SOLUTION: a rectangle is to be consructed with its length four times the size of its width. the area of the rectangle must be 100 sqaure centimeters. find the dmension of the rectangle.

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Question 693415: a rectangle is to be consructed with its length four times the size of its width. the area of the rectangle must be 100 sqaure centimeters. find the dmension of the rectangle.
Answer by RedemptiveMath(80) About Me  (Show Source):
You can put this solution on YOUR website!
It may be helpful to draw this as a diagram. I obviously cannot do it on here, so try to bear with me as I denote the sides of this rectangle. We have a rectangle that has a length and width. This problem wants its length to be four times its width. So, we can just call the width "w" and its length "4w" (since it is 4 times greater than the width). It would also be helpful to try to keep the variables the same when dealing with these types of problems. That is, we need to relate the sides in such a way that they both share the same variable. Normally the length would be given variable "l", but we can write it with a "w" due to the information we are given (4 times the width or 4w). Now we are given the area as information. Its area must be 100 cm^2. In order to get this value, we need to use the area formula for a rectangle, or length times width. Now that we have this information, we can start plugging in values and solve for a variable.

What variable you use first (l or w) determines what variable will be solved first. I will attempt to show both ways below. We need to first write the area formula for a rectangle since we know the area is 100 cm^2:

lw = 100 cm^2.

Since we know that l = 4w, we can plug that into the place of l:

(4w)w = 100 cm^2.

Now we just solve for w using algebra:

4w^2 = 100 cm^2 (multiply 4w and w)
w^2 = 25 cm^2 (divide by 4 on each side)
w = 5 cm (square root both sides).

We have found w to equal 5 cm. The final thing to do is plug what w equals into what l equals. We must use the value we used for l, or 4w:

l = 4w = 4(5) = 20 cm.

This is how you solve these problems. To check we simply ask questions from the information we are given. Is our length 4 times our width? Well, 20 is 4 times 5, so we have that right. If we find the area of our rectangle, does it equal 100 cm^2? Well, the area of a rectangle is length times width, and since our length is 20 cm and our width is 5 cm, the area would be 100 cm^2. We have met both requirements. Length = 20 cm and width = 5 cm.

We could also write the terms with respect to length. If we do so we have length simply as "l" and width as "l/4". (Since length is 4 times the width, width is 1/4 times the length.) Then we would solve for l first and plug what we find into what w equals (in this case l/4) to find w. We would get the same answers as above.