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Question Number 90948 by tw000001 last updated on 27/Apr/20

Commented by tw000001 last updated on 27/Apr/20

Find ∣PQ∣.

$$\mathrm{Find}\:\mid\mathrm{PQ}\mid. \\ $$

Commented by tw000001 last updated on 27/Apr/20

If △APQ is right triangle, ∣PQ∣ should be (√(38)). However, what if △APQ is not a right triangle?

$$\mathrm{If}\:\bigtriangleup\mathrm{APQ}\:\mathrm{is}\:\mathrm{right}\:\mathrm{triangle}, \\ $$$$\mid\mathrm{PQ}\mid\:\mathrm{should}\:\mathrm{be}\:\sqrt{\mathrm{38}}. \\ $$$$\mathrm{However},\:\mathrm{what}\:\mathrm{if}\:\bigtriangleup\mathrm{APQ}\:\mathrm{is}\:\mathrm{not}\:\mathrm{a}\:\mathrm{right}\:\mathrm{triangle}? \\ $$

Answered by MJS last updated on 27/Apr/20

∡ABC=(π/2) ⇒ AC=13 the radius of the excircle is ((√((a+b+c)(−a+b+c)(a−b+c)(a+b−c)))/(2(−a+b+c)))=3 PC=QC=15 PQ=((30)/(√(26)))

$$\measuredangle{ABC}=\frac{\pi}{\mathrm{2}}\:\Rightarrow\:{AC}=\mathrm{13} \\ $$$$\mathrm{the}\:\mathrm{radius}\:\mathrm{of}\:\mathrm{the}\:\mathrm{excircle}\:\mathrm{is} \\ $$$$\frac{\sqrt{\left({a}+{b}+{c}\right)\left(−{a}+{b}+{c}\right)\left({a}−{b}+{c}\right)\left({a}+{b}−{c}\right)}}{\mathrm{2}\left(−{a}+{b}+{c}\right)}=\mathrm{3} \\ $$$${PC}={QC}=\mathrm{15} \\ $$$${PQ}=\frac{\mathrm{30}}{\sqrt{\mathrm{26}}} \\ $$

Commented by tw000001 last updated on 27/Apr/20

That′s correct, thank you.

$$\mathrm{That}'\mathrm{s}\:\mathrm{correct},\:\mathrm{thank}\:\mathrm{you}. \\ $$

Answered by mr W last updated on 27/Apr/20

AC=(√(5^2 +12^2 ))=13 CB+BQ=CA+AP 12+BQ=13+AP ⇒BQ=1+AP BQ+AP=AB 1+AP+AP=5 ⇒AP=3 ⇒BQ=1+2=3 CP=CQ=12+3=15 PQ^2 =15^2 +15^2 −2×15×15×cos ∠C PQ^2 =2(1−((12)/(13)))×15^2 =((2×15^2 )/(13)) ⇒PQ=((15(√(26)))/(13))≈5.883

$${AC}=\sqrt{\mathrm{5}^{\mathrm{2}} +\mathrm{12}^{\mathrm{2}} }=\mathrm{13} \\ $$$${CB}+{BQ}={CA}+{AP} \\ $$$$\mathrm{12}+{BQ}=\mathrm{13}+{AP}\:\Rightarrow{BQ}=\mathrm{1}+{AP} \\ $$$${BQ}+{AP}={AB} \\ $$$$\mathrm{1}+{AP}+{AP}=\mathrm{5}\:\Rightarrow{AP}=\mathrm{3} \\ $$$$\Rightarrow{BQ}=\mathrm{1}+\mathrm{2}=\mathrm{3} \\ $$$${CP}={CQ}=\mathrm{12}+\mathrm{3}=\mathrm{15} \\ $$$${PQ}^{\mathrm{2}} =\mathrm{15}^{\mathrm{2}} +\mathrm{15}^{\mathrm{2}} −\mathrm{2}×\mathrm{15}×\mathrm{15}×\mathrm{cos}\:\angle{C} \\ $$$${PQ}^{\mathrm{2}} =\mathrm{2}\left(\mathrm{1}−\frac{\mathrm{12}}{\mathrm{13}}\right)×\mathrm{15}^{\mathrm{2}} =\frac{\mathrm{2}×\mathrm{15}^{\mathrm{2}} }{\mathrm{13}} \\ $$$$\Rightarrow{PQ}=\frac{\mathrm{15}\sqrt{\mathrm{26}}}{\mathrm{13}}\approx\mathrm{5}.\mathrm{883} \\ $$

Commented by tw000001 last updated on 27/Apr/20

Your answer is correct, too.

$$\mathrm{Your}\:\mathrm{answer}\:\mathrm{is}\:\mathrm{correct},\:\mathrm{too}. \\ $$

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