The Student Room Group
Could split it up into 5 isosceles triangles...
Reply 2
yea im stuck...

by splitting it up into 5 isoceles triangles, they have the angles- 72, 54 and 54 with the side between the two equal angles as 80m

where do i go from here?
generalebriety
Could split it up into 5 isosceles triangles...

Yep and a bit of trig (or maybe pythag) could give you the height of one of the triangles.

The it's easy to work out the area of the pentagon from there...


(I think you'd need to know the size of the interior angle of a regular pentagon though...do you know that? Or at least how to find it?)
Reply 4
how do i find the length between the centre and one of the 'corners'?
drunknmunky
how do i find the length between the centre and one of the 'corners'?

You don't need to find that length as it wouldn't help you work out the are directly.

What you do need to find it the height of each triangel (the perendiclr ditance from the centre of one side to the entre of the pentagon (see the blue line in the diagram). That can be done using trigonometry.
Add an extra line inside the pentagon dividing it into a triangle and a trapezium. Since it is a regular pentagon its internal angles are 108 degrees. Using this with the cosine rule you can calculate the length of the extra line. You can then find the area of the trapezium and triangle, add them together to get the area of the pentagon.
Reply 7
Divide the pentagon into 5 isoceles triangles.
For each triangle, the angle at the centre of the pentagon is 2pi/5.
The pentagon has 5 vertices, or corners. The sum of the angles at the vertices of an n-sided polygon is equal to pi(n-2).
You now have a triangle where you know all the angles of it and also you have the length of one of the sides. You can work out the area of such a triangle.