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Question Number 191364 by cortano12 last updated on 23/Apr/23

Commented by cortano12 last updated on 23/Apr/23

$$\mathrm{find}\mathrm{CD}.$$

Commented by mr W last updated on 24/Apr/23

$$isBEdiameter?otherwisethereis$$$$nouniquesolution.$$

Answered by a.lgnaoui last updated on 23/Apr/23

$$△\mathrm{ABC}\mathrm{rectangle}\mathrm{en}\mathrm{B};\left(\mathrm{AC}\mathrm{diametre}\right)$$$${\mathrm{AC}}^2={\mathrm{AB}}^2+{\mathrm{BC}}^2\Rightarrow \mathrm{AC}=\sqrt{36+16}=\sqrt{42}$$$$△\mathrm{BED}(\mathrm{E}\mathrm{tangente}\mathrm{au}\mathrm{cercle}$$$$\measuredangle \mathrm{AED}=90\mathrm{de}\mathrm{plus}\mathrm{AE}\mathrm{diametre}\mathrm{de}\mathrm{C}$$$$\Rightarrow \mathrm{BCE}=90\mathrm{donc}$$$$\Rightarrow \mathrm{BCE}\mathrm{et}\mathrm{CED}\mathrm{triangles}\mathrm{rectangles}$$$$\mathrm{CE}\mid \mid \mathrm{AB}\Rightarrow \mathrm{CE}=\mathrm{AB}=4$$$$\bullet △\mathrm{BED}{\mathrm{BD}}^2={\mathrm{BE}}^2+{\mathrm{DE}}^2\left(1\right)$$$$\mathrm{CD}=\mathrm{x}\mathrm{BD}=6+\mathrm{x}$$$$\mathrm{BE}=\mathrm{AC}=\sqrt{42}$$$$\bullet \mathrm{CED}{\mathrm{DE}}^2={\mathrm{CE}}^2+{\mathrm{CD}}^2$$$${\mathrm{DE}}^2=16+{\mathrm{x}}^2\left(2\right)$$$$\left(1\right)\mathrm{et}\left(2\right)\Rightarrow {\mathrm{DE}}^2={(6+\mathrm{x})}^2-42=16+{\mathrm{x}}^2$$$${\mathrm{x}}^2+12\mathrm{x}-6=16+{\mathrm{x}}^2$$$$12\mathrm{x}=22$$$$\Rightarrow \mathbf{CD}=\mathbf{x}=\frac{11}{6}$$$$$$

Commented by cortano12 last updated on 24/Apr/23

$$\mathrm{ans}\frac{8}{3}$$

Answered by mr W last updated on 24/Apr/23

$$BE=AC$$$$sayCD=x$$$$\frac{6+x}{ED}=\frac{ED}{x}\Rightarrow {ED}^2=x(6+x)$$$${ED}^2={(6+x)}^2-{BE}^2={(6+x)}^2-{AC}^2={(6+x)}^2-(4^2+6^2)$$$$x(6+x)={(6+x)}^2-(4^2+6^2)$$$$\Rightarrow x=\frac{8}{3}✓$$

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