SOLUTION: I have a photographic image on a 10.5-centimeter by 8.2-centimeter background. The overall area of the background is to be reduced to 80% of its original area by cutting off equal

Algebra ->  Algebra  -> Pythagorean-theorem -> SOLUTION: I have a photographic image on a 10.5-centimeter by 8.2-centimeter background. The overall area of the background is to be reduced to 80% of its original area by cutting off equal       Log On

Ad: Algebrator™ solves your algebra problems and provides step-by-step explanations!
Ad: Algebra Solved!™: algebra software solves algebra homework problems with step-by-step help!

   

Question 5681: I have a photographic image on a 10.5-centimeter by 8.2-centimeter background. The overall area of the background is to be reduced to 80% of its original area by cutting off equal strips on all four sides. What is the width of the strip that is cut from each side? The answer is 0.49 cen, what I need to know is how to set the problem up. Thanks!!!!
Answer by prince_abubu(198) About Me  (Show Source):
You can put this solution on YOUR website!
The first step would have to be set up a proportion. You know that 100% refers to the full 10.5 x 8.2. The problem is finding an expression for 80%.

So far we have the proportion +1%2F%28%2810.5%29%288.2%29%29+=+0.80%2FE+. We need to work on finding the expression for E - the expression for the new dimensions when equal width strips are taken off all four sides.

Let w = the width of the strip to be taken off all four sides so that the resulting (reduced) size is 80% of the original background.
The full length is 10.5, right? And we're chopping off a strip from BOTH sides of it, so the reduced length is 10.5 - 2w. The full width is 8.2, so chopping off from both sides reduced the width to 8.2 - 2w. The area of the 80% would have to be (10.5 - 2w)(8.2 - 2w). It's just that we have to find the right w value so that the resulting area of the remaining background be 80% of the old background. Here's the full proportion:
+1%2F%28%2810.5%29%288.2%29%29+=+0.80%2F%28%2810.5+-+2w%29%288.2+-+2w%29%29+

+0.80%2A10.5%2A8.2+=+%2810.5+-+2w%29%288.2+-+2w%29+ <---- Cross multiplied

+68.88+=+86.10+-+37.4x+%2B+4x%5E2+ <---- simplified and distributed.

+4x%5E2+-+37.4x+%2B+17.22+=+0+ <---- rearranged so that the right hand side is 0 so that we may solve for x.

Since the coefficients are not nice-looking numbers, we'll have to pull out our handy quadratic formula:

x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+

Let's evaluate first using the "positive" by plugging in the 4 for the a, the -37.4 for the b, and the 17.22 for the c:

x+=+%28-%28-37.4%29+%2B+sqrt%28+%28-37.4%29%5E2-4%2A4%2A17.22+%29%29%2F%282%2A4%29+ <---- once evaluated, this gives us x = 8.86 approximately.

Let's then compute the "negative":

x+=+%28-%28-37.4%29+-+sqrt%28+%28-37.4%29%5E2-4%2A4%2A17.22+%29%29%2F%282%2A4%29+ <----- once evaluated, this gives us x = 0.49 approximately.

Now, this proportion gave us two answers. Which one is right? I would say that the 0.49 cm would make more sense. It's impossible to "chop off" 8.87 centimeters from all sides of a background of only 10.5 x 8.2 cm^2.