Math Problem

[isaiahk9]

Hey, I looked at this problem for about an hour, puzzleing over how its done. The answers came immediately to me (9 and 12), but I have no way of putting that down. I was wondering if anybody knew how to solve this. I found it in the "Analyzing and Applying Quadratic Models" chapter in a grade 10 math textbook. Here's the question :

Nancy walks 15m 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.

The questions before and after it are nothing like it. Was just wondering if anybody knew anyway to solve this.
thanx

[A.J]

First of all, you must realize that the 15m is the diagonal of the rectangular field (Given). This means that if 'x' and 'y' are the length and the width of the rectangular field, then: x^2 + y^2 = 15^2 (Using Pythagoras) Let's call this equation #1.

We are also given that she walked a total distance of 36. So: x + y + 15 = 36. Lets call this equation #2.

from equation #2, a little rearranging gives us : x = 21 - y. Lets call this equation #3.

Subbing equation #3 into equation #1 we get:

(21 - y)^2 + y^2 = 225

(441 - 42y + y^2) + y^2 = 225

2y^2 - 42y + 441 = 225

2y^2 - 42y + 216 = 0

y^2 - 21y + 108 = 0

Factoring the above equation, we get : (y - 9)*(y -12) = 0

Therefore, y = 9 or y = 12.

Subbing y = 9 back into equation #2, we get x = 12.
Subbing y = 12 back into equation #2, we get x = 9.

Therefore, x = 9 and y = 12 or x = 12 and y = 9

[isaiahk9]

thanx so much. I forgot about substituting.

[A.J]

no problem

[dc116]

Apply the Pythagorean theorem
a^2+b^2=c^2
In this case, a and b are the sides of the rectangle, and c is the diagonal.
It is stated that c = 15.
"The total distance she walks is 36m" implies that a + b + c= 36, meaning that a + b = 36- c = 36-15 = 21.
then you might as well as let b = 21 - a
and sub those values into the Pythagorean theorem
a^2 + (21-a)^2 = 225
a^2 + 441 - 42a + a^2 = 225
2a^2 - 42 a + 216 = 0
Then simply solve the quadratic equation.