Length of a rectangle is 13 cm less than 5 times it's width. Its area is 28 square cm. Find dimensions of rectangle

L = 5W-13

LW = 28

(5W-13)W = 28
5W^2 - 13W - 28 = 0
(5W+7)(W-4) = 0

so, W=4
L = 5*4-13 = 7

Well, looks like this rectangle has quite the personality. Let's give it a name: Rectangle Randy.

Now, we know that Randy's length is 13 cm less than 5 times his width. That sounds like Randy really wants to make an impression, but ends up coming up a little short.

So, let's say Randy's width is "W" cm. According to Randy's statement, his length would be 5W - 13 cm. Now we have Randy's dimensions – he's getting clearer!

Next, we're told that the area of Randy is 28 square cm. The area of a rectangle is determined by multiplying its length by its width. So, we can use this information to set up an equation:

(W) * (5W - 13) = 28

Now it's time to put on our math hats and solve this equation for Randy's dimensions.

Expanding the equation, we get:
5W^2 - 13W = 28

Bringing everything to one side:
5W^2 - 13W - 28 = 0

Well, this doesn't look like the equation Randy was hoping for, but that's life sometimes. To solve this quadratic equation, we can use factoring, completing the square, or the quadratic formula.

After performing the calculations, the two possible values for Randy's width are approximately 3.17 cm and -1.77 cm.

Since dimensions can't be negative, we discard the negative value, and we're left with Randy's width: 3.17 cm.

To find Randy's length, we'll plug this value back into his statement:
Length = 5W - 13
Length = 5(3.17) - 13
Length ≈ 1.85 cm

So, Randy's dimensions are approximately 3.17 cm by 1.85 cm. He may come up a little short, but at least he has a good sense of humor!

Let's assign variables to the width and length of the rectangle to make it easier to solve the problem.

Let's say the width of the rectangle is 'w' cm.

According to the given information, the length of the rectangle is 13 cm less than 5 times its width, or 5w - 13 cm.

The area of a rectangle is calculated by multiplying its length and width. In this case, the area is given as 28 square cm.

So, we have the equation:

w * (5w - 13) = 28

To simplify the equation, let's expand it:

5w^2 - 13w = 28

Now, let's rearrange the equation into a quadratic form:

5w^2 - 13w - 28 = 0

To solve this quadratic equation, we can either factor it or use the quadratic formula. In this case, let's use the quadratic formula:

w = (-b ± √(b^2 - 4ac)) / (2a)

In the quadratic equation 5w^2 - 13w - 28 = 0, the coefficients are:

a = 5
b = -13
c = -28

Substituting these values into the quadratic formula, we get:

w = (-(-13) ± √((-13)^2 - 4 * 5 * -28)) / (2 * 5)
w = (13 ± √(169 + 560)) / 10
w = (13 ± √729) / 10
w = (13 ± 27) / 10

Now, let's calculate both solutions:

w1 = (13 + 27) / 10 = 40 / 10 = 4 cm
w2 = (13 - 27) / 10 = -14 / 10 = -1.4 cm

Since the width of a rectangle cannot be negative, we can only consider the solution w1 = 4 cm.

Now, let's find the length of the rectangle:

length = 5w - 13
length = 5 * 4 - 13
length = 20 - 13
length = 7 cm

Therefore, the dimensions of the rectangle are:

Width = 4 cm
Length = 7 cm

To find the dimensions of the rectangle, we can use the given information and set up equations to solve for the width and length.

Let's assume the width of the rectangle is represented by 'w'.

According to the given information, the length of the rectangle is 13 cm less than 5 times its width. So, the length can be expressed as (5w - 13).

The area of a rectangle is given by multiplying its length and width. In this case, the area is given as 28 square cm. So, we can set up the equation:

Area = Length * Width
28 = (5w - 13) * w

Now, let's solve this equation to find the width of the rectangle:

28 = (5w - 13) * w
28 = 5w^2 - 13w

Rearranging the equation to be in standard quadratic form:
5w^2 - 13w - 28 = 0

To solve this quadratic equation, we can either factor it or use the quadratic formula. Since the factors of 28 are not readily apparent, let's use the quadratic formula:

w = (-b ± √(b^2 - 4ac)) / (2a)

For our quadratic equation of 5w^2 - 13w - 28 = 0, the coefficient of w^2 is 'a' (5), the coefficient of w is 'b' (-13), and the constant term is 'c' (-28). Plugging these values into the quadratic formula:

w = (-(-13) ± √((-13)^2 - 4 * 5 * -28)) / (2 * 5)
w = (13 ± √(169 + 560)) / 10
w = (13 ± √729) / 10
w = (13 ± 27) / 10

Now, we have two possible values for the width of the rectangle:
w1 = (13 + 27) / 10 = 40 / 10 = 4
w2 = (13 - 27) / 10 = -14 / 10 = -1.4

Since the width of a rectangle cannot be negative, we discard the negative value, and the width of the rectangle is 4 cm.

To find the length, we substitute this value of the width back into the equation for the length:

Length = 5w - 13
Length = 5*4 - 13
Length = 20 - 13
Length = 7 cm

Therefore, the dimensions of the rectangle are width = 4 cm and length = 7 cm.