**Math Help: Solving Width? - ice skating uniforms**

A skating rink is in the form of a rectangle 30 meters long and 20 meters wide. 1 page is against the side of a building. attempts to extend the runway Getting thin and uniform ice on the outside of the track, but u can not off the wall strip. Find the width of the strip, even if the surface of the original length is 2 times the size of the ice strip

## 3 comments:

The surface of the original route is 600m ², so that the area of the strip of ice is 300m ². Let x = width of the strip.

There are two ways to solve this problem: Calculate the surface of the tape only, and comparing the area of the original titles in the expanded definition.

If you calculate the surface of the strip, and is divided into three rectangles by extending the line does not contain the longest side of the track against the original building divided. This leaves you with a rectangle on each side of the width and length x 20, and a rectangle the length of long side with a width of x and a length of 2 x 30 Draw a diagram to visualize if it helps, too.

However, the sum of the total surface area of these three rectangles 300m ² are to:

20x 20 x + x (30 x 2) = 300

2x ² 70 x 300 = 0

² x 35 x 150 = 0

Use the quadratic formula:

x = (-35 ± √ 1825) / 2 = (-35 ± 5 √ 73) / 2

Now is √ 73 on 8,54, then 5 √ 73 is pretty close to 43. Of course, X can be negative, we reject the negative rootING:

x = (43-35) / 2 = approx. 4m

The band of ice, Articles 3, will be held 2 into account, and ends with a long side.

If we believe that in parts 2 with a width "W" and a length of 20 to have added:

2 (20W)

The part on the longer side has a surface area of:

w * (30 + 2W)

Thus, the entire area to be added:

2 (20W) + w * (30 + 2W)

40W + 30W + 2W ^ 2

2V ^ 2 + 70W

However, this amount should be at half the original length, format, ie:

2V ^ 2 + 70 = ½ (30 * 20)

2V ^ 2 + 70 = 300

w ^ 2 + 35W - 150 = 0 with the quadratic formula I get:

(w - 3.85) (38.85 W) = 0

w = 3.85, -38.85

Since there can be negative, the width is 3.85 meters

3.860009363 meters. I used the 600 = 140x + 4x ^ 2 and simplifying there started. I hope it is good ..

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