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 214040 by ajfour last updated on 25/Nov/24

Commented by ajfour last updated on 25/Nov/24

The △ is equilateral. Find its side s.

$${The}\:\bigtriangleup\:{is}\:{equilateral}.\:{Find}\:{its}\:{side}\:\boldsymbol{{s}}. \\ $$

Answered by mr W last updated on 25/Nov/24

y=x^2   (dy/dx)=2x=tan 30°=(1/( (√3)))  ⇒x=(1/(2(√3)))=p  r=p^2 −r cos 30°=p^2 −(((√3)r)/2)  ⇒r=(p^2 /(1+((√3)/( 2))))  x_C =p+r sin 30°=p+(r/2)=p+(p^2 /(2+(√3)))  s=2x_C =2p+((2p^2 )/(2+(√3)))=2×(1/(2(√3)))+(2/(2+(√3)))×(1/((2(√3))^2 ))  ⇒s=((2+(√3))/( 6)) ✓

$${y}={x}^{\mathrm{2}} \\ $$$$\frac{{dy}}{{dx}}=\mathrm{2}{x}=\mathrm{tan}\:\mathrm{30}°=\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}} \\ $$$$\Rightarrow{x}=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{3}}}={p} \\ $$$${r}={p}^{\mathrm{2}} −{r}\:\mathrm{cos}\:\mathrm{30}°={p}^{\mathrm{2}} −\frac{\sqrt{\mathrm{3}}{r}}{\mathrm{2}} \\ $$$$\Rightarrow{r}=\frac{{p}^{\mathrm{2}} }{\mathrm{1}+\frac{\sqrt{\mathrm{3}}}{\:\mathrm{2}}} \\ $$$${x}_{{C}} ={p}+{r}\:\mathrm{sin}\:\mathrm{30}°={p}+\frac{{r}}{\mathrm{2}}={p}+\frac{{p}^{\mathrm{2}} }{\mathrm{2}+\sqrt{\mathrm{3}}} \\ $$$${s}=\mathrm{2}{x}_{{C}} =\mathrm{2}{p}+\frac{\mathrm{2}{p}^{\mathrm{2}} }{\mathrm{2}+\sqrt{\mathrm{3}}}=\mathrm{2}×\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{3}}}+\frac{\mathrm{2}}{\mathrm{2}+\sqrt{\mathrm{3}}}×\frac{\mathrm{1}}{\left(\mathrm{2}\sqrt{\mathrm{3}}\right)^{\mathrm{2}} } \\ $$$$\Rightarrow{s}=\frac{\mathrm{2}+\sqrt{\mathrm{3}}}{\:\mathrm{6}}\:\checkmark \\ $$

Commented by mr W last updated on 25/Nov/24

Commented by ajfour last updated on 25/Nov/24

Thank you Sir.

$${Thank}\:{you}\:{Sir}. \\ $$

Commented by nikif99 last updated on 25/Nov/24

Is there a unique solution, even in  relation to r?

$${Is}\:{there}\:{a}\:{unique}\:{solution},\:{even}\:{in} \\ $$$${relation}\:{to}\:{r}? \\ $$

Commented by nikif99 last updated on 25/Nov/24

Commented by mr W last updated on 25/Nov/24

two sides of the equilateral should  be the normals of the parabola,  therefore there is an unique   solution.

$${two}\:{sides}\:{of}\:{the}\:{equilateral}\:{should} \\ $$$${be}\:{the}\:{normals}\:{of}\:{the}\:{parabola}, \\ $$$${therefore}\:{there}\:{is}\:{an}\:{unique}\: \\ $$$${solution}. \\ $$

Commented by nikif99 last updated on 25/Nov/24

Thank you, Sir.

$${Thank}\:{you},\:{Sir}. \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com