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Question Number 203716 by professorleiciano last updated on 26/Jan/24

Answered by deleteduser1 last updated on 26/Jan/24

$$LetAPandDCmeetatF,then△BPA\approx △CPF$$$$\Rightarrow \frac{BA}{CF}=\frac{BP}{CP}\Rightarrow \frac{s}{CF}=\frac{8}{s-8}\Rightarrow CF=\frac{s(s-8)}{8}...\left(i\right)$$$$in△AFD;AQbisects\mathrm{\angle }FAD\Rightarrow \frac{AD}{DQ}=\frac{AF}{FQ}$$$$\Rightarrow \frac{s}{9}=\frac{\sqrt{s^2+{(s+CF)}^2}}{CF+s-9}\Rightarrow \frac{s}{9}=\frac{\sqrt{s^2+{(s+\frac{s^2-8s}{8})}^2}}{\frac{s^2-8s}{8}+s-9}$$$$\Rightarrow \frac{s}{9}=\frac{\sqrt{s^2+\frac{s^4}{64}}}{\frac{s^2-72}{8}}\Rightarrow \frac{1}{9}=\frac{8\sqrt{(1+\frac{s^2}{64})}}{s^2-72}\Rightarrow s=15$$$$PQ=\sqrt{{(s-8)}^2+{(s-9)}^2}=\sqrt{85}cm$$

Answered by professorleiciano last updated on 26/Jan/24

Commented by mr W last updated on 26/Jan/24

$$you{didn}^{\prime }tshowwhyAP=8+9=17.$$

Commented by esmaeil last updated on 26/Jan/24

$$profesor,¿por{que}^{{}^{\prime }}norespondi\stackrel{{}^{\prime }}{o}la$$$$preguntadenadie?$$

Answered by mr W last updated on 26/Jan/24

Commented by mr W last updated on 26/Jan/24

$$\mathrm{\angle }AEB=\mathrm{\angle }AQD=\gamma +\alpha =\mathrm{\angle }EAB$$$$\Rightarrow PA=PE=9+8=17$$$$AB=\sqrt{{17}^2-8^2}=15$$$$PC=15-8=7$$$$QC=15-9=6$$$$PQ=\sqrt{7^2+6^2}=\sqrt{85}$$

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