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Question-76108

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Question Number 76108 by Maclaurin Stickker last updated on 23/Dec/19
Commented by Maclaurin Stickker last updated on 23/Dec/19
in the figure, the circumferences have radius 8 cm and 6 cm and the distance between their centers is 12 cm. If QP=PR, find QP.
$${in}\:{the}\:{figure},\:{the}\:{circumferences} \\ $$$${have}\:{radius}\:\mathrm{8}\:{cm}\:{and}\:\mathrm{6}\:{cm}\:{and}\:{the} \\ $$$${distance}\:{between}\:{their}\:{centers}\:{is}\:\mathrm{12}\:{cm}.\: \\ $$$${If}\:{QP}={PR},\:{find}\:{QP}. \\ $$
Answered by mr W last updated on 23/Dec/19
Commented by mr W last updated on 23/Dec/19
r_1 =8, r_2 =6 let AP=2b (√(8^2 −b^2 ))+(√(6^2 −b^2 ))=12 ((√(8^2 −b^2 )))^2 =(12−(√(6^2 −b^2 )))^2 0=116−24(√(36−b^2 )) 29=6(√(36−b^2 )) (((29)/6))^2 =36−b^2 ⇒b=(√(36−(((29)/6))^2 ))=((√(455))/6) AP=2b=((√(455))/3) α=cos^(−1) (b/r_1 )=cos^(−1) ((√(455))/(48)) β=cos^(−1) (b/r_2 )=cos^(−1) ((√(455))/(36)) let QP=PR=x sin ∠QAP=(x/(2r_1 ))=(x/(16)) sin ∠PAR=(x/(2r_2 ))=(x/(12)) ∠QAP+∠PAR=π−(α+β) cos (∠QAP+∠PAR)=−cos (α+β) ((√((16^2 −x^2 )(12^2 −x^2 )))/(16×12))−(x^2 /(16×12))=((43×29)/(48×36))−((455)/(36×48)) (√((16^2 −x^2 )(12^2 −x^2 )))−x^2 =88 (16^2 −x^2 )(12^2 −x^2 )=(88+x^2 )^2 16^2 ×12^2 −88^2 =(16^2 +12^2 +2×88)x^2 455=9x^2 ⇒x=((√(455))/3)
$${r}_{\mathrm{1}} =\mathrm{8},\:{r}_{\mathrm{2}} =\mathrm{6} \\ $$$${let}\:{AP}=\mathrm{2}{b} \\ $$$$\sqrt{\mathrm{8}^{\mathrm{2}} −{b}^{\mathrm{2}} }+\sqrt{\mathrm{6}^{\mathrm{2}} −{b}^{\mathrm{2}} }=\mathrm{12} \\ $$$$\left(\sqrt{\mathrm{8}^{\mathrm{2}} −{b}^{\mathrm{2}} }\right)^{\mathrm{2}} =\left(\mathrm{12}−\sqrt{\mathrm{6}^{\mathrm{2}} −{b}^{\mathrm{2}} }\right)^{\mathrm{2}} \\ $$$$\mathrm{0}=\mathrm{116}−\mathrm{24}\sqrt{\mathrm{36}−{b}^{\mathrm{2}} } \\ $$$$\mathrm{29}=\mathrm{6}\sqrt{\mathrm{36}−{b}^{\mathrm{2}} } \\ $$$$\left(\frac{\mathrm{29}}{\mathrm{6}}\right)^{\mathrm{2}} =\mathrm{36}−{b}^{\mathrm{2}} \\ $$$$\Rightarrow{b}=\sqrt{\mathrm{36}−\left(\frac{\mathrm{29}}{\mathrm{6}}\right)^{\mathrm{2}} }=\frac{\sqrt{\mathrm{455}}}{\mathrm{6}} \\ $$$${AP}=\mathrm{2}{b}=\frac{\sqrt{\mathrm{455}}}{\mathrm{3}} \\ $$$$\alpha=\mathrm{cos}^{−\mathrm{1}} \frac{{b}}{{r}_{\mathrm{1}} }=\mathrm{cos}^{−\mathrm{1}} \frac{\sqrt{\mathrm{455}}}{\mathrm{48}} \\ $$$$\beta=\mathrm{cos}^{−\mathrm{1}} \frac{{b}}{{r}_{\mathrm{2}} }=\mathrm{cos}^{−\mathrm{1}} \frac{\sqrt{\mathrm{455}}}{\mathrm{36}} \\ $$$${let}\:{QP}={PR}={x} \\ $$$$\mathrm{sin}\:\angle{QAP}=\frac{{x}}{\mathrm{2}{r}_{\mathrm{1}} }=\frac{{x}}{\mathrm{16}} \\ $$$$\mathrm{sin}\:\angle{PAR}=\frac{{x}}{\mathrm{2}{r}_{\mathrm{2}} }=\frac{{x}}{\mathrm{12}} \\ $$$$\angle{QAP}+\angle{PAR}=\pi−\left(\alpha+\beta\right) \\ $$$$\mathrm{cos}\:\left(\angle{QAP}+\angle{PAR}\right)=−\mathrm{cos}\:\left(\alpha+\beta\right) \\ $$$$\frac{\sqrt{\left(\mathrm{16}^{\mathrm{2}} −{x}^{\mathrm{2}} \right)\left(\mathrm{12}^{\mathrm{2}} −{x}^{\mathrm{2}} \right)}}{\mathrm{16}×\mathrm{12}}−\frac{{x}^{\mathrm{2}} }{\mathrm{16}×\mathrm{12}}=\frac{\mathrm{43}×\mathrm{29}}{\mathrm{48}×\mathrm{36}}−\frac{\mathrm{455}}{\mathrm{36}×\mathrm{48}} \\ $$$$\sqrt{\left(\mathrm{16}^{\mathrm{2}} −{x}^{\mathrm{2}} \right)\left(\mathrm{12}^{\mathrm{2}} −{x}^{\mathrm{2}} \right)}−{x}^{\mathrm{2}} =\mathrm{88} \\ $$$$\left(\mathrm{16}^{\mathrm{2}} −{x}^{\mathrm{2}} \right)\left(\mathrm{12}^{\mathrm{2}} −{x}^{\mathrm{2}} \right)=\left(\mathrm{88}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} \\ $$$$\mathrm{16}^{\mathrm{2}} ×\mathrm{12}^{\mathrm{2}} −\mathrm{88}^{\mathrm{2}} =\left(\mathrm{16}^{\mathrm{2}} +\mathrm{12}^{\mathrm{2}} +\mathrm{2}×\mathrm{88}\right){x}^{\mathrm{2}} \\ $$$$\mathrm{455}=\mathrm{9}{x}^{\mathrm{2}} \\ $$$$\Rightarrow{x}=\frac{\sqrt{\mathrm{455}}}{\mathrm{3}} \\ $$

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