Mike would like to plant some trees so he needs to use a portion of his existing yard. The perimeter of this rectangular portion needs to be 14 yards and the diagonal is 5 yards. Can you help Mike determine the length and width of this new portion? Mike requested that you show him how you arrive at the answers, so please show your complete work.

An old Greek did this already.

5^2=w^2+L^2 I hope you recognize that formula...

but 2w+2L=14 or w=7-L
25=(7-L)^2 + L^2
solve for L and you have it.

To find the length and width of the rectangular portion of the yard, we can use algebraic equations based on the given information.

Let's assume the length of the rectangular portion is "L" and the width is "W".

We are given two pieces of information:
1) The perimeter of the rectangular portion is 14 yards.
2) The diagonal of the rectangular portion is 5 yards.

1) Perimeter equation:
The perimeter of a rectangle is given by the formula: 2L + 2W = Perimeter.
In this case, the perimeter is 14 yards, so we substitute the values into the equation:
2L + 2W = 14.

2) Diagonal equation:
Using the Pythagorean theorem, we know that the diagonal (5 yards) is the hypotenuse of a right triangle formed by the length and width of the rectangle. The Pythagorean theorem states that the square of the length of the hypotenuse is equal to the sum of the squares of the other two sides. Therefore, we have:
L^2 + W^2 = Diagonal^2
L^2 + W^2 = 5^2
L^2 + W^2 = 25.

Now, we have a system of two equations with two variables. We can solve this system to find the values of L and W.

Using the first equation (2L + 2W = 14), we can simplify it by dividing both sides by 2:
L + W = 7.

Now, we have two equations:
L + W = 7
L^2 + W^2 = 25.

To solve this system, we can use substitution or elimination. Let's use substitution.

Rearranging the first equation (L + W = 7) to solve for L, we get:
L = 7 - W.

Now, substitute this expression for L in the second equation (L^2 + W^2 = 25):
(7 - W)^2 + W^2 = 25.

Expanding and simplifying this equation, we get:
49 - 14W + W^2 + W^2 = 25
2W^2 - 14W + 49 - 25 = 0
2W^2 - 14W + 24 = 0.

Now, we have a quadratic equation. We can solve this equation by factoring, completing the square, or using the quadratic formula. In this case, it is easier to factor out a 2 from the equation:

2(W^2 - 7W + 12) = 0.

Now, we have the equation in factored form:
2(W - 3)(W - 4) = 0.

Setting each factor equal to zero and solving for W, we get:
W - 3 = 0 or W - 4 = 0

Solving these equations, we find:
W = 3 or W = 4.

If W = 3, we can substitute this value back into the first equation (L + W = 7) to find L:
L + 3 = 7
L = 7 - 3
L = 4.

So, if W = 3, then L = 4.

If W = 4, we can substitute this value back into the first equation (L + W = 7) to find L:
L + 4 = 7
L = 7 - 4
L = 3.

So, if W = 4, then L = 3.

In conclusion, there are two possible solutions for the length and width of the rectangular portion of Mike's yard:
1) If the width (W) is 3 yards, then the length (L) is 4 yards.
2) If the width (W) is 4 yards, then the length (L) is 3 yards.