The length of a rectangle is 1cm longer than its width. If the diagonal of the rectangle is 4cm, what are the dimensions(the length and the width)of the rectangle?
L=W-1
Diagonal = sqrt(L^2 + W^2)
substitutuet for L in the second equation, solve for W.
Let's solve for the width of the rectangle first.
Given: L = W - 1
Substituting this value of L into the formula for the diagonal:
Diagonal = √(L^2 + W^2)
4 = √((W - 1)^2 + W^2)
16 = (W - 1)^2 + W^2
Expanding and simplifying the equation:
16 = W^2 - 2W + 1 + W^2
0 = 2W^2 - 2W + 1 - 16
0 = 2W^2 - 2W - 15
Now we can solve this quadratic equation for W using the quadratic formula:
W = (-b ± √(b^2 - 4ac)) / (2a)
Plugging in the values:
W = (-(-2) ± √((-2)^2 - 4(2)(-15))) / (2(2))
W = (2 ± √(4 + 120)) / 4
W = (2 ± √124) / 4
W = (2 ± 2√31) / 4
Simplifying further:
W = (1 ± √31) / 2
Since the width cannot be negative, we discard the negative value.
W = (1 + √31) / 2
Now we can find the length (L) by substituting this value of W back into the expression L = W - 1:
L = (1 + √31) / 2 - 1
L = (1 + √31 - 2) / 2
L = (√31 - 1) / 2
Therefore, the dimensions of the rectangle are:
Width (W) = (1 + √31) / 2
Length (L) = (√31 - 1) / 2
To solve for the dimensions of the rectangle, let's start by using the given information to create an equation.
According to the problem, the length (L) of the rectangle is 1cm longer than its width (W). So we can express the relationship between L and W as:
L = W - 1
Now, we can use the Pythagorean theorem to relate the diagonal of the rectangle (4cm) to its length and width. The formula is:
Diagonal = sqrt(L^2 + W^2)
Substituting for L in the formula using the relationship we found earlier:
Diagonal = sqrt((W - 1)^2 + W^2)
Since we know that the diagonal is 4cm, we can substitute Diagonal with 4:
4 = sqrt((W - 1)^2 + W^2)
Now, we can solve this equation to find the value of W, which represents the width of the rectangle.
Squaring both sides of the equation:
16 = (W - 1)^2 + W^2
Expanding:
16 = W^2 - 2W + 1 + W^2
Combining like terms:
16 = 2W^2 - 2W + 1
Rearranging the equation:
2W^2 - 2W - 15 = 0
Now we have a quadratic equation. We can solve for W by factoring, completing the square, or using the quadratic formula.
Factoring the quadratic equation is not possible, so let's solve it using the quadratic formula:
W = (-b ± sqrt(b^2 - 4ac)) / (2a)
In this equation, a = 2, b = -2, and c = -15. Substituting these values:
W = (2 ± sqrt((-2)^2 - 4(2)(-15))) / (2(2))
Simplifying:
W = (2 ± sqrt(4 + 120)) / 4
W = (2 ± sqrt(124)) / 4
W = (2 ± 2sqrt(31)) / 4
Now we have two possible values for W. Evaluating both options:
When W = (2 + 2sqrt(31)) / 4, simplifying:
W = (1 + sqrt(31)) / 2
When W = (2 - 2sqrt(31)) / 4, simplifying:
W = (1 - sqrt(31)) / 2
These are the possible values for the width of the rectangle.
To find the length (L), we can substitute each value of W into the equation L = W - 1:
When W = (1 + sqrt(31)) / 2, substituting into L = W - 1:
L = (1 + sqrt(31)) / 2 - 1
L = (1 + sqrt(31)) / 2 - 2/2
L = (1 + sqrt(31) - 2) / 2
L = (sqrt(31) - 1) / 2
When W = (1 - sqrt(31)) / 2, substituting into L = W - 1:
L = (1 - sqrt(31)) / 2 - 1
L = (1 - sqrt(31)) / 2 - 2/2
L = (1 - sqrt(31) - 2) / 2
L = (-1 - sqrt(31)) / 2
Therefore, the dimensions of the rectangle are:
Width (W):
- W = (1 + sqrt(31)) / 2
- W = (1 - sqrt(31)) / 2
Length (L):
- L = (sqrt(31) - 1) / 2
- L = (-1 - sqrt(31)) / 2