the grade 11 math class is building a tennis court measuring 45m by 15m they want to put a grass border completely around the tennis court,

Question

find the width of the grass so that the border is 200m squared.

Answer 1

Hello Maggie! Raymond B did a great job solving the problem, but you may still be a little confused as to how he solved the problem. I will explain a little more about what each part of the equation means. He also made a small error near the end, but Raymond did the important parts correct. I hope you enjoy my explanation!

Imagine a 45 by 15 rectangle (below):

(tennis court only)

_______

|______| (15m)

(45m)

Now add a grass border:

____________

| ________ |

| |_______| | (15+2x)m

|___________|

(45+2x)m

Let x be the distance from the original rectangle to the new edge, also the "Width added" to each side of the rectangle. Notice that because there are 2 ends of a rectangle in both horizontal and vertical directions, the rectangle lengths each increased by 2 times of x.

x=width added to the rectangle

Returning back to the problem, we know the grass border is the space occupied around the original rectangle (the tennis court). Imagine we took the big rectangle (the tennis court with grass border) and subtract away the smaller rectangle (only tennis court), then we would result in only the grass border (the thing we are solving for).

As can be seen, 45+2x is the length and 15+2x is the width. Therefore, the big rectangle area would be (45+2x)*(15+2x). The smaller rectangle would just be 45*15. To solve for just the grass, we take the big rectangle and subtract the smaller rectangle from it. In mathematical terms, it would be (45+2x)(15+2x)-(45*15).

We are also given that the grass area is equal to 200m^2.

Therefore, (45+2x)(15+2x)-(45*15) = 200

Now by expanding using the first, outer, inner, last method,

First = 45*15 = 675

Outer = 45*2x = 90x

Inner = 2x*15 = 30x

Last = 2x*2x = 4x^2

we get 675+90x+30x+4x^2 - (45*15) = 200 (expanded, and now we simplify)

675+ 120x + 4x^2 -675 = 200 (added the 90x and 30x and multiplied 45*15)

120x + 4x^2 = 200 (675-675=0)

30x + x^2 = 50 (divide both sides by 4)

x^2 + 30x - 50 = 0 (minus 50 from both sides)

Now we are in the quadratic form of ax^2 + bx + c. Also, we have the right side of the equation equal to 0. Therefore, we can use the quadradic formula. By comparing the quadratic form to the equation x^2 + 30x -50 = 0, we find that

a = 1

b = 30

c = -50

The quadratic formula is (-b +- sqrt(b^2-4ac) )/2a

Plug in the values for a, b, and c, and there occurs 2 solutions, x = ~1.58312 and x= ~ −31.5831.

However, x = width added to the rectangle, which makes x a length. Length cannot be negative, and thus, x = ~ −31.5831 would not be a possible solution.

After removing the x = ~ −31.5831 solution, there is only one remaining solution for x:

x (width added to the tennis court) = ~1.58312 meters (unit given in the problem)

Answer 2

area of the grass border = the total area of the border plus the court minus the court

(45 +2x)(15+2x)-45(15) = 200

120x + 4x^2 = 200

x^2 + 30x - 50 = 0

x^2 + 30x + 225 = 50+225 = 275

complete the square

(x+15)^2 = 25(11)

x+15 = 5sqr11

x = 5sqr11-15 = width of the grass border

= about 1.583 m