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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

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

Answered by MJS last updated on 27/Apr/20

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

Commented by tw000001 last updated on 27/Apr/20

$${\mathrm{That}}^{\prime }\mathrm{s}\mathrm{correct},\mathrm{thank}\mathrm{you}.$$

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

$$AC=\sqrt{5^2+{12}^2}=13$$$$CB+BQ=CA+AP$$$$12+BQ=13+AP\Rightarrow BQ=1+AP$$$$BQ+AP=AB$$$$1+AP+AP=5\Rightarrow AP=3$$$$\Rightarrow BQ=1+2=3$$$$CP=CQ=12+3=15$$$${PQ}^2={15}^2+{15}^2-2\times 15\times 15\times \cos \mathrm{\angle }C$$$${PQ}^2=2(1-\frac{12}{13})\times {15}^2=\frac{2\times {15}^2}{13}$$$$\Rightarrow PQ=\frac{15\sqrt{26}}{13}\approx 5.883$$

Commented by tw000001 last updated on 27/Apr/20

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

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