A plant is designed to be in the shape of a regular pentagon with 92.5m on each side. A security fence surrounds the building to form a circle and each corner of the building is to be 25m from the closest point on the fence. How much fencing is required?
To solve this, we would break the pentagon up into triangle, like one I diagrammed in the attachment. We need to find the red line I diagrammed. To find this, we need to consider the pentagon would be broken up into 5 triangles. That means each central angle is 72 degrees. So, given one half of the triangle, or a right triangle, the top angle is 36 degrees. So, given that, we can find the red line with trig:
sin 36 = 46.25/x
x = 46.25/sin 36
x = 78.69 m
Given that the fence is 25 meters from this, we would add 25 meters to this value for the radius of the fence:
r = 78.69 + 25 = 103.69 m
So, to find the amount of fencing, we would find the circumference:
C = 2*3.14*103.69 = 651.14 m
Let O be the centre of centre of the pentagon. A and B are two vertices of the pentagone. Thus `/_AOB=72^o` and `OA=46.25xxcosec(46.25^o) approx 64.026` m.
Thus radius of the circle =( 64.026+25 )m
The circumference of the circle= `2pixx89.026=559.366m`
Thus required fencing = 559.366m.
Let O be the centre of centre of the pentagon. A and B are two vertices of the pentagone. Thus `/_AOB=72^o` and
`OA=46.25xxcosec(36^0) approx78.685` m.
Thus radius of the circle =( 78.685 +25 )m
The circumference of the circle=`2pixx103.685=651.473m`
Thus required fencing = 651.473m.