A rectangular skating rink measuring 30m by 20m is doubled in area by adding a strip at one end of the rink and a strip fo the same width along one side of the rink. Find the width of the strips if the enlarged rink is still rectangular.
A = (30 + X) (20 + X) = 2(30 * 20),
X^2 + 30X + 20X + 600 = 1200 m,
X^2 + 50X + 600 = 1200,
X^2 + 50X + 600 - 1200 = 0,
X^2 + 50X - 600 = 0,
Factor the Eq:
(X - 10) (X + 60) = 0,
X - 10 = 0,
X = 10.
X + 60 = 0,
X = -60.
Use the positive value of x:
X = 10 m = width of strips added,
Ah, the case of the expanding skating rink! Let's put on our math skates and glide into the solution.
We know that the original rink measures 30m by 20m, giving us an area of 600 square meters. Now, in order to double the area, we need to add two strips to the rink.
Let's call the width of the strip along one side "x" meters. Since we're adding this strip along the length of the rink, the length of the new enlarged rink will remain the same at 30m.
The strip added at the end will have a width of "x" meters. This will increase the length of the rink, making it 20 + x meters.
To double the area, we multiply the new length by the new width and make it equal to double the original area (600 x 2 = 1200 square meters):
(30) * (20 + x) = 1200
Now, let's solve for x. Ready? 1, 2, 3, go!
600 + 30x = 1200
30x = 600
x = 20
So, the width of the strip along one side is 20 meters, and the width of the strip at the end is also 20 meters.
Voila! We have solved the mystery of the expanding rink! Now you can skate into twice the fun. Enjoy!
P.S. Just make sure the clowns don't use those extra strips for their silly shenanigans!
To find the width of the strips, we need to determine the new dimensions of the enlarged rectangular rink.
Let's assume the width of each strip is "x" meters.
When the strip is added at one end, the length of the rink increases by "x" meters and becomes 30 + x meters.
When the strip is added along one side, the width of the rink increases by "x" meters and becomes 20 + x meters.
The area of the enlarged rink is doubled, so the new area is twice the original area, which can be expressed as:
(30 + x) * (20 + x) = 2 * (30 * 20)
Simplifying the equation, we have:
(30 + x) * (20 + x) = 2 * 600
Expanding the multiplication, we get:
600 + 30x + 20x + x^2 = 1200
Combining like terms, the equation becomes:
x^2 + 50x + 600 - 1200 = 0
x^2 + 50x - 600 = 0
To solve this quadratic equation, we can factor it or use the quadratic formula. Let's use the quadratic formula.
The quadratic formula is given by:
x = (-b ± sqrt(b^2 - 4ac)) / (2a)
Here, a = 1, b = 50, and c = -600.
Substituting these values into the quadratic formula, we have:
x = (-50 ± sqrt(50^2 - 4*1*(-600))) / (2*1)
Simplifying further,
x = (-50 ± sqrt(2500 + 2400)) / 2
x = (-50 ± sqrt(4900)) / 2
x = (-50 ± 70) / 2
This gives two possible values for x:
x1 = (-50 + 70) / 2 = 20 / 2 = 10
x2 = (-50 - 70) / 2 = -120 / 2 = -60
Since the width cannot be negative, the width of the strips is 10 meters.
To find the width of the strips, we can start by calculating the original area of the skating rink. The original area is given by multiplying the length and width of the rectangle:
Original Area = Length × Width = 30m × 20m = 600m²
According to the question, the enlarged rink is double the area of the original rink. Therefore, the new area is:
New Area = 2 × Original Area = 2 × 600m² = 1200m²
Let's denote the width of the added strip along the side of the rink as 'x', and the width of the added strip at the end as 'y'. Since the rink is still rectangular, we can set up an equation based on the new area:
(30 + y) × (20 + x) = 1200
Simplifying this equation, we have:
600 + 30x + 20y + xy = 1200
Rearranging the equation, we get:
xy + 30x + 20y = 600
To proceed, we need to factor the left side of the equation. Since we know the answer will be a positive value, we will look for factors with a positive coefficient for both 'x' and 'y'.
We observe that xy, 30x, and 20y are all divisible by 10, so we can divide both sides of the equation by 10:
(xy + 30x + 20y)/10 = 600/10
This simplifies to:
xy/10 + 3x + 2y = 60
To factor the left side, we look for pairs of factors of 'xy/10' such that their sum is equal to the coefficient of either 'x' or 'y'. In this case, the factors of 'xy/10' are 'x/10' and 'y/10', and their sum is 'x/10 + y/10 = (x + y)/10'.
So, we can factor the equation as:
((x + y)/10)(10 + 30) = 60
Simplifying further, we have:
(x + y)(4) = 60
Dividing both sides by 4, we get:
x + y = 15
We now have a system of two equations:
xy + 30x + 20y = 600 (equation 1)
x + y = 15 (equation 2)
Given these equations, we can solve for the width of the strips. Let's rearrange equation 2 to express 'x' in terms of 'y':
x = 15 - y
Substituting this into equation 1, we get:
(15 - y)y + 30(15 - y) + 20y = 600
Expanding and simplifying the equation:
15y - y² + 450 - 30y + 20y = 600
- y² + 5y - 150 = 0
To solve this quadratic equation, we can factor it:
(-y + 15)(y - 10) = 0
Setting each factor equal to zero and solving for 'y', we obtain two possible solutions:
1) -y + 15 = 0 => y = 15
2) y - 10 = 0 => y = 10
Since the width cannot be negative, we discard the first solution, y = 15.
Therefore, the width of the added strip at the end of the rink is 10m.
To find the width of the strip added to the side of the rink (x), we can substitute the value of y into equation 2 and solve for x:
x + 10 = 15
x = 15 - 10
x = 5
Thus, the width of the added strip along the side of the rink is 5m.
In conclusion, the width of the added strip at the end of the rink is 10m, while the width of the added strip along the side of the rink is 5m.