A leach free area for dogs is to be created in a field behind a recreation centre. what length of fencing will be required, if the plans show an irregular pentagonal area with vertices at (2,0), (1,6), (8,9), (10,7)and (6,0)?( one unit represents 10m)
Find the distance of all sides using the distance formula :
sqr root[(x2-x1)^2 - (y2-y1)^2]
Then once you have the distances, multiply each one of the distances by 10, then add that together.
To find the length of fencing required for the irregular pentagonal area, we need to calculate the perimeter of the shape.
First, let's plot the given vertices on a coordinate plane:
(2,0), (1,6), (8,9), (10,7), and (6,0)
Now, we can calculate the distance between each consecutive pair of vertices using the distance formula:
The distance between (2,0) and (1,6) is calculated as:
d1 = √((x2 - x1)² + (y2 - y1)²)
= √((1 - 2)² + (6 - 0)²)
= √((-1)² + 6²)
= √(1 + 36)
= √37
The distance between (1,6) and (8,9) is calculated as:
d2 = √((x2 - x1)² + (y2 - y1)²)
= √((8 - 1)² + (9 - 6)²)
= √(7² + 3²)
= √(49 + 9)
= √58
The distance between (8,9) and (10,7) is calculated as:
d3 = √((x2 - x1)² + (y2 - y1)²)
= √((10 - 8)² + (7 - 9)²)
= √(2² + (-2)²)
= √(4 + 4)
= √8
The distance between (10,7) and (6,0) is calculated as:
d4 = √((x2 - x1)² + (y2 - y1)²)
= √((6 - 10)² + (0 - 7)²)
= √((-4)² + (-7)²)
= √(16 + 49)
= √65
The distance between (6,0) and (2,0) is calculated as:
d5 = √((x2 - x1)² + (y2 - y1)²)
= √((2 - 6)² + (0 - 0)²)
= √((-4)² + 0²)
= √(16 + 0)
= √16
= 4
Next, we need to sum up all the distances to calculate the perimeter:
Perimeter = d1 + d2 + d3 + d4 + d5
= √37 + √58 + √8 + √65 + 4
Finally, since each unit represents 10 meters, we multiply the perimeter by 10 to get the total length of fencing required in meters:
Fencing Length = Perimeter * 10
Using this method, we can find the length of fencing required for the leach-free area for dogs in the field behind the recreation center.