How would you work this out? I know how to do it by trial and improvement but what is the method involving quadratics?|||Hi,
Assuming the field is rectangular, let x = its width. Then the length is 110 - x for the length.
The area is found by:
A = L x W
2500 = (110 - x)x
2500 = 110x - x虏
x虏 - 110x + 2500 = 0
................________________
-(-110) 卤 鈭?-110)虏 - 4(1)(2500)
--------------------------------------... = x
..................2(1)
............_____
110 卤 鈭?100
--------------------- = x
........2
x = 32.09 metres
110 - x = 77.91 metres
The field is 77.91 metres x 32,09 metres. %26lt;==ANSWER
I hope that helps!! :-)|||You are presuming a rectangle for the field?
I have a field which has 6 sides. One %26#039;corner%26#039; has an interior angle of about 270 degrees. None of the corners are of 90 degrees. The fences enclose three acres (say, 10,000 sq meters). I challenge anyone to calculate the length of fence that I need to maintain!|||This kind of problem is a multi-step process.
Step 1
Call the perimeter of the area P.
Assuming that the area is a rectangle, the formula for the perimeter is
P = 2L + 2W.
Now, they give us P = 220. Put that into the equation.
220 = 2L + 2W
Solve for W.
Subtract 2L from both sides.
220 - 2L = 2L + 2W - 2L
220 - 2L = 2W
Divide by 2.
(220 - 2L)/2 = 2W/2
(220 - 2L)/2 = W
****************************
Step 2
The formula for the area of a rectangle is
A = LW.
We know what W looks like from Step 1. Put that expression into the equation for the area.
A = L (220 - 2L)/2
Distribute the L across the parentheses.
A = (220L - 2L^2)/2
They told us that the area is 2500. Put that value into the equation.(I%26#039;m ignoring all of the meter stuff right now. We%26#039;ll put that in at the end.)
2500 = (220L - 2L^2)/2
Multiply both sides by 2.
2(2500) = 2 [ (220L - 2L^2)/2 ]
5000 = 220L - 2L^2
Add 2L^2 to both sides.
5000 + 2L^2 = 220L - 2L^2 + 2L^2
5000 + 2L^2 = 220L
Subtract 220L from both sides.
5000 + 2L^2 - 220L = 220L - 220L
5000 + 2L^2 - 220L = 0
Rewrite for ease of solving a quadratic.
2L^2 - 220L + 5000 = 0
This next series of steps are optional, but will make the problem simpler to solve.
Factor 2 out of the terms on the left hand side.
2 (L^2 - 110L + 2500) = 0
Divide both sides by 2.
[2 (L^2 - 110L + 2500)] / 2 = 0 / 2
L^2 - 110L + 2500 = 0
*************************************
Step 3
Use the Quadratic Formula to solve for L.
L = [ -B +- sq_rt (B^2 - 4AC) ] / 2A
Here, A = 1, B = -110, C = 2500 and +- means %26quot;plus or minus.%26quot; So...
L = [ -(-110) +- sq_rt ( (-110)^2 - 4(1)(2500) ] / 2(1)
L = [ 110 +- sq_rt ( 12100 - 10000 ] / 2
L = [ 110 +- sq_rt ( 12100 - 10000 ] / 2
L = [ 110 +- sq_rt ( 2100 ) ] / 2
Now, sq_rt ( 2100 ) = sq_rt [ (21)(100 ) ]
= 10 sq_rt (21).
Now we have...
L = [ 110 +- 10 sq_rt (21) ] / 2
Factor out a 2 from the terms in the numerator. It will cancel out the 2 in the denominator.
L = 2 [ 55 +- 5 sq_rt (21) ] / 2
L = 55 +- 5 sq_rt (21)
We have 2 possibilities.
L = 55 + 5 sq_rt (21)
or
L = 55 - 5 sq_rt (21)
Note: 5 sq_rt (21) = 5(4.58) = 22.9, approximately.
We now have...
L = 55 + 22.9 = 77.9
or
L = 55 - 22.9 = 32.1.
***********************************
Step 4
If L = 55 + 22.9 = 77.9,
then
A = LW means that 2500 = 77.9 W.
Divide by 77.9 on both sides.
2500/77.9 = 77.9 W / 77.9
32.09 = W.
If L = 55 - 22.9 = 32.1,
then
A = LW means that 2500 = 32.1 W.
Divide both sides by 32.1.
2500/32.1 = 32.1 W / 32.1
77.88 = W.
Note that the pairs of values are very close. (77.88 is almost 77.9 and
32.09 is almost 32.1.)
L = 77.9 W = 32.09
L = 32.1 W = 77.88.
Since length is understood to be larger than width,
I%26#039;m not sure whether your teacher wants to see decimal approximations or not, but I%26#039;d take the following as the answer:
L = 77.9 W = 32.09
.
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