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Question Number 191143 by Mingma last updated on 19/Apr/23

Answered by HeferH last updated on 21/Apr/23

$$\mathrm{Say}\mathrm{CD}=\mathrm{a},\mathrm{DE}=\mathrm{b},\mathrm{AC}=\mathrm{c},\mathrm{AE}=\mathrm{d}$$$$\ast \frac{1}{\mathrm{a}}+\frac{1}{\mathrm{b}}=\frac{1}{5}\Rightarrow$$$$5(\mathrm{b}-\mathrm{a})=\mathrm{ab}$$$$\ast \mathrm{cd}=\mathrm{ab}+64$$$$\mathrm{Since}\mathrm{AD}\mathrm{is}\mathrm{bisector}:$$$${\mathrm{AD}}^2=\mathrm{cd}-\mathrm{ab}$$$${\mathrm{AD}}^2=(\mathrm{ab}+64)-\mathrm{ab}=64$$$$\mathrm{AD}=8$$$$$$$$\mathrm{Also}:$$$${\mathrm{AB}}^2=(\mathrm{AB}+\mathrm{b})(\mathrm{AB}-\mathrm{a})$$$${\mathrm{AB}}^2={\mathrm{AB}}^2-\mathrm{AB}\cdot \mathrm{a}+\mathrm{AB}\cdot \mathrm{b}-\mathrm{ab}$$$$\mathrm{ab}=\mathrm{AB}(\mathrm{b}-\mathrm{a})$$$$$$$$\mathrm{but}:\mathrm{ab}=5(\mathrm{b}-\mathrm{a})\Rightarrow$$$$\mathrm{AB}=5=\mathrm{DB}$$$$\therefore \mathrm{Area}=\sqrt{9(9-5)(9-5)(9-8)}=3\times 4=12{\mathrm{u}}^2$$

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