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?
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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.
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