The length of a rectangle is 2 cm more than the width. If the area is 5 cm^2, find the dimensions
w(w+2) = 5
w^2 + 2 w - 5 = 0
(
w = [ -2 +/- sqrt (4 + 20) ]/2
w = -1 +/- .5 sqrt 24
have to use the + sign
w = 1.45
L = 3.45
Let's assume the width of the rectangle is "x" cm.
Given that the length is 2 cm more than the width, the length would be "x + 2" cm.
To find the area of the rectangle, we multiply the length by the width:
Area = Length × Width
Substituting the given values, we have:
5 cm^2 = (x + 2) cm × x cm
To solve for x, we can simplify the equation:
5 cm^2 = x^2 + 2x cm^2
Rearranging the equation:
x^2 + 2x cm^2 - 5 cm^2 = 0
We have a quadratic equation. Let's solve it using factoring:
(x + 5)(x - 1) = 0
Setting each factor equal to zero:
x + 5 = 0 or x - 1 = 0
Solving for x in each case:
x = -5 or x = 1
Since a negative length is not possible, we discard the solution x = -5.
Therefore, the width of the rectangle is x = 1 cm.
The length of the rectangle is x + 2 = 1 + 2 = 3 cm.
Hence, the dimensions of the rectangle are 1 cm × 3 cm.
To find the dimensions of the rectangle, we'll use algebraic expressions.
Let's assume that the width of the rectangle is "w" cm.
According to the given information, the length of the rectangle is 2 cm more than the width, so it can be represented as "w + 2" cm.
The formula for the area of a rectangle is length * width. In this case, the area is given as 5 cm^2, so we can set up the equation:
(w + 2) * w = 5
Now, we can solve this equation to find the value of "w" (width).
Expanding the equation, we get:
w^2 + 2w = 5
Rearranging the equation in standard quadratic form:
w^2 + 2w - 5 = 0
Now, we can solve this quadratic equation using factoring, completing the square, or the quadratic formula. In this case, using the quadratic formula is the most convenient.
The quadratic formula states that for an equation in the form of ax^2 + bx + c = 0, the solution for x is given by:
x = (-b ± √(b^2 - 4ac)) / 2a
In our equation, a = 1, b = 2, and c = -5. Plugging these values into the quadratic formula:
w = (-2 ± √(2^2 - 4(1)(-5))) / 2(1)
Simplifying:
w = (-2 ± √(4 + 20)) / 2
w = (-2 ± √24) / 2
w = (-2 ± 2√6) / 2
Now, simplifying further:
w = -1 ± √6
Since width cannot be negative, we discard the negative solution.
So, the width of the rectangle is w = √6 - 1 cm.
To find the length, substitute the value of "w" into the expression for the length:
Length = w + 2 = (√6 - 1) + 2 = √6 + 1 cm
Therefore, the dimensions of the rectangle are approximately width: √6 - 1 cm and length: √6 + 1 cm.