Area of a pentagon

K

K‡rbyroth

New member
Joined
Jan 23, 2009
Messages
15
pentagon.jpg


I got from this the hypo(r)= 3.4 longleg(a) = 2 the short leg should = 1.7

A= 1/2aP a = apothem P = perimeter
A= 1/2(2*17)
A = 17

I thinking the apothem is wrong here. I know the triangle is wrong unless i missing something
 
Try again: you have 5 identical isosceles triangles (equal sides = r) and height = a.
Hint: look for right triangles...
 
Refigure your short leg using the Pythagorean Theorem. I get 2.7495.....
 
K‡rbyroth said:
pentagon.jpg


I got from this the hypo(r)= 3.4 longleg(a) = 2 the short leg should = 1.7<<< that is only true for 60-30-90 triangles - this one is not.

A= 1/2aP a = apothem P = perimeter
A= 1/2(2*17)
A = 17

I thinking the apothem is wrong here. I know the triangle is wrong unless i missing something
Click to expand...
 
ok thats what trowing me off i thought it was a 60 30 90 triangle i was looking at so i didn't try and find the top angle. So i figured that out now its a 72 90 18 triangle but of course the pic wont represent that.
 
K‡rbyroth said:
ok thats what trowing me off i thought it was a 60 30 90 triangle i was looking at so i didn't try and find the top angle. So i figured that out now its a 72 90 18 triangle but of course the pic wont represent that.
Click to expand...

For this problem - angles really don't matter. Like Denis suggested - use Pythagorus's theorem.
 
i know but because i thought it was 60 30 90 i assumed the side was equal to the hypo and if i knew one of the other angles i wouldn't have done that.
 
So now do you know how to finish the problem...
 
Side length of a regular pentagon, s, in terms of radius and apothem:

s = 2(r^2 –a^2)^.5

Area of a regular pentagon:

A = (5/2)sa

Note: The values a = 2 and r = 3.4 given in the problem are not compatible with a regular pentagon.
 
You must log in or register to reply here.
Top