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Question Number 66667 by naka3546 last updated on 18/Aug/19

Commented by naka3546 last updated on 18/Aug/19

(([PBQ])/([PQCA])) = ?

$$\frac{\left[{PBQ}\right]}{\left[{PQCA}\right]}\:\:=\:\:? \\ $$

Commented by MJS last updated on 18/Aug/19

−(4/9)+(2/9)(√(13))

$$−\frac{\mathrm{4}}{\mathrm{9}}+\frac{\mathrm{2}}{\mathrm{9}}\sqrt{\mathrm{13}} \\ $$

Commented by naka3546 last updated on 18/Aug/19

Please show your working

$${Please}\:\:{show}\:\:{your}\:\:{working} \\ $$

Answered by mr W last updated on 18/Aug/19

let BQ=x 2^2 =1^2 +x^2 −2×1×x cos 60° ⇒x^2 −x−3=0 ⇒x=((1+(√(13)))/2) AB=AC=BC=((1+(√(13)))/2)+3=((7+(√(13)))/2) A_(ΔBPQ) =(1/2)×1×((1+(√(13)))/2)×((√3)/2)=(((√3)(1+(√(13))))/8) A_(ΔBAC) =((√3)/4)(((7+(√(13)))/2))^2 =(((√3)(31+7(√(13))))/8) A_(PQCA) =(((√3)(31+7(√(13))))/8)−(((√3)(1+(√(13))))/8)=((6(√3)(5+(√(13))))/8) (A_(ΔBPQ) /A_(PQCA) )=(((√3)(1+(√(13))))/8)×(8/(6(√3)(5+(√(13))))) ⇒(A_(ΔBPQ) /A_(PQCA) )=((1+(√(13)))/(6(5+(√(13)))))=(((√(13))−2)/(18))≈0.089≈(1/(11.2))

$${let}\:{BQ}={x} \\ $$$$\mathrm{2}^{\mathrm{2}} =\mathrm{1}^{\mathrm{2}} +{x}^{\mathrm{2}} −\mathrm{2}×\mathrm{1}×{x}\:\mathrm{cos}\:\mathrm{60}° \\ $$$$\Rightarrow{x}^{\mathrm{2}} −{x}−\mathrm{3}=\mathrm{0} \\ $$$$\Rightarrow{x}=\frac{\mathrm{1}+\sqrt{\mathrm{13}}}{\mathrm{2}} \\ $$$${AB}={AC}={BC}=\frac{\mathrm{1}+\sqrt{\mathrm{13}}}{\mathrm{2}}+\mathrm{3}=\frac{\mathrm{7}+\sqrt{\mathrm{13}}}{\mathrm{2}} \\ $$$${A}_{\Delta{BPQ}} =\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{1}×\frac{\mathrm{1}+\sqrt{\mathrm{13}}}{\mathrm{2}}×\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}=\frac{\sqrt{\mathrm{3}}\left(\mathrm{1}+\sqrt{\mathrm{13}}\right)}{\mathrm{8}} \\ $$$${A}_{\Delta{BAC}} =\frac{\sqrt{\mathrm{3}}}{\mathrm{4}}\left(\frac{\mathrm{7}+\sqrt{\mathrm{13}}}{\mathrm{2}}\right)^{\mathrm{2}} =\frac{\sqrt{\mathrm{3}}\left(\mathrm{31}+\mathrm{7}\sqrt{\mathrm{13}}\right)}{\mathrm{8}} \\ $$$${A}_{{PQCA}} =\frac{\sqrt{\mathrm{3}}\left(\mathrm{31}+\mathrm{7}\sqrt{\mathrm{13}}\right)}{\mathrm{8}}−\frac{\sqrt{\mathrm{3}}\left(\mathrm{1}+\sqrt{\mathrm{13}}\right)}{\mathrm{8}}=\frac{\mathrm{6}\sqrt{\mathrm{3}}\left(\mathrm{5}+\sqrt{\mathrm{13}}\right)}{\mathrm{8}} \\ $$$$\frac{{A}_{\Delta{BPQ}} }{{A}_{{PQCA}} }=\frac{\sqrt{\mathrm{3}}\left(\mathrm{1}+\sqrt{\mathrm{13}}\right)}{\mathrm{8}}×\frac{\mathrm{8}}{\mathrm{6}\sqrt{\mathrm{3}}\left(\mathrm{5}+\sqrt{\mathrm{13}}\right)} \\ $$$$\Rightarrow\frac{{A}_{\Delta{BPQ}} }{{A}_{{PQCA}} }=\frac{\mathrm{1}+\sqrt{\mathrm{13}}}{\mathrm{6}\left(\mathrm{5}+\sqrt{\mathrm{13}}\right)}=\frac{\sqrt{\mathrm{13}}−\mathrm{2}}{\mathrm{18}}\approx\mathrm{0}.\mathrm{089}\approx\frac{\mathrm{1}}{\mathrm{11}.\mathrm{2}} \\ $$

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