SOLUTION: Hi, Here is my problem Nancy walks 15 m diagonally across a rectangular field. She then returns to her starting position along the outside of the field. The total distance she w

Question 169672: Hi, Here is my problem
Nancy walks 15 m diagonally across a rectangular field. She then returns to her starting position along the outside of the field. The total distance she walks is 36m. What are the dimensions of the field.
I have a good understanding of how to solve quadratic equations, but the word problems really confuse me. I made a diagram and used the pythagorean theorem to identify an equation, but it didn't factor in the end...Please help, if you could walk me through the steps that would be great
thanks
Found 2 solutions by Mathtut, jim_thompson5910:
Answer by Mathtut(3670) (Show Source): You can put this solution on YOUR website!
ok ........so when she walks diagonally across the field that ends up being the hypothenuse of the right triange which is 15 meters. what we dont know is the other 2 legs...but we have a hint since the perimeter of the triangel is 36m. So now we know that both legs add up to 21----->(36-15)....so we will call one leg a and the other 21-a.
:

:

:
divide by 2
:

:
using the quadratic we find that or
each leg can be either value. The dimensions of the field is 9m by 12m

Solved by pluggable solver: SOLVE quadratic equation with variable

Quadratic equation (in our case ) has the following solutons:

For these solutions to exist, the discriminant should not be a negative number.

First, we need to compute the discriminant : .

Discriminant d=9 is greater than zero. That means that there are two solutions: .

Quadratic expression can be factored:

Again, the answer is: 12, 9. Here's your graph:

Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!
Let x=length of field and y=width of field

Remember, the perimeter of any triangle is simply the sum of its sides. In this case, two of the sides are the length and width (x and y) and the third side is the diagonal of the field

So the perimeter is

Now plug in to get . After solving for y, we get

Now since the diagonal splits the field into two triangles, where the length and width form the legs and the diagonal forms the hypotenuse, this means that we can use Pythagorean's Theorem

Start with Pythagorean's Theorem

Plug in , (the length and width) and (the diagonal)

Square 15 to get 225

Plug in (the previous isolated equation)

FOIL

Subtract 225 from both sides

Combine like terms.

Notice we have a quadratic equation in the form of where , , and

Let's use the quadratic formula to solve for x

Start with the quadratic formula

Plug in , , and

Negate to get .

Square to get .

Multiply to get

Subtract from to get

Multiply and to get .

Take the square root of to get .

or Break up the expression.

or Combine like terms.

or Simplify.

So the answers are or

===========================================

Answer:

So the dimensions of the field are 12 m by 9 m
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