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Question Number 58756 by Tawa1 last updated on 29/Apr/19

Answered by Kunal12588 last updated on 29/Apr/19

Commented by Kunal12588 last updated on 29/Apr/19

$$PR=\sqrt{5^2+{12}^2}=13cm$$$$$$$$\mathrm{\angle }RBQ=\mathrm{\angle }RQP\left[each90°\right]$$$$\mathrm{\angle }QRB=\mathrm{\angle }PRQ\left[common\right]$$$$\therefore △BQR\sim △QPR$$$$\Rightarrow \frac{BR}{QR}=\frac{QR}{PR}$$$$\Rightarrow \frac{BR}{5}=\frac{5}{13}$$$$\Rightarrow BR=\frac{25}{13}cm$$$$AB=PR-2BR$$$$\Rightarrow AB=13-\frac{50}{13}=\frac{119}{13}cm=9.1538...cm$$

Commented by Tawa1 last updated on 29/Apr/19

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}$$

Commented by maxmathsup by imad last updated on 30/Apr/19

$$wehaveP\overrightarrow{R}.S\overrightarrow{Q}=P\stackrel{-}{R}.A\stackrel{-}{B}(orthogonalprojectiononline\left(PQ\right)$$$$=PR.ABalsowehaveP\overrightarrow{R}.S\overrightarrow{Q}=P\overrightarrow{R}.(S\overrightarrow{P}+P\overrightarrow{Q})$$$$=P\overrightarrow{R}.S\overrightarrow{P}+P\overrightarrow{R}.P\overrightarrow{Q}butP\overrightarrow{R}.S\overrightarrow{P}=S\stackrel{-}{P}.P\overrightarrow{S}=-{SP}^2=-5^2=-25(orth.proj.on\left(SP\right)$$$$P\overrightarrow{R}.P\overrightarrow{Q}=P\stackrel{-}{Q}.P\stackrel{-}{Q}={PQ}^2={12}^2=144\Rightarrow PR.AB=144-25=119$$$$\Rightarrow AB=\frac{119}{PR}andpythagoretheoremgivePR=\sqrt{{12}^2+5^2}=\sqrt{144+25}$$$$=\sqrt{169}=13\Rightarrow AB=\frac{119}{13}.$$

Commented by maxmathsup by imad last updated on 30/Apr/19

$$hereihaveusedscalarproduct.$$

Answered by tanmay last updated on 29/Apr/19

$$\sqrt{{12}^2+5^2}=13$$$$5\times 12=2\left(\frac{1}{2}\times 13\times h\right)$$$$h=\frac{60}{13}$$$$AB+2x=13$$$$x^2+h^2=5^2$$$$x^2=25-\frac{3600}{169}$$$$x^2=\frac{{\left(5\times 13\right)}^2-{\left(60\right)}^2}{{13}^2}$$$$x^2=\frac{125\times 5}{{13}^2}$$$$x=\frac{25}{13}$$$$AB+2x=13$$$$AB=13-2\times \frac{25}{13}$$$$AB=\frac{169-50}{13}=\frac{119}{13}$$

Commented by Tawa1 last updated on 29/Apr/19

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}$$

Answered by ajfour last updated on 29/Apr/19

$$\mathrm{diagonal}\mathrm{d}=13$$$$12\mathrm{cm}=\mathrm{dcos}\theta 5\mathrm{cm}=\mathrm{dsin}\theta$$$$\tan \theta =\frac{5}{12}$$$$\mathrm{AB}=\mathrm{d}-2\times \left(5\mathrm{cm}\right)\sin \theta$$$$=13\mathrm{cm}-2\times 5\mathrm{cm}\times \left(\frac{5}{13}\right)$$$$=\frac{169\mathrm{cm}-50\mathrm{cm}}{13}=9\frac{2}{13}\mathrm{cm}.$$

Commented by Tawa1 last updated on 29/Apr/19

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}$$

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