A person decides to build a horse corral using a barn for one side. Has has 30m of fencing materials and wants the corral to hav
1 answer:
Two possible answers:
- A 10 m by 10 m corral
- A 5 m by 20 m corral
See the diagrams below.
============================================================
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
----------------
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.
----------------
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.
You might be interested in