Math

QuestionA farmer has 250 ft of fencing for 3125 sq ft in two adjoining squares. Find the side lengths xx and yy.

Studdy Solution

STEP 1

Assumptions1. The total length of fencing available is250 feet. . The total area to be enclosed is3125 square feet.
3. The region to be enclosed is in the shape of two adjoining squares.
4. The larger square has sides of length xx.
5. The smaller square has sides of length yy.
6. The lengths xx and yy can be any real numbers.

STEP 2

First, we need to set up an equation for the perimeter of the two squares. The total length of the fencing is equal to the sum of the perimeters of the two squares.
Perimeter=4x+4yPerimeter =4x +4y

STEP 3

Now, plug in the given value for the total length of the fencing to get an equation in terms of xx and yy.
250=x+y250 =x +y

STEP 4

implify the equation by dividing through by4.
62.=x+y62. = x + y

STEP 5

Next, we need to set up an equation for the area of the two squares. The total area to be enclosed is equal to the sum of the areas of the two squares.
Area=x2+y2Area = x^2 + y^2

STEP 6

Now, plug in the given value for the total area to get another equation in terms of xx and yy.
3125=x2+y23125 = x^2 + y^2

STEP 7

We now have a system of two equations in two variables. We can solve this system by substitution or elimination. Here, we will use substitution. From the first equation, we can express yy in terms of xx.
y=62.5xy =62.5 - x

STEP 8

Substitute this expression for yy into the second equation.
3125=x2+(62.5x)23125 = x^2 + (62.5 - x)^2

STEP 9

implify the equation by expanding the square and combining like terms.
3125=x2+3906.25125x+x23125 = x^2 +3906.25 -125x + x^2

STEP 10

Combine the x2x^2 terms and move the constant term to the other side of the equation.
2x2125x+781.25=02x^2 -125x +781.25 =0

STEP 11

This is a quadratic equation in the form ax+bx+c=0ax^ + bx + c =0. We can solve it by factoring, completing the square, or using the quadratic formula. Here, we will use the quadratic formula.
x=b±b4acax = \frac{-b \pm \sqrt{b^ -4ac}}{a}

STEP 12

Plug in the values for aa, bb, and cc from our quadratic equation.
x=125±(125)242781.2522x = \frac{125 \pm \sqrt{(-125)^2 -4*2*781.25}}{2*2}

STEP 13

implify under the square root.
x=125±156253125x = \frac{125 \pm \sqrt{15625 -3125}}{}

STEP 14

Further simplify under the square root.
x=125±125004x = \frac{125 \pm \sqrt{12500}}{4}

STEP 15

Calculate the square root.
x=125±111.84x = \frac{125 \pm111.8}{4}

STEP 16

We have two possible solutions for xx. Calculate each one.
x=125+111.84=59.45x = \frac{125 +111.8}{4} =59.45x2=125111.84=3.3x2 = \frac{125 -111.8}{4} =3.3

STEP 17

Substitute each solution for xx back into the equation y=62.5xy =62.5 - x to find the corresponding values for yy.
y=62.559.45=3.05y =62.5 -59.45 =3.05y2=62.53.3=59.2y2 =62.5 -3.3 =59.2The values of xx and yy are (59.45,3.05) and (3.3,59.2). Since xx is the length of the larger square, we take the larger value for xx and the corresponding value for yy. Therefore, x=59.45x =59.45 and y=3.05y =3.05.

Was this helpful?

Studdy solves anything!

banner

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

ParentsInfluencer programContactPolicyTerms
TwitterInstagramFacebookTikTokDiscord