The playing surface in the game of curling is a rectangular sheet of ice with an area of about 225 m^2. The width is about 40m less than the length. Find the approximate dimensions of the playing surface.
How would i solve this?
Length = L meters.
Width = (L-40) Meters.
L * (L-40) = 225m^2,
L^2 - 40L = 225,
L^2 - 40L - 225 = 0,
(L-5)(L+45) = 0,
Length = L meters.
Width = (L-40) Meters.
L * (L-40) = 225m^2,
L^2 - 40L = 225,
L^2 - 40L - 225 = 0,
(L+5)(L-45) = 0,
L+5 = 0,
L = -5.
L-45 = 0,
L=45.
Choose positive value of L:
L = 45m = Length.
Width = L-40 = 45-40 = 5m.
To solve this problem, we can set up an equation based on the information given.
Let's assume that the length of the playing surface is L meters. Since the width is 40 meters less than the length, we can represent the width as (L - 40) meters.
The area of a rectangle is given by the formula: Area = Length × Width.
Using this formula, we can set up the equation:
225 = L × (L - 40)
To solve this equation, we can rearrange it and solve for L:
L^2 - 40L - 225 = 0
This is a quadratic equation, so we can use factoring, completing the square, or the quadratic formula to solve for L. In this case, using factoring is the most straightforward method.
We can factor the equation as follows:
(L - 45)(L + 5) = 0
This gives us two possible solutions: L - 45 = 0 or L + 5 = 0.
If we solve for L in each case, we get:
L = 45 or L = -5
Since length cannot be negative, we discard the second solution.
Therefore, the approximate dimension of the playing surface is a length of 45 meters and a width of (45 - 40) = 5 meters.
So, the approximate dimensions of the playing surface are 45 meters by 5 meters.