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Solving Proportions: More Examples (page 6 of 6)

Sections: Ratios, Proportions, Checking proportionality, Solving proportions


Another category of proportion problem is that of "similar figures". This is a geometric term, referring to geometric shapes that are the same, except that one is larger than the other. Think of what happens when you use the "enlarge" or "reduce" setting on a copier, and you'll have the right idea. If you've used a graphics program, think "aspect ratio". The point is that, in "similar" figures, the corresponding sides are proportional. For instance:

similar triangles

In these triangles, corresponding pairs of sides are proportional. That is, A : a = B : b = C : c. This can be used to find the missing side of a figure.

  • In the above triangle, A = 48mm, B = 81mm, C = 68mm, and a = 21mm. Find the lengths of b and c, rounded to the nearest whole number.

    I'll set up the proportions and solve. First, I'll find the length of b.

       A/a  =  B/b
      .48/21 = 81/b
      48b = 1701
      b = 35.4375

    Now I'll find the length of c. Copyright © Elizabeth Stapel 2006-2008 All Rights Reserved

       A/a   =  C/c
      .48/2168/c
      48c = 1428
      c = 29.75

    Rounded to the nearest whole number, b = 35mm and c = 30mm.

  • A picture measuring 3.5" high by 5" wide is to be enlarged so that the width is now 9 inches. How tall will the picture be?

    In other words, they will be maintaining the aspect ratio; the rectangles will be similar figures. Set up the proportion and solve:

      .(height)/(width) :  3.5/5  =  h/9
      .3.5/5   =  h/9
      31.5 = 5h
      6.3 = h

    The picture will be 6.3 inches high.

  • A building casts a 103-foot shadow at the same time that a 32-foot flagpole casts as 34.5-foot shadow.  How tall is the building? (Round your answer to the nearest tenth.)

    To set up this problem, I first draw the building, and flagpole, and the lines indicating the sunlight's path:

    building and flagpole

    The idea here is that the sun is so far away that the angle formed by the sun's rays is the same for both the building and the flagpole. So the heights of the building and the flagpole, the shadows along the ground, and the sunlight lines not only form right triangles, but they form similar triangles. Since the triangles are similar, I can set up a proportion and solve:

      similar triangles

       

      .(height)/(shadow) h/103   =  32/34.5
      h/103   =  32/34.5
      34.5h = 3296
      h = 95.5362318841...

    The building is 95.5 feet tall.


There is another type of problem that occasionally crops up, and you don't even think of it as being a "ratios and proportions" kind of problem. Sometimes when you are mixing something (such as concrete), the measurements are given in terms of "parts", rather than cups or gallons or milliliters. For instance:

  • The instructions for mixing a certain type of concrete call for 1 part cement, 2 parts sand, and 3 parts gravel. You have four cubic feet of sand. How much cement and gravel should you mix with this sand?

    The ratio of cement to sand is 1/2. I have four cubic feet of sand, so I set up the proportion and solve for the amount of cement:

      .(cement)/(sand) :  1/2  =  c/4
      .1/2   =  c/4
      1×4 = c×2
      4 = 2c
      2 = c

    Don't forget your units! The answer here is not "2", but "two cubic feet of cement". Now I'll solve for the amount of gravel. The ratio of sand to gravel is 2/3, and I have four cubic feet of sand, so I set up the proportion and solve for the amount of gravel:

      .(sand)/(gravel) 2/3  =   4/g
      .2/3   =  4/g
      g = 4×3
      2g = 12
      g = 6

    Remembering the units, the answer is not "g = 6", but is "six cubic feet of gravel". Then the complete answer is:

    I need two cubic feet of cement and six cubic feet of gravel.

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Cite this article as:

Stapel, Elizabeth. "Solving Proportions: More Examples." Purplemath. Available from
    http://www.purplemath.com/modules/ratio6.htm. Accessed
 

 

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