Sunday, December 4, 2011

Plane Geometric Figure Problem 3 Solution

This is the solution for the Algebra plane geometric figure problem as asked by an anonymous visitor which read, "catrina and tom want to buy a rug for a room that s 14 by 15 feet. They want to leave an even strip of flooring uncovered around the edges of the room. How wide a strip if they buy a rug with an area of 110 square feet?"

It's easier to picture this.  Draw a rectangle that's almost a square 14 x 15 feet.  Within in draw another such figure but smaller.   The space between that figure and the 14 x 15 figure is x.  Now the smaller square has an area of 110 square feet.  We now have an equation :


(15 -x ) (14 -x ) = 110

210 + x^2 -14x - 15x = 110

x^2 - 29x + 210 = 110

x^2 -29x + 100 = 0

(x - 25)(x - 4) = 0

x = 4 or 25 however using 25 would result in negative dimensions which does not make sense.  So 4 is the answer

Check:  (15 - 4) (14 - 4) = 110  = 11*10 = 110 

9 comments:

  1. Were do you get the 210 from?

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  2. 15 * 14 = 210

    (15 -x ) (14 -x ) = 110 <-- You are multiplying what's in the brackets together. 15*14 -x*14 -x*15 -x*-x = 210 - 14x - 15x + x^2 = x^2 - 29x + 210

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  3. wat is the unit of measure in the final answer?

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  4. Wouldn't the strip be 2 since the length/width of the rug is 2x less than the room?

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  5. Didn't you skip a step? You went straight from (X^-29x + 100 = 0) to the quadratic equation of (x-25)(x-4)= 0. How do you reverse the simplification of a quadratic equation? In other words, how did you separate -29x into -25X and -4x?

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  6. I'm pretty confident you solved for the wrong unknown. You solved not for the width of the strip but for twice the width of the strip. You stated: "The space between that figure and the 14 x 15 figure is x." but if it is the case that x is the rugless border, then the problem to be solved is:
    (14-2x)(15-2x)=110
    thus
    210-58x+4x**2=110
    4x**2-58x-100=0
    x**2-(29/2)x-25=0
    (x-2)(x-25/2)=0
    so x=2 is the realistic answer if x is the rugless border.

    The arithmetic works nicer your way, but in that case you have to say you are going to make x TWICE THE RUGLESS BORDER.

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  7. Ya I forgot to add the extra x -- its x on both sides of the width and length for the rug-- I will update. I just did it for one side in a hurry.

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  8. Thanks for pointing that out...

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