Question-203187

Question Number 203187 by Amidip last updated on 12/Jan/24

Answered by mr W last updated on 12/Jan/24

supposed: AB⊥AC, AD⊥BC ((45)/x)=((x+48)/(45)) ⇒x^2 +48x−45^2 =0 ⇒x=−24+(√(24^2 +45^2 ))=27 (x/y)=(y/(48)) ⇒y=(√(48x))=(√(48×27))=36 (z/(48))=((45)/y) ⇒z=((45×48)/y)=((45×48)/(36))=60

$${supposed}:\:{AB}\bot{AC},\:{AD}\bot{BC} \\ $$$$\frac{\mathrm{45}}{{x}}=\frac{{x}+\mathrm{48}}{\mathrm{45}}\: \\ $$$$\Rightarrow{x}^{\mathrm{2}} +\mathrm{48}{x}−\mathrm{45}^{\mathrm{2}} =\mathrm{0} \\ $$$$\Rightarrow{x}=−\mathrm{24}+\sqrt{\mathrm{24}^{\mathrm{2}} +\mathrm{45}^{\mathrm{2}} }=\mathrm{27} \\ $$$$ \\ $$$$\frac{{x}}{{y}}=\frac{{y}}{\mathrm{48}}\: \\ $$$$\Rightarrow{y}=\sqrt{\mathrm{48}{x}}=\sqrt{\mathrm{48}×\mathrm{27}}=\mathrm{36} \\ $$$$ \\ $$$$\frac{{z}}{\mathrm{48}}=\frac{\mathrm{45}}{{y}} \\ $$$$\Rightarrow{z}=\frac{\mathrm{45}×\mathrm{48}}{{y}}=\frac{\mathrm{45}×\mathrm{48}}{\mathrm{36}}=\mathrm{60} \\ $$

Answered by Rasheed.Sindhi last updated on 12/Jan/24

x^2 +y^2 =45^2 .....(i) z^2 −y^2 =48^2 .....(ii) (x+48)^2 −z^2 =45^2 ...(iii) (i)+(ii): x^2 +z^2 =45^2 +48^2 ...(iv) (iii)+(iv): (x+48)^2 +x^2 =2(45)^2 +48^2 2x^2 +96x=2(45)^2 x^2 +48x−45^2 =0 x=((−48±(√(48^2 +4(45)^2 )))/2) x=((−48±102)/2)=27^(✓) ,−75^(×) (i):y^2 =45^2 −27^2 ⇒y=36 (ii): z^2 =48^2 +36^2 ⇒z=60

$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} =\mathrm{45}^{\mathrm{2}} …..\left({i}\right) \\ $$$${z}^{\mathrm{2}} −{y}^{\mathrm{2}} =\mathrm{48}^{\mathrm{2}} …..\left({ii}\right) \\ $$$$\left({x}+\mathrm{48}\right)^{\mathrm{2}} −{z}^{\mathrm{2}} =\mathrm{45}^{\mathrm{2}} …\left({iii}\right) \\ $$$$\left({i}\right)+\left({ii}\right):\:{x}^{\mathrm{2}} +{z}^{\mathrm{2}} =\mathrm{45}^{\mathrm{2}} +\mathrm{48}^{\mathrm{2}} …\left({iv}\right) \\ $$$$\left({iii}\right)+\left({iv}\right):\:\left({x}+\mathrm{48}\right)^{\mathrm{2}} +{x}^{\mathrm{2}} =\mathrm{2}\left(\mathrm{45}\right)^{\mathrm{2}} +\mathrm{48}^{\mathrm{2}} \\ $$$$\mathrm{2}{x}^{\mathrm{2}} +\mathrm{96}{x}=\mathrm{2}\left(\mathrm{45}\right)^{\mathrm{2}} \\ $$$${x}^{\mathrm{2}} +\mathrm{48}{x}−\mathrm{45}^{\mathrm{2}} =\mathrm{0} \\ $$$${x}=\frac{−\mathrm{48}\pm\sqrt{\mathrm{48}^{\mathrm{2}} +\mathrm{4}\left(\mathrm{45}\right)^{\mathrm{2}} }}{\mathrm{2}} \\ $$$${x}=\frac{−\mathrm{48}\pm\mathrm{102}}{\mathrm{2}}=\overset{\checkmark} {\mathrm{27}},−\overset{×} {\mathrm{75}} \\ $$$$\left({i}\right):{y}^{\mathrm{2}} =\mathrm{45}^{\mathrm{2}} −\mathrm{27}^{\mathrm{2}} \Rightarrow{y}=\mathrm{36} \\ $$$$\left({ii}\right):\:{z}^{\mathrm{2}} =\mathrm{48}^{\mathrm{2}} +\mathrm{36}^{\mathrm{2}} \Rightarrow{z}=\mathrm{60} \\ $$

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