Two possible answers:
- A 10 m by 10 m corral
- A 5 m by 20 m corral
See the diagrams below.
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Explanation:
x = length
y = width
Let side x be parallel to the barn wall. This means we have 1 copy of x and 2 copies of y to form this partial rectangle. The fourth side is the barn itself, so we don't need this as the fencing.
The perimeter of the fencing is x+y+y = x+2y
Set this equal to the 30 meters of fencing and solve for x.
x+2y = 30
x = -2y+30
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area = length*width
area = x*y
area = (-2y+30)*y
area = -2y^2+30y
Set this equal to the given area of 100 m^2 and solve for y
-2y^2+30y = 100
-2y^2+30y-100 = 0
-2(y^2-15y+50) = 0
y^2-15y+50 = 0
(y-10)(y-5) = 0
y-10 = 0 or y-5 = 0
y = 10 or y = 5
Side note: The quadratic formula is an alternative path you can take.
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If y = 10, then x would be...
x = -2y+30
x = -2*10+30
x = -20+30
x = 10
This produces a 10 meter by 10 meter enclosure.
Or, if y = 5, then,
x = -2y+30
x = -2*5+30
x = -10+30
x = 20
The corral is 5 by 20 in this case.
Unfortunately there are two possible answers. Without more info, we cannot nail down which one to go with.
The diagram of each is shown below.