+33661 You can use the following relationship for a regular pentagon:
\(s=\sqrt{4\frac{A}{5}\tan{\frac{\pi}{5}}}\)
where s is a (single) side length (multiply by 5 to get the perimeter) and A is the area.
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+118687 Just put it in exacly as Alan has written it. Replace A with the Area you ahve been given.
Exactly what buttons to push depends on what calc you are using.
\(s=\sqrt{4\frac{A}{5}\tan{\frac{\pi}{5}}}\)
A= 387
sqrt(4*387/5*tan(pi/5)) = 14.9979187446241353 = 15
which is the same I got below.
+118687 what is the perimeter of a regular pentagon whose area is 387cm2
\(Area \;of \;triangle=0.5abSinC\\ 77.4=0.5r^2Sin72\\ \frac{77.4}{0.5Sin72}=r^2\\ r=\sqrt{\frac{77.4}{0.5Sin72}}\)
r= sqr(77.4/(0.5*sin(72))) = 12.757991703716932
\(L^2=a^2+b^2-2abcosC\\ L^2=2r^2-2r^2*cos 72\\ L^2=a^2+b^2-2abcosC\\ L^2=2r^2-2r^2cos 72\\ L^2=2*12.757991703716932^2(1-cos(72))\)
sqrt(2*12.757991703716932^2(1-cos(72))) = 14.997918744628501
5*14.997918744628501 = 74.989593723142505
I get the perimeter to be 75cm long
