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

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

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

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

$$x^2+y^2={45}^2.....\left(i\right)$$$$z^2-y^2={48}^2.....\left(ii\right)$$$${(x+48)}^2-z^2={45}^2...\left(iii\right)$$$$\left(i\right)+\left(ii\right):x^2+z^2={45}^2+{48}^2...\left(iv\right)$$$$\left(iii\right)+\left(iv\right):{(x+48)}^2+x^2=2{\left(45\right)}^2+{48}^2$$$$2x^2+96x=2{\left(45\right)}^2$$$$x^2+48x-{45}^2=0$$$$x=\frac{-48\pm \sqrt{{48}^2+4{\left(45\right)}^2}}{2}$$$$x=\frac{-48\pm 102}{2}=\check{27},-\stackrel{\times }{75}$$$$\left(i\right):y^2={45}^2-{27}^2\Rightarrow y=36$$$$\left(ii\right):z^2={48}^2+{36}^2\Rightarrow z=60$$

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