Return to the Purplemath home page

 The Purplemath Forums
Helping students gain understanding
and self-confidence in algebra


powered by FreeFind

 

Return to the Lessons Index  | Do the Lessons in Order  |  Get "Purplemath on CD" for offline use  |  Print-friendly page

Geometry Word Problems: Complex Examples (page 4 of 6)

Sections: Introduction, Basic examples, Triangle formulas, Complex examples, The Box Problem & the Goat Problem, Max / Min problems


  • You work for a fencing company. A customer called this morning, wanting to fence in his 1,320 square-foot garden. He ordered 148 feet of fencing, but you forgot to ask him for the width and length of the garden. Because he wants a nicer grade of fence along the narrow street-facing side of his plot, these dimensions will determine some of the details of the order, so you do need the information. But you don't want the customer to think that you're an idiot, so you need to figure out the length and width from the information the customer has already given you. What are the dimensions?

    The perimeter P of this rectangular area with (as-yet unknown) length L and width W is given by:

     

    ADVERTISEMENT

     

      2L + 2W = 148

    The area A is given by:

      LW = 1320

    I will divide my "perimeter" equation above by 2, so I am dealing with smaller numbers. This gives me the following system (or "set") of equations:

      L + W = 74
      LW = 1320

    I can solve either one of these equations for either one of the variables, and then plug this into the other equation. I think I'll solve the addition equation, and plug the result into the multiplication equation:

      L = 74 W   (solving the first equation for L)
      (74 W) W = 1320    (substituting into the second equation)
      74W W 2 = 1320
      0 = W 2 74W + 1320

      0 = (W 30)(W 44)

      W = 30
        or  W = 44

    Once again, I've come up with two valid solutions. If W = 30, then L = 74 W = 74 30 = 44.  If W = 44, then L = 74 W = 74 44 = 30. The important point is that the shorter side (whether I refer to it as the "width" or the "length") is across the front of the lot.

      The garden is 44 feet by 30 feet, with the 30-ft length along the front.

Note that we cannot say which of the dimensions is the length or the width, since no information was provided regarding which was longer. So "this by that" is as accurate an answer as we can give.

  • Three times the width of a certain rectangle exceeds twice its length by three inches, and four times its length is twelve more than its perimeter. Find the dimensions of the rectangle.

    The first statement, "three times the width exceeds twice its length by three inches", compares the length L and the width W. I'll start by doing things orderly, with clear and complete labelling:

      the width: W
      three times the width: 3W
      twice its length: 2L
      exceeds by three inches, meaning "is three inches greater than": + 3
      equation: 3W = 2L + 3

    The second statement, "four times its length is twelve more than its perimeter", compares the length L and the perimeter P. I will again be complete with my labelling:

      four times its length: 4L
      perimeter: P = 2L + 2W     (this is just the perimeter formula for rectangles)
      twelve more than: + 12
      equation: 4L = P + 12, or 4L = (2L + 2W) + 12  (by substitution)

    So now I have my two equations:   Copyright Elizabeth Stapel 2000-2011 All Rights Reserved

      3W = 2L + 3
      4L = 2L + 2W + 12

    There are various ways of solving this; the way I do it (below) just happens to be what I thought of first. I'll take the first equation and solve for W:

      3W = 2L + 3
      W = ( 2/3 )L + 1

    Now I'll simplify the second equation, and then plug in this above expression for W:

      4L = 2L + 2W + 12
      2L = 2W + 12

      2L = 2[ ( 2/3 )L + 1 ] + 12
          (by substitution from above)
      2L = ( 4/3 )L + 2 + 12
      2L = ( 4/3 )L + 14

      2L ( 4/3 )L = 14

      ( 6/3 )L ( 4/3 )L = 14

      ( 2/3 )L = 14

      L = (14)( 3/2 ) = 21

    Then:

      W = ( 2/3 )L + 1
          = ( 2/3 )(21) + 1
          = 14 + 1 = 15

    The question didn't ask me to "Find the values of the variables L and W". It asked me to "Find the dimensions of the rectangle," so the actual answer is:

      The length is 21 inches and the width is 15 inches.

<< Previous  Top  |  1 | 2 | 3 | 4 | 5 | 6  |  Return to Index  Next >>

Cite this article as:

Stapel, Elizabeth. "Geometry Word Problems: Complex Examples." Purplemath.
    Available from 
http://www.purplemath.com/modules/perimetr4.htm.
    Accessed
 

 



Purplemath:
  Linking to this site
  Printing pages
  School licensing


Reviews of
Internet Sites:
   Free Help
   Practice
   Et Cetera

The "Homework
   Guidelines"

Study Skills Survey

Tutoring from Purplemath
Find a local math tutor


This lesson may be printed out for your personal use.

Content copyright protected by Copyscape website plagiarism search

  Copyright 2000-2012  Elizabeth Stapel   |   About   |   Terms of Use

 

 Feedback   |   Error?