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Question Number 87911 by jagoll last updated on 07/Apr/20

Commented by jagoll last updated on 07/Apr/20

$$\mathrm{dear}\mathrm{mr}\mathrm{W}.$$$$\mathrm{i}\mathrm{forgot}\mathrm{your}\mathrm{method}.$$$$\mathrm{please}\mathrm{remember}\mathrm{me}\mathrm{for}\mathrm{this}$$$$\mathrm{question}$$

Commented by john santu last updated on 07/Apr/20

$$\frac{{\mathrm{t}}_1}{8\mathrm{x}}=\frac{{\mathrm{t}}_2+12\mathrm{z}}{6\mathrm{y}}$$$$\frac{{\mathrm{t}}_1+8\mathrm{x}}{6\mathrm{y}}=\frac{{\mathrm{t}}_2}{12\mathrm{z}}$$$$[\mathrm{let}8\mathrm{x}=\mathrm{a},6\mathrm{y}=\mathrm{b},12\mathrm{z}=\mathrm{c}]$$$$\Rightarrow {\mathrm{t}}_1=\frac{\mathrm{a}}{\mathrm{b}}({\mathrm{t}}_2+\mathrm{c})\left(\mathrm{i}\right)$$$$\Rightarrow {\mathrm{t}}_1+\mathrm{a}=\frac{\mathrm{b}}{\mathrm{c}}{\mathrm{t}}_2,{\mathrm{t}}_1=\frac{{\mathrm{bt}}_2}{\mathrm{c}}-\mathrm{a}\left(\mathrm{ii}\right)$$$$\frac{\mathrm{a}}{\mathrm{b}}({\mathrm{t}}_2+\mathrm{c})=\frac{\mathrm{b}}{\mathrm{c}}{\mathrm{t}}_2-\mathrm{a}$$$${\mathrm{t}}_2(\frac{\mathrm{a}}{\mathrm{b}}-\frac{\mathrm{b}}{\mathrm{c}})=-\mathrm{a}-\frac{\mathrm{ac}}{\mathrm{b}}$$$${\mathrm{t}}_2\left(\frac{{\mathrm{b}}^2-\mathrm{ac}}{\mathrm{bc}}\right)=\frac{\mathrm{a}(\mathrm{b}+\mathrm{c})}{\mathrm{b}}$$$${\mathrm{t}}_2=\frac{\mathrm{ac}(\mathrm{b}+\mathrm{c})}{{\mathrm{b}}^2-\mathrm{ac}}.\mathrm{now}$$$${\mathrm{t}}_1=\frac{{\mathrm{bt}}_2-\mathrm{ac}}{\mathrm{c}}=\frac{\left(\frac{\mathrm{abc}(\mathrm{b}+\mathrm{c})}{{\mathrm{b}}^2-\mathrm{ac}}\right)-\mathrm{ac}}{\mathrm{c}}$$$${\mathrm{t}}_1=\frac{{\mathrm{ab}}^2\mathrm{c}+{\mathrm{abc}}^2-{\mathrm{ab}}^2\mathrm{c}+{\left(\mathrm{ac}\right)}^2}{\mathrm{c}({\mathrm{b}}^2-\mathrm{ac})}$$$${\mathrm{t}}_1=\frac{{\mathrm{ac}}^2(\mathrm{b}+\mathrm{a})}{\mathrm{c}({\mathrm{b}}^2-\mathrm{ac})}$$$$\mathrm{the}\mathrm{area}\mathrm{we}\mathrm{required}={\mathrm{t}}_1+{\mathrm{t}}_2$$$$=\frac{{\mathrm{ac}}^2(\mathrm{a}+\mathrm{b})}{\mathrm{c}({\mathrm{b}}^2-\mathrm{ac})}+\frac{\mathrm{ac}(\mathrm{b}+\mathrm{c})}{{\mathrm{b}}^2-\mathrm{ac}}$$$$=\frac{{\mathrm{ac}}^2(\mathrm{a}+2\mathrm{b}+\mathrm{c})}{\mathrm{c}({\mathrm{b}}^2-\mathrm{ac})}=\frac{\mathrm{ac}(\mathrm{a}+2\mathrm{b}+\mathrm{c})}{{\mathrm{b}}^2-\mathrm{ac}}$$

Commented by jagoll last updated on 07/Apr/20

$$=\frac{96\mathrm{xz}(8\mathrm{x}+12\mathrm{y}+12\mathrm{z})}{{36\mathrm{y}}^2-96\mathrm{xz}}$$$$=\frac{96\mathrm{xz}(2\mathrm{x}+3\mathrm{y}+3\mathrm{z})}{{9\mathrm{y}}^2-24\mathrm{xz}}$$

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