#3**+2 **

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.

.

Alan Oct 7, 2017

#6**+1 **

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.

Melody
Oct 8, 2017

#5**+2 **

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

Melody Oct 8, 2017