=\frac{60-12\sqrt{10}-2(35+14\sqrt{10}-10\sqrt{10}-40)}{3}$$$$=\frac{70-20\sqrt{10}}{3}=\frac{10}{3}(7-2\sqrt{10})$$$$G[\frac{5+2\sqrt{10}}{3},\frac{10}{3}(7-2\sqrt{10})].$$

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