We want to find the perimeter of a rectangle given that we know the area and length of the diagonal of the rectangle.
We will see that the perimeter is equal to 42 meters.
We know that for a rectangle of length L and width W, the area is given by:
A = L*W
And the length of the diagonal is given by:
d = √(L^2 + W^2)
From the given information we have that:
A = 108m^2 = L*W
d = 15m = √(L^2 + W^2)
From the first equation we can write:
L = (108m^2)/W
Now we can replace this in the other equation to get:
15m = √(((108m^2)/W)^2 + W^2)
(15m)^2 = ((108m^2)/W)^2 + W^2
Multiplying both sides by W^2 we get:
(15m)^2*W^2 = (108m^2)^2 + W^4
Replacing W^2 = x we get:
(15m)^2*x = (108m^2)^2 + x^2
Now we can rewrite this as:
x^2 + 11,664 m^4 - (225m^2)*x = 0
And solve it using Bhaskara's formula:
[tex]x = \frac{+225m^2 \pm \sqrt{(225m^2)^2 - 4*1*(11,664 m^4)} }{2} \\\\x = \frac{225m^2 \pm 63m^2 }{2}[/tex]
Because X is the width squared, we take the positive solution:
X = (225m^2 + 63m^2)/2 = 144m^2
W = √(144m^2) = 12m
Then we can replace this in the area equation:
A = 108m^2 = L*12m
108m^2/12m = L = 9m
Now that we know the length and width, we can write the perimeter as:
P = 2*(L + W) = 2*(12m + 9m) = 42m
If you want to learn more, you can read:
https://brainly.com/question/18793958
