Question-203716

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

Answered by AST last updated on 26/Jan/24

Let AP and DC meet at F,then △BPA≈△CPF ⇒((BA)/(CF))=((BP)/(CP))⇒(s/(CF))=(8/(s−8))⇒CF=((s(s−8))/8)...(i) in △AFD;AQ bisects ∠FAD⇒((AD)/(DQ))=((AF)/(FQ)) ⇒(s/9)=((√(s^2 +(s+CF)^2 ))/(CF+s−9))⇒(s/9)=((√(s^2 +(s+((s^2 −8s)/8))^2 ))/(((s^2 −8s)/8)+s−9)) ⇒(s/9)=((√(s^2 +(s^4 /(64))))/((s^2 −72)/8))⇒(1/9)= ((8(√((1+(s^2 /(64))))))/(s^2 −72))⇒s=15 PQ=(√((s−8)^2 +(s−9)^2 ))=(√(85))cm

$${Let}\:{AP}\:{and}\:{DC}\:{meet}\:{at}\:{F},{then}\:\bigtriangleup{BPA}\approx\bigtriangleup{CPF} \\ $$$$\Rightarrow\frac{{BA}}{{CF}}=\frac{{BP}}{{CP}}\Rightarrow\frac{{s}}{{CF}}=\frac{\mathrm{8}}{{s}−\mathrm{8}}\Rightarrow{CF}=\frac{{s}\left({s}−\mathrm{8}\right)}{\mathrm{8}}…\left({i}\right) \\ $$$${in}\:\bigtriangleup{AFD};{AQ}\:{bisects}\:\angle{FAD}\Rightarrow\frac{{AD}}{{DQ}}=\frac{{AF}}{{FQ}} \\ $$$$\Rightarrow\frac{{s}}{\mathrm{9}}=\frac{\sqrt{{s}^{\mathrm{2}} +\left({s}+{CF}\right)^{\mathrm{2}} }}{{CF}+{s}−\mathrm{9}}\Rightarrow\frac{{s}}{\mathrm{9}}=\frac{\sqrt{{s}^{\mathrm{2}} +\left({s}+\frac{{s}^{\mathrm{2}} −\mathrm{8}{s}}{\mathrm{8}}\right)^{\mathrm{2}} }}{\frac{{s}^{\mathrm{2}} −\mathrm{8}{s}}{\mathrm{8}}+{s}−\mathrm{9}} \\ $$$$\Rightarrow\frac{{s}}{\mathrm{9}}=\frac{\sqrt{{s}^{\mathrm{2}} +\frac{{s}^{\mathrm{4}} }{\mathrm{64}}}}{\frac{{s}^{\mathrm{2}} −\mathrm{72}}{\mathrm{8}}}\Rightarrow\frac{\mathrm{1}}{\mathrm{9}}=\:\frac{\mathrm{8}\sqrt{\left(\mathrm{1}+\frac{{s}^{\mathrm{2}} }{\mathrm{64}}\right)}}{{s}^{\mathrm{2}} −\mathrm{72}}\Rightarrow{s}=\mathrm{15} \\ $$$${PQ}=\sqrt{\left({s}−\mathrm{8}\right)^{\mathrm{2}} +\left({s}−\mathrm{9}\right)^{\mathrm{2}} }=\sqrt{\mathrm{85}}{cm} \\ $$

Answered by professorleiciano last updated on 26/Jan/24

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

$${you}\:{didn}'{t}\:{show}\:{why}\:{AP}=\mathrm{8}+\mathrm{9}=\mathrm{17}. \\ $$

Commented by esmaeil last updated on 26/Jan/24

profesor,¿por que^(′ ) no respondio^′ la pregunta de nadie?

$${profesor},¿{por}\:{que}^{'\:} {no}\:{respondi}\overset{'} {{o}}\:{la} \\ $$$${pregunta}\:{de}\:{nadie}? \\ $$

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

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

∠AEB=∠AQD=γ+α=∠EAB ⇒PA=PE=9+8=17 AB=(√(17^2 −8^2 ))=15 PC=15−8=7 QC=15−9=6 PQ=(√(7^2 +6^2 ))=(√(85))

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

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