Question and Answers Forum

All Questions      Topic List

Geometry Questions

Previous in All Question      Next in All Question      

Previous in Geometry      Next in Geometry      

Question Number 170073 by mr W last updated on 15/May/22

Commented by mr W last updated on 15/May/22

Find ∣CD∣_(min) =?, ∣CD∣_(max) =?

$${Find}\:\mid{CD}\mid_{{min}} =?,\:\mid{CD}\mid_{{max}} =? \\ $$

Commented by cortano1 last updated on 16/May/22

max = 5−((4(√3))/3) = (1/3)(15−4(√3) )?

$$\:{max}\:=\:\mathrm{5}−\frac{\mathrm{4}\sqrt{\mathrm{3}}}{\mathrm{3}}\:=\:\frac{\mathrm{1}}{\mathrm{3}}\left(\mathrm{15}−\mathrm{4}\sqrt{\mathrm{3}}\:\right)?\: \\ $$

Commented by mr W last updated on 16/May/22

no.

$${no}. \\ $$

Answered by ajfour last updated on 16/May/22

Commented by ajfour last updated on 16/May/22

centre of circle P (not A) P((2/( (√3))), 2) [B origin] cos 30°=(2/r) ⇒ r=(4/( (√3))) CP^( 2) =(5−((2(√3))/3))^2 +2^2 =29+(4/3)−((20(√3))/3) =((91−20(√3))/3) CD_(min) =CP−r

$${centre}\:{of}\:{circle}\:{P}\:\left({not}\:{A}\right) \\ $$$${P}\left(\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}},\:\mathrm{2}\right)\:\:\:\:\:\:\:\left[{B}\:{origin}\right] \\ $$$$\mathrm{cos}\:\mathrm{30}°=\frac{\mathrm{2}}{{r}}\:\:\Rightarrow\:\:{r}=\frac{\mathrm{4}}{\:\sqrt{\mathrm{3}}} \\ $$$${CP}^{\:\mathrm{2}} =\left(\mathrm{5}−\frac{\mathrm{2}\sqrt{\mathrm{3}}}{\mathrm{3}}\right)^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:=\mathrm{29}+\frac{\mathrm{4}}{\mathrm{3}}−\frac{\mathrm{20}\sqrt{\mathrm{3}}}{\mathrm{3}} \\ $$$$\:\:\:\:\:\:\:=\frac{\mathrm{91}−\mathrm{20}\sqrt{\mathrm{3}}}{\mathrm{3}} \\ $$$${CD}_{{min}} ={CP}−{r} \\ $$

Commented by mr W last updated on 16/May/22

thanks sir!

$${thanks}\:{sir}! \\ $$

Commented by ajfour last updated on 16/May/22

let me think..

$${let}\:{me}\:{think}.. \\ $$

Commented by Tawa11 last updated on 08/Oct/22

Great sir

$$\mathrm{Great}\:\mathrm{sir} \\ $$

Answered by mr W last updated on 16/May/22

Commented by ajfour last updated on 16/May/22

yeah, sure guess m not focussing..

$${yeah},\:{sure}\:{guess}\:{m}\:{not}\:{focussing}.. \\ $$

Commented by mr W last updated on 16/May/22

r=(2/(cos 30°))=((4(√3))/( 3)) CE=(√((5−(2/( (√3))))^2 +2^2 ))=((√(273−60(√3)))/3) CE′=(√((5+(2/( (√3))))^2 +2^2 ))=((√(273+60(√3)))/3) CD_(min) =CE−r=(((√(273−60(√3)))−4(√3))/3) CD_(max) =CE′+r=(((√(273+60(√3)))+4(√3))/3)

$${r}=\frac{\mathrm{2}}{\mathrm{cos}\:\mathrm{30}°}=\frac{\mathrm{4}\sqrt{\mathrm{3}}}{\:\mathrm{3}} \\ $$$${CE}=\sqrt{\left(\mathrm{5}−\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}\right)^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}} }=\frac{\sqrt{\mathrm{273}−\mathrm{60}\sqrt{\mathrm{3}}}}{\mathrm{3}} \\ $$$${CE}'=\sqrt{\left(\mathrm{5}+\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}\right)^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}} }=\frac{\sqrt{\mathrm{273}+\mathrm{60}\sqrt{\mathrm{3}}}}{\mathrm{3}} \\ $$$${CD}_{{min}} ={CE}−{r}=\frac{\sqrt{\mathrm{273}−\mathrm{60}\sqrt{\mathrm{3}}}−\mathrm{4}\sqrt{\mathrm{3}}}{\mathrm{3}} \\ $$$${CD}_{{max}} ={CE}'+{r}=\frac{\sqrt{\mathrm{273}+\mathrm{60}\sqrt{\mathrm{3}}}+\mathrm{4}\sqrt{\mathrm{3}}}{\mathrm{3}} \\ $$

Commented by ajfour last updated on 16/May/22

fabulous.

$${fabulous}. \\ $$

Answered by cortano1 last updated on 17/May/22

Commented by mr W last updated on 17/May/22

good!

$${good}! \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com