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Question Number 158069 by zainaltanjung last updated on 30/Oct/21

Answered by Rasheed.Sindhi last updated on 30/Oct/21

Let the side of old hexagon=a ▲_(old) =6(√(((3a)/2)(((3a)/2)−a)^3 )) ▲_(old) =6(√(((3a)/2)((a/2))^3 )) =6(√(3((a/2))^4 )) =6((a/2))^2 (√3)=6((a^2 /4))(√3)=((3a^2 (√3))/2) Side of new 6-gon=3a ▲_(new) =6(√(((9a)/2)(((9a)/2)−3a)^3 )) =6(√(((9a)/2)(((3a)/2))^3 )) =6(√(3(((3a)/2))^4 )) =6(((3a)/2))^2 (√3) =6(((9a^2 )/4))(√3) =((27a^2 (√3))/2) (▲_(new) /▲_(old) )=(((27a^2 (√3))/2)/((3a^2 (√3))/2))=((27a^2 (√3))/(3a^2 (√3)))=(9/1)=9:1

$${Let}\:{the}\:{side}\:{of}\:\:{old}\:\:{hexagon}={a} \\ $$$$\blacktriangle_{{old}} =\mathrm{6}\sqrt{\frac{\mathrm{3}{a}}{\mathrm{2}}\left(\frac{\mathrm{3}{a}}{\mathrm{2}}−{a}\right)^{\mathrm{3}} } \\ $$$$\blacktriangle_{{old}} =\mathrm{6}\sqrt{\frac{\mathrm{3}{a}}{\mathrm{2}}\left(\frac{{a}}{\mathrm{2}}\right)^{\mathrm{3}} }\:=\mathrm{6}\sqrt{\mathrm{3}\left(\frac{{a}}{\mathrm{2}}\right)^{\mathrm{4}} } \\ $$$$\:\:\:\:\:\:\:\:\:=\mathrm{6}\left(\frac{{a}}{\mathrm{2}}\right)^{\mathrm{2}} \sqrt{\mathrm{3}}=\mathrm{6}\left(\frac{{a}^{\mathrm{2}} }{\mathrm{4}}\right)\sqrt{\mathrm{3}}=\frac{\mathrm{3}{a}^{\mathrm{2}} \sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$${Side}\:{of}\:{new}\:\mathrm{6}-{gon}=\mathrm{3}{a} \\ $$$$\blacktriangle_{{new}} =\mathrm{6}\sqrt{\frac{\mathrm{9}{a}}{\mathrm{2}}\left(\frac{\mathrm{9}{a}}{\mathrm{2}}−\mathrm{3}{a}\right)^{\mathrm{3}} } \\ $$$$\:\:\:\:\:\:\:\:\:\:=\mathrm{6}\sqrt{\frac{\mathrm{9}{a}}{\mathrm{2}}\left(\frac{\mathrm{3}{a}}{\mathrm{2}}\right)^{\mathrm{3}} }\:=\mathrm{6}\sqrt{\mathrm{3}\left(\frac{\mathrm{3}{a}}{\mathrm{2}}\right)^{\mathrm{4}} } \\ $$$$\:\:\:\:\:\:\:\:\:=\mathrm{6}\left(\frac{\mathrm{3}{a}}{\mathrm{2}}\right)^{\mathrm{2}} \sqrt{\mathrm{3}}\:=\mathrm{6}\left(\frac{\mathrm{9}{a}^{\mathrm{2}} }{\mathrm{4}}\right)\sqrt{\mathrm{3}} \\ $$$$\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{27}{a}^{\mathrm{2}} \sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$$\frac{\blacktriangle_{{new}} }{\blacktriangle_{{old}} }=\frac{\frac{\mathrm{27}{a}^{\mathrm{2}} \sqrt{\mathrm{3}}}{\mathrm{2}}}{\frac{\mathrm{3}{a}^{\mathrm{2}} \sqrt{\mathrm{3}}}{\mathrm{2}}}=\frac{\mathrm{27}{a}^{\mathrm{2}} \sqrt{\mathrm{3}}}{\mathrm{3}{a}^{\mathrm{2}} \sqrt{\mathrm{3}}}=\frac{\mathrm{9}}{\mathrm{1}}=\mathrm{9}:\mathrm{1} \\ $$

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