Respuesta :
Answer:
Step-by-step explanation:
Perimeter = 2L + 2W
2L + 2W = 25
Area = L * W
L * W = 25
W = 25 / L
2L + 2(25 / L) = 25
2L^2 + 50 = 25L
2L^2 - 25L + 50 = 0
L = (-b ± √(b^2 - 4ac)) / (2a)
For the equation 2L^2 - 25L + 50 = 0, a = 2, b = -25, and c = 50.
L = (-(-25) ± √((-25)^2 - 4 * 2 * 50)) / (2 * 2)
L = (25 ± √(625 - 400)) / 4
L = (25 ± √225) / 4
L = (25 ± 15) / 4
L1 = (25 + 15) / 4 = 40 / 4 = 10
L2 = (25 - 15) / 4 = 10 / 4 = 2.5
Since the length of a side cannot be negative, we discard the solution L2 = 2.5
Perimeter = 2L + 2W
2L + 2W = 25
Area = L * W
L * W = 25
W = 25 / L
2L + 2(25 / L) = 25
2L^2 + 50 = 25L
2L^2 - 25L + 50 = 0
L = (-b ± √(b^2 - 4ac)) / (2a)
For the equation 2L^2 - 25L + 50 = 0, a = 2, b = -25, and c = 50.
L = (-(-25) ± √((-25)^2 - 4 * 2 * 50)) / (2 * 2)
L = (25 ± √(625 - 400)) / 4
L = (25 ± √225) / 4
L = (25 ± 15) / 4
L1 = (25 + 15) / 4 = 40 / 4 = 10
L2 = (25 - 15) / 4 = 10 / 4 = 2.5
Since the length of a side cannot be negative, we discard the solution L2 = 2.5.
Therefore, the length of the longest side, L, is 10 meters.
Answer:
10m
Step-by-step explanation:
Let's denote the length of the rectangle as [tex] L [/tex] and the width as [tex] W [/tex]. The perimeter ([tex] P [/tex]) of a rectangle is given by the formula:
[tex] P = 2L + 2W [/tex]
In this case, the perimeter is given as 25 meters:
[tex] 25 = 2L + 2W [/tex]
The area ([tex] A [/tex]) of a rectangle is given by the formula:
[tex] A = L \times W [/tex]
In this case, the area is given as 25 square meters:
[tex] 25 = L \times W [/tex]
Now, we have a system of two equations:
[tex] \begin{cases} 2L + 2W = 25 \\ L \times W = 25 \end{cases} [/tex]
Let's solve this system. One way to proceed is to isolate one of the variables in one of the equations and substitute it into the other equation. I'll solve it for [tex] L [/tex]:
From the first equation, [tex] 2L + 2W = 25 [/tex], isolate [tex] L [/tex]:
[tex] 2L = 25 - 2W [/tex]
[tex] L = \dfrac{25 - 2W}{2} [/tex]
Now, substitute this expression for [tex] L [/tex] into the second equation:
[tex] \dfrac{25 - 2W}{2} \times W = 25 [/tex]
Multiply both sides by 2 to get rid of the fraction:
[tex] (25 - 2W) \times W = 50 [/tex]
Expand and rearrange:
[tex] 25W - 2W^2 = 50 [/tex]
This is a quadratic equation. Rearrange it to standard form:
[tex] 2W^2 - 25W + 50 = 0 [/tex]
Now, we can solve for [tex] W [/tex] using the quadratic formula:
[tex] W = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a} [/tex]
In this case, comparing with ax²+bx +c = 0, we get [tex] a = 2, b = -25, [/tex] and [tex] c = 50 [/tex].
[tex] W = \dfrac{25 \pm \sqrt{(-25)^2 - 4(2)(50)}}{2(2)} [/tex]
[tex] W = \dfrac{25 \pm \sqrt{625 - 400}}{4} [/tex]
[tex] W = \dfrac{25 \pm \sqrt{225}}{4} [/tex]
[tex] W = \dfrac{25 \pm 15}{4} [/tex]
This gives two possible values for [tex] W [/tex]. Discard the negative value (as length cannot be negative in this context):
[tex] W = \dfrac{25 + 15}{4} = \dfrac{40}{4} = 10 [/tex]
Now that we have the width ([tex] W [/tex]), substitute it back into the equation for [tex] L [/tex]:
[tex] L = \dfrac{25 - 2(10)}{2} = \dfrac{5}{2} [/tex]
So, the dimensions of the rectangle are [tex] L = \dfrac{5}{2} = 2.5 [/tex] meters and [tex] W = 10 [/tex] meters.
The longest side is the 10m.
