Question Number 90948 by tw000001 last updated on 27/Apr/20
Commented by tw000001 last updated on 27/Apr/20
$$\mathrm{Find}\:\mid\mathrm{PQ}\mid. \\ $$
Commented by tw000001 last updated on 27/Apr/20
$$\mathrm{If}\:\bigtriangleup\mathrm{APQ}\:\mathrm{is}\:\mathrm{right}\:\mathrm{triangle}, \\ $$$$\mid\mathrm{PQ}\mid\:\mathrm{should}\:\mathrm{be}\:\sqrt{\mathrm{38}}. \\ $$$$\mathrm{However},\:\mathrm{what}\:\mathrm{if}\:\bigtriangleup\mathrm{APQ}\:\mathrm{is}\:\mathrm{not}\:\mathrm{a}\:\mathrm{right}\:\mathrm{triangle}? \\ $$
Answered by MJS last updated on 27/Apr/20
$$\measuredangle{ABC}=\frac{\pi}{\mathrm{2}}\:\Rightarrow\:{AC}=\mathrm{13} \\ $$$$\mathrm{the}\:\mathrm{radius}\:\mathrm{of}\:\mathrm{the}\:\mathrm{excircle}\:\mathrm{is} \\ $$$$\frac{\sqrt{\left({a}+{b}+{c}\right)\left(−{a}+{b}+{c}\right)\left({a}−{b}+{c}\right)\left({a}+{b}−{c}\right)}}{\mathrm{2}\left(−{a}+{b}+{c}\right)}=\mathrm{3} \\ $$$${PC}={QC}=\mathrm{15} \\ $$$${PQ}=\frac{\mathrm{30}}{\:\sqrt{\mathrm{26}}} \\ $$
Commented by tw000001 last updated on 27/Apr/20
$$\mathrm{That}'\mathrm{s}\:\mathrm{correct},\:\mathrm{thank}\:\mathrm{you}. \\ $$
Answered by mr W last updated on 27/Apr/20
$${AC}=\sqrt{\mathrm{5}^{\mathrm{2}} +\mathrm{12}^{\mathrm{2}} }=\mathrm{13} \\ $$$${CB}+{BQ}={CA}+{AP} \\ $$$$\mathrm{12}+{BQ}=\mathrm{13}+{AP}\:\Rightarrow{BQ}=\mathrm{1}+{AP} \\ $$$${BQ}+{AP}={AB} \\ $$$$\mathrm{1}+{AP}+{AP}=\mathrm{5}\:\Rightarrow{AP}=\mathrm{3} \\ $$$$\Rightarrow{BQ}=\mathrm{1}+\mathrm{2}=\mathrm{3} \\ $$$${CP}={CQ}=\mathrm{12}+\mathrm{3}=\mathrm{15} \\ $$$${PQ}^{\mathrm{2}} =\mathrm{15}^{\mathrm{2}} +\mathrm{15}^{\mathrm{2}} −\mathrm{2}×\mathrm{15}×\mathrm{15}×\mathrm{cos}\:\angle{C} \\ $$$${PQ}^{\mathrm{2}} =\mathrm{2}\left(\mathrm{1}−\frac{\mathrm{12}}{\mathrm{13}}\right)×\mathrm{15}^{\mathrm{2}} =\frac{\mathrm{2}×\mathrm{15}^{\mathrm{2}} }{\mathrm{13}} \\ $$$$\Rightarrow{PQ}=\frac{\mathrm{15}\sqrt{\mathrm{26}}}{\mathrm{13}}\approx\mathrm{5}.\mathrm{883} \\ $$
Commented by tw000001 last updated on 27/Apr/20
$$\mathrm{Your}\:\mathrm{answer}\:\mathrm{is}\:\mathrm{correct},\:\mathrm{too}. \\ $$