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Question Number 100644 by I want to learn more last updated on 27/Jun/20

Commented by ghiniboss last updated on 27/Jun/20

$$\mathrm{looking}\mathrm{for}\mathrm{x}?$$

Answered by mr W last updated on 28/Jun/20

$$let\mathrm{\angle }ABP=\alpha ,\mathrm{\angle }CBP=\beta$$$$\alpha +\beta =90°$$$${15}^2=x^2+9^2-2x\times 9\cos \alpha$$$$\Rightarrow \cos \alpha =\frac{x^2-144}{18x}$$$${12}^2=x^2+9^2-2x\times 9\cos \beta$$$$\Rightarrow \cos \beta =\frac{x^2-63}{18x}=\sin \alpha$$$${\left(\frac{x^2-144}{18x}\right)}^2+{\left(\frac{x^2-63}{18x}\right)}^2=1$$$$2x^4-738x^2+24705=0$$$$x^2=\frac{738\pm 54\sqrt{119}}{4}$$$$x=\sqrt{\frac{738+54\sqrt{119}}{4}}=18.214$$$$$$$$thereisanothersolution,ifitis$$$$allowedthatPcanlieoutsidethe$$$$triangle:$$$$x=\sqrt{\frac{738-54\sqrt{119}}{4}}=6.101$$

Commented by I want to learn more last updated on 28/Jun/20

$$\mathrm{Thanks}\mathrm{sir}$$

Answered by 1549442205 last updated on 28/Jun/20

Commented by 1549442205 last updated on 28/Jun/20

$$\mathrm{From}\mathrm{the}\mathrm{hypothesis}\mathrm{we}\mathrm{have}:$$$$\{\begin{array}{l}\widehat{\mathrm{ABC}}=90°\\ \mathrm{BP}=9\mathrm{cm}\\ \mathrm{PC}=12\mathrm{cm}\\ \mathrm{PA}=15\mathrm{cm}\end{array}\mathrm{we}\mathrm{need}\mathrm{to}\mathrm{find}\mathrm{x}=\mathrm{AB}=\mathrm{BC}$$$$\mathrm{Denoting}\mathrm{by}\mathrm{D},\mathrm{E}\mathrm{be}\mathrm{projections}\mathrm{of}\mathrm{P}\mathrm{on}$$$$\mathrm{AB}\mathrm{and}\mathrm{BC}\mathrm{respectively}.\mathrm{Putting}\mathrm{BD}=\mathrm{m}$$$$\mathrm{AD}=\mathrm{p},\mathrm{BE}=\mathrm{n},\mathrm{CE}=\mathrm{q},\mathrm{we}\mathrm{have}{9}^2-{\mathrm{n}}^2={\mathrm{PE}}^2={12}^2-{\mathrm{q}}^2\left(1\right)$$$$9^2-{\mathrm{m}}^2={15}^2-{\mathrm{p}}^2\left(2\right).\mathrm{Since},\mathrm{m}+\mathrm{p}=\mathrm{n}+\mathrm{q}=\mathrm{x}$$$$,\mathrm{from}\left(1\right),\left(2\right)\mathrm{we}\mathrm{get}63=\mathrm{x}(\mathrm{q}-\mathrm{n}).\mathrm{From}\mathrm{that}$$$$\mathrm{we}\mathrm{get}\mathrm{the}\mathrm{system}:\{\begin{array}{l}\mathrm{q}+\mathrm{n}=\mathrm{x}\\ \mathrm{q}-\mathrm{n}=\frac{63}{\mathrm{x}}\end{array}\Rightarrow 2\mathrm{n}=\mathrm{x}-\frac{63}{\mathrm{x}}$$$$\Rightarrow \mathrm{n}=\frac{{\mathrm{x}}^2-63}{2\mathrm{x}}.\mathrm{Similarly},\mathrm{we}\mathrm{get}\mathrm{m}=\frac{{\mathrm{x}}^2-144}{2\mathrm{x}}$$$$\mathrm{On}\mathrm{the}\mathrm{other}\mathrm{hands},{\mathrm{m}}^2+{\mathrm{n}}^2=9^2,\mathrm{so}\mathrm{we}\mathrm{have}$$$${\left(\frac{{\mathrm{x}}^2-63}{2\mathrm{x}}\right)}^2+{\left(\frac{{\mathrm{x}}^2-144}{2\mathrm{x}}\right)}^2=81\iff {2\mathrm{x}}^4-{414\mathrm{x}}^2+24705={324\mathrm{x}}^2$$$$\iff {2\mathrm{x}}^4-{738\mathrm{x}}^2+24705=0.\iff {\mathrm{x}}^2\in \{331.7676;37.2323\}$$$$\iff \mathbf{x}\in \{18.21\mathbf{cm};6.10\mathbf{cm}\}$$$$$$

Commented by I want to learn more last updated on 28/Jun/20

$$\mathrm{Thanks}\mathrm{sir}$$

Answered by mr W last updated on 28/Jun/20

$$A(0,a)$$$$B(0,0)$$$$C(a,0)$$$$P(k,h)$$$$k^2+h^2=9^2...\left(i\right)$$$$k^2+{(a-h)}^2={15}^2...\left(ii\right)$$$${(a-k)}^2+h^2={12}^2...\left(iii\right)$$$$$$$$\left(ii\right)-\left(i\right):$$$$a(a-2h)=144$$$$\Rightarrow h=\frac{1}{2}(a-\frac{144}{a})...\left(iv\right)$$$$\left(iii\right)-\left(i\right):$$$$a(a-2k)=63$$$$\Rightarrow k=\frac{1}{2}(a-\frac{63}{a})...\left(v\right)$$$$put\left(iv\right)and\left(v\right)into\left(i\right):$$$$\frac{1}{4}{(a-\frac{144}{a})}^2+\frac{1}{4}{(a-\frac{63}{a})}^2=81$$$${(a-\frac{144}{a})}^2+{(a-\frac{63}{a})}^2=324$$$$2a^4-738a^2+24705=0$$$$a^2=\frac{738\pm 54\sqrt{119}}{4}$$$$\Rightarrow x=a=\sqrt{\frac{738\pm 54\sqrt{119}}{4}}=18.214,6.101$$

Commented by I want to learn more last updated on 28/Jun/20

$$\mathrm{Thanks}\mathrm{sir}$$

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