Question and Answers Forum

All Questions      Topic List

Mensuration Questions

Previous in All Question      Next in All Question      

Previous in Mensuration      Next in Mensuration      

Question Number 143332 by bramlexs22 last updated on 13/Jun/21

Commented by soumyasaha last updated on 13/Jun/21

 x = (√((3+(√5))/2))

$$\:\mathrm{x}\:=\:\sqrt{\frac{\mathrm{3}+\sqrt{\mathrm{5}}}{\mathrm{2}}} \\ $$

Commented by mr W last updated on 13/Jun/21

(√((3+(√5))/2))=(√((6+2(√5))/4))=(√((((√5)+1)^2 )/2^2 ))=(((√5)+1)/2)

$$\sqrt{\frac{\mathrm{3}+\sqrt{\mathrm{5}}}{\mathrm{2}}}=\sqrt{\frac{\mathrm{6}+\mathrm{2}\sqrt{\mathrm{5}}}{\mathrm{4}}}=\sqrt{\frac{\left(\sqrt{\mathrm{5}}+\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{2}^{\mathrm{2}} }}=\frac{\sqrt{\mathrm{5}}+\mathrm{1}}{\mathrm{2}} \\ $$

Answered by nimnim last updated on 13/Jun/21

cosθ=((x^2 +1^2 −((√(√5)))^2 )/(2.1.x))=((x^2 +1−(√5))/(2x))  also cos(180−θ)=((x^2 +1^2 −((√5))^2 )/(2.x.1))           −cosθ=((x^2 −4)/(2x))⇒cosθ=((−x^2 +4)/(2x))  ∴((x^2 +1−(√5))/(2x))=((−x^2 +4)/(2x))  ⇒2x^2 =3+(√5) ⇔4x^2 =6+2(√5)  ⇒x^2 =(((1+(√5))^2 )/4)  ⇔  x=±((1+(√5))/2)  ⇒  x=((1+(√5))/2)=ϕ  (as length x>0)

$${cos}\theta=\frac{{x}^{\mathrm{2}} +\mathrm{1}^{\mathrm{2}} −\left(\sqrt{\sqrt{\mathrm{5}}}\right)^{\mathrm{2}} }{\mathrm{2}.\mathrm{1}.{x}}=\frac{{x}^{\mathrm{2}} +\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{2}{x}} \\ $$$${also}\:{cos}\left(\mathrm{180}−\theta\right)=\frac{{x}^{\mathrm{2}} +\mathrm{1}^{\mathrm{2}} −\left(\sqrt{\mathrm{5}}\right)^{\mathrm{2}} }{\mathrm{2}.{x}.\mathrm{1}} \\ $$$$\:\:\:\:\:\:\:\:\:−{cos}\theta=\frac{{x}^{\mathrm{2}} −\mathrm{4}}{\mathrm{2}{x}}\Rightarrow{cos}\theta=\frac{−{x}^{\mathrm{2}} +\mathrm{4}}{\mathrm{2}{x}} \\ $$$$\therefore\frac{{x}^{\mathrm{2}} +\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{2}{x}}=\frac{−{x}^{\mathrm{2}} +\mathrm{4}}{\mathrm{2}{x}} \\ $$$$\Rightarrow\mathrm{2}{x}^{\mathrm{2}} =\mathrm{3}+\sqrt{\mathrm{5}}\:\Leftrightarrow\mathrm{4}{x}^{\mathrm{2}} =\mathrm{6}+\mathrm{2}\sqrt{\mathrm{5}} \\ $$$$\Rightarrow{x}^{\mathrm{2}} =\frac{\left(\mathrm{1}+\sqrt{\mathrm{5}}\right)^{\mathrm{2}} }{\mathrm{4}}\:\:\Leftrightarrow\:\:{x}=\pm\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$$$\Rightarrow\:\:{x}=\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}=\varphi\:\:\left({as}\:{length}\:{x}>\mathrm{0}\right) \\ $$

Commented by nimnim last updated on 13/Jun/21

Answered by mr W last updated on 13/Jun/21

Commented by mr W last updated on 13/Jun/21

1 is a median of triangle with  sides (√(√5)), (√5) and 2x.  4×1^2 =2((√(√5)))^2 +2((√5))^2 −(2x)^2   ((√5)+1)^2 =(2x)^2   ⇒x=(((√5)+1)/2)

$$\mathrm{1}\:{is}\:{a}\:{median}\:{of}\:{triangle}\:{with} \\ $$$${sides}\:\sqrt{\sqrt{\mathrm{5}}},\:\sqrt{\mathrm{5}}\:{and}\:\mathrm{2}{x}. \\ $$$$\mathrm{4}×\mathrm{1}^{\mathrm{2}} =\mathrm{2}\left(\sqrt{\sqrt{\mathrm{5}}}\right)^{\mathrm{2}} +\mathrm{2}\left(\sqrt{\mathrm{5}}\right)^{\mathrm{2}} −\left(\mathrm{2}{x}\right)^{\mathrm{2}} \\ $$$$\left(\sqrt{\mathrm{5}}+\mathrm{1}\right)^{\mathrm{2}} =\left(\mathrm{2}{x}\right)^{\mathrm{2}} \\ $$$$\Rightarrow{x}=\frac{\sqrt{\mathrm{5}}+\mathrm{1}}{\mathrm{2}} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com