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avatar+171 

what is the perimeter whose area is 387cm2

BOSEOK  Oct 7, 2017
 #1
avatar+92629 
+1

What is the shape?

Melody  Oct 7, 2017
 #2
avatar+171 
+1

The shape is pentagon.

BOSEOK  Oct 7, 2017
 #3
avatar+26722 
+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
edited by Alan  Oct 7, 2017
 #4
avatar+171 
0

How to use calculator to solve this?

BOSEOK  Oct 7, 2017
 #6
avatar+92629 
+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
avatar+92629 
+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

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