what-is-the-perimeter-of-a-regular-dodecagon-12-sides-whose-area-is-24-12-3-

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

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

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

$$\mathrm{good} \\ $$

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