Jim has designed a rectangle with an area of 100 square feet and a perimeter of 401 feet.
What are the dimensions of Jim's rectangle?please show me how u got the answer.
A = LW
Let's see which factors of 100 would add up to 401.
10 * 10
2 * 50
4 * 25
5 * 20
P = 2L + 2W
None of those come close to a perimeter of 401.
Are you sure you copied the problem correctly?
Yes and there is only one way
lw = 100
and 2l + 2w = 401
from the 1st .... l = 100/w
into the 2nd:
2(100/w) + 2w = 401
times w
200 + 2w^2 = 401w
2w^2 - 401w + 200 = 0
by the formula:
w = (401 ± √159201)/4
= (401 ± 399)/4
= 200 or .5
or , it factors to
(w-200)(2w - 1) = 0
w = 200 or w = 1/2
if w=200, the l = 100/200 = 1/2
if w = 1/2, the l = 100/(1/2) = 200
The rectangle is 200 feet by 1/2 ft
check:
area = (1/2)(200) = 100 , checks!
perimeter = 200 + 200 + 1/2 + 1/2 = 401, checks!
To find the dimensions of Jim's rectangle, we can set up a system of equations using the given information. Let's assume the length of the rectangle is 'l' and the width is 'w'.
We know that the area of a rectangle is given by the formula: Area = Length * Width.
In this case, the area is 100 square feet, so we can write the equation: l * w = 100.
We also know that the perimeter of a rectangle is the sum of all four sides. In this case, the perimeter is 401 feet. Since a rectangle has two equal sides (length and width) we can write the equation: 2*l + 2*w = 401.
Now we have a system of equations:
l * w = 100,
2 * l + 2 * w = 401.
To solve this system, we can use the substitution or elimination method. Let's solve it using the substitution method:
From the first equation, we can express 'l' in terms of 'w' by dividing both sides by 'w':
l = 100/w.
Now substitute this value of 'l' into the second equation:
2 * (100/w) + 2 * w = 401.
Multiply through by 'w' to get rid of the fraction:
200/w + 2w = 401.
Multiply every term by 'w' to eliminate the fraction:
200 + 2w^2 = 401w.
Rearrange the equation:
2w^2 - 401w + 200 = 0.
Now we have a quadratic equation. We can solve it by factoring or using the quadratic formula. Let's solve it using the quadratic formula:
w = (-b ± sqrt(b^2 - 4ac)) / 2a,
where a = 2, b = -401, and c = 200.
Substituting the values, we get:
w = (-(-401) ± sqrt((-401)^2 - 4 * 2 * 200)) / (2 * 2).
w = (401 ± sqrt(160801 - 1600)) / 4.
w = (401 ± sqrt(159201)) / 4.
Calculating the square root of 159201, we get approximately 399.001. Therefore:
w = (401 ± 399.001) / 4.
Case 1: w = (401 + 399.001) / 4 = 800.001 / 4 = 200.00025.
Case 2: w = (401 - 399.001) / 4 = 1.999 / 4 = 0.49975.
Since the width cannot be negative, we discard case 2. Thus, the width of Jim's rectangle is approximately 0.49975 feet.
Now we can find the length by substituting the width into the first equation:
l * 0.49975 = 100.
l = 100 / 0.49975.
l ≈ 200.10004 feet.
Therefore, the dimensions of Jim's rectangle are approximately 200.10004 feet by 0.49975 feet.