Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

Question Number 97051 by john santu last updated on 06/Jun/20

what is the perimeter of a regular   dodecagon (12 sides) whose   area is 24+12(√3) ?

$$\mathrm{what}\:\mathrm{is}\:\mathrm{the}\:\mathrm{perimeter}\:\mathrm{of}\:\mathrm{a}\:\mathrm{regular}\: \\ $$$$\mathrm{dodecagon}\:\left(\mathrm{12}\:\mathrm{sides}\right)\:\mathrm{whose}\: \\ $$$$\mathrm{area}\:\mathrm{is}\:\mathrm{24}+\mathrm{12}\sqrt{\mathrm{3}}\:?\: \\ $$

Commented by bemath last updated on 06/Jun/20

let the sides of dodecagon is a  area = 12×(r^2 /2)×sin (((360^o )/(12)))  24+12(√3) =6r^2 ×sin 30^o   24+12(√3) = 3r^2  ⇒ r = (√(8+4(√3)))  using Cosine of law   a = (√(2r^2 −2r^2 ×cos 30^o ))  a =(√(2r^2 (1−(1/2)(√3))))  a = (√(2(8+4(√3))(((2−(√3))/2))))  a = (√((8+4(√3))(2−(√3))))   a = (√(16−8(√3)+8(√3)−12)) = 2  so the perimeter of dodecagon  equal to 24 .

$$\mathrm{let}\:\mathrm{the}\:\mathrm{sides}\:\mathrm{of}\:\mathrm{dodecagon}\:\mathrm{is}\:{a} \\ $$$${area}\:=\:\mathrm{12}×\frac{{r}^{\mathrm{2}} }{\mathrm{2}}×\mathrm{sin}\:\left(\frac{\mathrm{360}^{{o}} }{\mathrm{12}}\right) \\ $$$$\mathrm{24}+\mathrm{12}\sqrt{\mathrm{3}}\:=\mathrm{6}{r}^{\mathrm{2}} ×\mathrm{sin}\:\mathrm{30}^{{o}} \\ $$$$\mathrm{24}+\mathrm{12}\sqrt{\mathrm{3}}\:=\:\mathrm{3}{r}^{\mathrm{2}} \:\Rightarrow\:{r}\:=\:\sqrt{\mathrm{8}+\mathrm{4}\sqrt{\mathrm{3}}} \\ $$$${using}\:{C}\mathrm{osine}\:\mathrm{of}\:\mathrm{law}\: \\ $$$${a}\:=\:\sqrt{\mathrm{2}{r}^{\mathrm{2}} −\mathrm{2}{r}^{\mathrm{2}} ×\mathrm{cos}\:\mathrm{30}^{{o}} } \\ $$$${a}\:=\sqrt{\mathrm{2}{r}^{\mathrm{2}} \left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\mathrm{3}}\right)} \\ $$$${a}\:=\:\sqrt{\mathrm{2}\left(\mathrm{8}+\mathrm{4}\sqrt{\mathrm{3}}\right)\left(\frac{\mathrm{2}−\sqrt{\mathrm{3}}}{\mathrm{2}}\right)} \\ $$$${a}\:=\:\sqrt{\left(\mathrm{8}+\mathrm{4}\sqrt{\mathrm{3}}\right)\left(\mathrm{2}−\sqrt{\mathrm{3}}\right)}\: \\ $$$${a}\:=\:\sqrt{\mathrm{16}−\mathrm{8}\sqrt{\mathrm{3}}+\mathrm{8}\sqrt{\mathrm{3}}−\mathrm{12}}\:=\:\mathrm{2} \\ $$$${so}\:{the}\:{perimeter}\:{of}\:{dodecagon} \\ $$$${equal}\:{to}\:\mathrm{24}\:.\: \\ $$

Commented by bobhans last updated on 06/Jun/20

good

$$\mathrm{good} \\ $$

Answered by 1549442205 last updated on 06/Jun/20

The angle at the center imtercept sides of  dodecagon equal  to ((360°)/(12))=30° .Hence,the length of  side of dodecagon is  a=2Rsin15°=2R×((1−cos30°)/2)=R(1−((√3)/2)).  It follows that the perimeter of  dodecagon  equal to 12×R(1−((√3)/2))(1).The area of   dodecagon equal to 6R^(2 ) sin30°=24+12(√3)  ⇒R=(√(8+4(√3) )) =(√2) ×(√(((√3)+1)^2 )) =(√6) +(√2)  Replace into (1) we get the perimeter of  dodecagon is C=6((√6)+(√2))(2−(√3))=  6((√6)−(√2))  Thus,C=6((√6)−(√2))

$$\mathrm{The}\:\mathrm{angle}\:\mathrm{at}\:\mathrm{the}\:\mathrm{center}\:\mathrm{imtercept}\:\mathrm{sides}\:\mathrm{of}\:\:\mathrm{dodecagon}\:\mathrm{equal} \\ $$$$\mathrm{to}\:\frac{\mathrm{360}°}{\mathrm{12}}=\mathrm{30}°\:.\mathrm{Hence},\mathrm{the}\:\mathrm{length}\:\mathrm{of}\:\:\mathrm{side}\:\mathrm{of}\:\mathrm{dodecagon}\:\mathrm{is} \\ $$$$\mathrm{a}=\mathrm{2Rsin15}°=\mathrm{2R}×\frac{\mathrm{1}−\mathrm{cos30}°}{\mathrm{2}}=\mathrm{R}\left(\mathrm{1}−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right). \\ $$$$\mathrm{It}\:\mathrm{follows}\:\mathrm{that}\:\mathrm{the}\:\mathrm{perimeter}\:\mathrm{of}\:\:\mathrm{dodecagon} \\ $$$$\mathrm{equal}\:\mathrm{to}\:\mathrm{12}×\mathrm{R}\left(\mathrm{1}−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)\left(\mathrm{1}\right).\mathrm{The}\:\mathrm{area}\:\mathrm{of}\: \\ $$$$\mathrm{dodecagon}\:\mathrm{equal}\:\mathrm{to}\:\mathrm{6R}^{\mathrm{2}\:} \mathrm{sin30}°=\mathrm{24}+\mathrm{12}\sqrt{\mathrm{3}} \\ $$$$\Rightarrow\mathrm{R}=\sqrt{\mathrm{8}+\mathrm{4}\sqrt{\mathrm{3}}\:}\:=\sqrt{\mathrm{2}}\:×\sqrt{\left(\sqrt{\mathrm{3}}+\mathrm{1}\right)^{\mathrm{2}} }\:=\sqrt{\mathrm{6}}\:+\sqrt{\mathrm{2}} \\ $$$$\mathrm{Replace}\:\mathrm{into}\:\left(\mathrm{1}\right)\:\mathrm{we}\:\mathrm{get}\:\mathrm{the}\:\mathrm{perimeter}\:\mathrm{of} \\ $$$$\mathrm{dodecagon}\:\mathrm{is}\:\mathrm{C}=\mathrm{6}\left(\sqrt{\mathrm{6}}+\sqrt{\mathrm{2}}\right)\left(\mathrm{2}−\sqrt{\mathrm{3}}\right)= \\ $$$$\mathrm{6}\left(\sqrt{\mathrm{6}}−\sqrt{\mathrm{2}}\right) \\ $$$$\mathrm{Thus},\boldsymbol{\mathrm{C}}=\mathrm{6}\left(\sqrt{\mathrm{6}}−\sqrt{\mathrm{2}}\right) \\ $$

Commented by bobhans last updated on 06/Jun/20

sin^2 (15^o ) = (1/2)−(1/2)cos 30^o   not sin 15^o  = ((1−cos 30^o )/2)

$$\mathrm{sin}\:^{\mathrm{2}} \left(\mathrm{15}^{\mathrm{o}} \right)\:=\:\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cos}\:\mathrm{30}^{\mathrm{o}} \\ $$$$\mathrm{not}\:\mathrm{sin}\:\mathrm{15}^{\mathrm{o}} \:=\:\frac{\mathrm{1}−\mathrm{cos}\:\mathrm{30}^{\mathrm{o}} }{\mathrm{2}}\: \\ $$

Commented by 1549442205 last updated on 07/Jun/20

Thank you,i am sorry done a mistake.  It must be sin15°=(√((1−cos30°)/2))=(√((2−(√3))/4))  =(√((((√3)−1)^2 )/8))=(((√3)−1)/(2(√2)))=(((√6)−(√2))/2).Hence,  a=R((√6) −(√2) )⇒C=12R((√6) −(√2) )  =12((√6)+(√2))((√6)−(√2))=48

$$\mathrm{Thank}\:\mathrm{you},\mathrm{i}\:\mathrm{am}\:\mathrm{sorry}\:\mathrm{done}\:\mathrm{a}\:\mathrm{mistake}. \\ $$$$\mathrm{It}\:\mathrm{must}\:\mathrm{be}\:\mathrm{sin15}°=\sqrt{\frac{\mathrm{1}−\mathrm{cos30}°}{\mathrm{2}}}=\sqrt{\frac{\mathrm{2}−\sqrt{\mathrm{3}}}{\mathrm{4}}} \\ $$$$=\sqrt{\frac{\left(\sqrt{\mathrm{3}}−\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{8}}}=\frac{\sqrt{\mathrm{3}}−\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}=\frac{\sqrt{\mathrm{6}}−\sqrt{\mathrm{2}}}{\mathrm{2}}.\mathrm{Hence}, \\ $$$$\mathrm{a}=\mathrm{R}\left(\sqrt{\mathrm{6}}\:−\sqrt{\mathrm{2}}\:\right)\Rightarrow\mathrm{C}=\mathrm{12R}\left(\sqrt{\mathrm{6}}\:−\sqrt{\mathrm{2}}\:\right) \\ $$$$=\mathrm{12}\left(\sqrt{\mathrm{6}}+\sqrt{\mathrm{2}}\right)\left(\sqrt{\mathrm{6}}−\sqrt{\mathrm{2}}\right)=\mathrm{48} \\ $$

Commented by john santu last updated on 07/Jun/20

but the answer is 24 sir

$$\mathrm{but}\:\mathrm{the}\:\mathrm{answer}\:\mathrm{is}\:\mathrm{24}\:\mathrm{sir} \\ $$

Answered by john santu last updated on 07/Jun/20

Commented by bemath last updated on 07/Jun/20

i agree sir. great

$$\mathrm{i}\:\mathrm{agree}\:\mathrm{sir}.\:\mathrm{great} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com