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

Commented by cortano12 last updated on 23/Apr/23

find CD.

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

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

is BE diameter? otherwise there is no unique solution.

$${is}\:{BE}\:{diameter}?\:{otherwise}\:{there}\:{is} \\ $$$${no}\:{unique}\:{solution}. \\ $$

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

△ABC rectangle en B ;(AC diametre) AC^2 =AB^2 +BC^2 ⇒ AC=(√(36+16)) =(√(42)) △BED (E tangente au cercle ∡AED=90 de plus AE diametre de C ⇒ BCE=90 donc ⇒BCE et CED triangles rectangles CE∣∣AB ⇒CE=AB=4 •△BED BD^2 =BE^2 +DE^2 (1) CD=x BD=6+x BE=AC=(√(42)) • CED DE^2 =CE^2 +CD^2 DE^2 =16+x^2 (2) (1) et (2)⇒DE^2 =(6+x)^2 −42=16+x^2 x^2 +12x−6=16+x^2 12x =22 ⇒ CD=x= ((11)/6)

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

Commented by cortano12 last updated on 24/Apr/23

ans (8/3)

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

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

BE=AC say CD=x ((6+x)/(ED))=((ED)/x) ⇒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 ) ⇒x=(8/3) ✓

$${BE}={AC} \\ $$$${say}\:{CD}={x} \\ $$$$\frac{\mathrm{6}+{x}}{{ED}}=\frac{{ED}}{{x}}\:\Rightarrow{ED}^{\mathrm{2}} ={x}\left(\mathrm{6}+{x}\right) \\ $$$${ED}^{\mathrm{2}} =\left(\mathrm{6}+{x}\right)^{\mathrm{2}} −{BE}^{\mathrm{2}} =\left(\mathrm{6}+{x}\right)^{\mathrm{2}} −{AC}^{\mathrm{2}} =\left(\mathrm{6}+{x}\right)^{\mathrm{2}} −\left(\mathrm{4}^{\mathrm{2}} +\mathrm{6}^{\mathrm{2}} \right) \\ $$$${x}\left(\mathrm{6}+{x}\right)=\left(\mathrm{6}+{x}\right)^{\mathrm{2}} −\left(\mathrm{4}^{\mathrm{2}} +\mathrm{6}^{\mathrm{2}} \right) \\ $$$$\Rightarrow{x}=\frac{\mathrm{8}}{\mathrm{3}}\:\checkmark \\ $$

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