# Can anyone help with this algebra problem? A rectangle is 4 times as long as it is wide. A second rectangle is 5 centimeters longer and 2 centimeters wider than the first. The area of the second...

Can anyone help with this algebra problem?

A rectangle is 4 times as long as it is wide. A second rectangle is 5 centimeters longer and 2 centimeters wider than the first. The area of the second rectangle is 270 square centimeters greater than the first. What are the dimensions of the original rectangle?

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### 2 Answers

The idea when answering word problems is to use numbers and symbols to represent the words in the problem. The first thing I always like to do is figure out what "X" is because that allows me to begin to build an equation.

What is "x" in this problem? What don't we know? We don't know how long or wide it is. But we do know that it is 4 times the length. If we make "X" the width of the rectangle, then we know that the length must be "4X."

Now, we can express the second rectangle in terms of the first triangle because the information we are given is based on the dimensions of the first rectangle. The width of the second rectangle is 2 cm wider than the width of the first rectangle. If the width of the first rectangle is "X," then the width of the second rectangle must be "X + 2." If the length of the first rectangle is "4X," then the second rectangle, being 5 cm longer, must be "4X + 5."

Now that you have both rectangles expressed in terms of one variable, "X," you can start working on an equation you can solve. What do we know that will help us put together this equation? We know the area of the second rectangle is 270 cm more than the area of the first rectangle. This means that on one side of the equation, we can set the area of the first rectangle as a product of its width and length, and we can set the area of the second rectangle as being a product of width times length plus 270 more cm. The area of the first rectangle equals the area of the second rectangle plus 270. Now you have an equation you can solve. Good luck!

We assume the first rectangle measures **x cm** wide. Then by given details it is 4x cm long .Then its algebraic area = 4x^2 cm^2. (1)

By the given facts the second rectangle measures 4x+5cm long and x+2cm wide. Therefore its area algebraically = (4x+5)(x+2) cm^2 (2)

By the given facts,the ralation between the areas of rectangles is that the area of second rectangle is greater than that of the first by 270 cm ^2.

Therefore,(2)-(1) gives us:** (4x+5)(x+2)-4x^2=270 ** (3).

So,we have set up the equation of this word probem. Now we solve for x to get the width, x and then the length, 4x of the original rectangle from the equation ,(3):

4x^2+(4x)(2)+5x+10-4x^2=270

13x+10=270

13x=270-10

13x=260

13x/13=260/13

x=20cm, width of the original rectangle

4x=80cm, length of the original rectangle.

Tally:

Area of the first(original) rectangle =80*20=1600cm^2

Area of the 2nd rectangle:(80+5)(20+2)=85*22=1870.

1870-1600=270. So, the fact that the 2nd rectangle is greater by 270cm^2 in area is verified.