Please help:
The length of a rectangle is 5m greater than twice its width, and its area is 33m^2 (the m is squared). Find the dimensions.
width: w
length: 2w+5
area: w(2w+5) = 33
2w^2 + 5w - 33 = 0
(w-3)(2w+11)
w=3 or -11/2
width: 3
length: 2(3)+5 = 11
of course, that solution would have been suggested by the fact that 33=3*11, without even solving for w.
Well, well, well, we have ourselves a math problem! Let's solve it, shall we?
Let's assume that the width of the rectangle is W meters. According to our problem, the length is 5 meters greater than twice the width, which means the length is 2W + 5 meters.
Now, let's use the formula for the area of a rectangle: Area = Length × Width. In our case, we know that the area is 33 square meters, so we can write:
33 = (2W + 5) × W
Now let's simplify this equation:
33 = 2W^2 + 5W
To solve the equation, let's set it equal to zero:
2W^2 + 5W - 33 = 0
Now we have a quadratic equation. We can use the quadratic formula to solve it:
W = (-b ± √(b^2 - 4ac)) / (2a)
In our case, a = 2, b = 5, and c = -33. Plugging these values in, we get:
W = (-5 ± √(5^2 - 4 × 2 × -33)) / (2 × 2)
W = (-5 ± √(25 + 264)) / 4
W = (-5 ± √(289)) / 4
W = (-5 ± 17) / 4
W = (-5 + 17) / 4 or W = (-5 - 17) / 4
W = 12 / 4 or W = -22 / 4
W = 3 or W = -5.5
Since the width cannot be negative, we discard the value of W = -5.5 and take W = 3.
Now we can find the length by substituting W into the expression we derived earlier:
Length = 2W + 5 = 2 × 3 + 5 = 6 + 5 = 11 meters
So, the dimensions of the rectangle are 3 meters by 11 meters. Tada!
To find the dimensions of the rectangle, we need to set up some equations based on the given information.
Let's assume the width of the rectangle is W meters.
According to the problem, the length of the rectangle is 5m greater than twice its width, so it can be expressed as (2W + 5) meters.
The area of a rectangle is given by the formula: Area = length * width.
Substituting the given values, we have: 33 = (2W + 5) * W.
Now, we can solve this quadratic equation for W to find the width of the rectangle.
First, let's simplify the equation:
33 = 2W^2 + 5W.
Rearranging the terms, we have:
2W^2 + 5W - 33 = 0.
Now, we can either factorize the quadratic equation or use the quadratic formula to solve for W. Let's use the quadratic formula.
The quadratic formula states that for an equation in the form of ax^2 + bx + c = 0, the solutions for x (in this case, W) are given by:
W = (-b ± sqrt(b^2 - 4ac)) / 2a.
For our equation, a = 2, b = 5, and c = -33.
Plugging in the values, we get:
W = (-5 ± sqrt(5^2 - 4 * 2 * -33)) / (2 * 2).
Simplifying further:
W = (-5 ± sqrt(25 + 264)) / 4.
W = (-5 ± sqrt(289)) / 4.
Now, finding the square root of 289, we have:
W = (-5 ± 17) / 4.
This gives us two possible values for W:
1. W = (-5 + 17) / 4 = 12 / 4 = 3.
2. W = (-5 - 17) / 4 = -22 / 4 = -5.5.
Since the width cannot be negative, the only valid value for the width of the rectangle is W = 3 meters.
Now, we can find the length of the rectangle:
Length = 2W + 5 = 2(3) + 5 = 6 + 5 = 11 meters.
Therefore, the dimensions of the rectangle are:
Width = 3 meters
Length = 11 meters.
To solve this problem, we can use the information given to set up an equation and solve for the dimensions.
Let's start by assigning variables to the width and length of the rectangle.
Let:
W = width of the rectangle
L = length of the rectangle
We are given two pieces of information:
1. "The length of a rectangle is 5m greater than twice its width." This can be expressed as L = 2W + 5.
2. "The area of the rectangle is 33m^2." The area of a rectangle is given by the formula A = L * W.
Now, we can substitute the value of L from the first equation into the second equation to get:
33 = (2W + 5) * W
Expanding the equation:
33 = 2W^2 + 5W
Rearranging the equation to get a quadratic form:
2W^2 + 5W - 33 = 0
Now, we can solve the quadratic equation for W using factoring, completing the square, or using the quadratic formula. Let's use the quadratic formula:
W = [-b ± √(b^2 - 4ac)] / (2a)
From the equation 2W^2 + 5W - 33 = 0, we have:
a = 2, b = 5, c = -33
Plugging these values into the quadratic formula:
W = [-5 ± √(5^2 - 4 * 2 * -33)] / (2 * 2)
Simplifying the equation:
W = [-5 ± √(25 + 264)] / 4
W = [-5 ± √289] / 4
W = [-5 ± 17] / 4
Now we have two possible values for W:
1. W = (-5 + 17) / 4 = 12 / 4 = 3
2. W = (-5 - 17) / 4 = -22 / 4 = -5.5 (Since negative width is not possible, we discard this solution.)
Therefore, the width of the rectangle is 3 meters.
Now, we can substitute this value of W back into the equation L = 2W + 5 to find the length:
L = 2 * 3 + 5 = 6 + 5 = 11
So, the dimensions of the rectangle are:
Width = 3 meters
Length = 11 meters