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Triangle and Trapezium Ratio


Date: 10/26/98 at 17:09:33
From: Jacky  DUVILLARD et SEB
Subject: Geometry problem

ABCD is a square with sides of length 4 cm. M is a point that is placed 
anywhere on BC. Fix M so that the proportion from the area of the 
triangle ABM to the area of the trapezium ADCM is equal to 1/3.

Jacky


Date: 10/31/98 at 00:51:18
From: Doctor Teeple
Subject: Re: Geometry problem

Hi Jacky,

The first step to solving this problem is to draw a picture. We'll 
start with just the square ABCD:

         4
   B +--------+ C
     |        |
   4 |        | 4
     |        | 
     |        |
   A +--------+ D
         4 

Place an arbitrary point M on the line BC:

         4
   B +---.----+ C
     |   M    |
   4 |        | 4
     |        | 
     |        |
   A +--------+ D
         4 

If the length of BM is x, the length of MC is 4-x because the length 
of BC is 4. So now we have: 

       x  4-x
   B +---.----+ C
     |   /M   |
   4 |  /     | 4
     | /      | 
     |/       |
   A +--------+ D
         4 

Next we need to find the areas of triangle ABM and trapezium ADCM. 
Since the area of a triangle is 1/2 * base * height, the area of ABM 
is:

   1/2 * x * 4 = 2x

To find the area of ADCM, we could divide the trapezium into a triangle 
and rectangle, find the individual areas, and add them together. We 
could also use the formula for the area of a trapezium, which is:

   (MC + AD) 
   --------- * CD
       2  

You can find more about the area of a trapezium in the Dr. Math 
Frequently Asked Questions (FAQ) (look for the word trapezoid) at: 

 http://mathforum.org/dr.math/faq/formulas/faq.quad.html   

An easier way, though, is to notice that we have the area of the 
trapezium when we start with the area of the square and take away the 
area of the triangle. We already have the area of the triangle. Since 
the area of the square is 4^2 = 16, the area of ADCM is:

   16 - 2x

Now all that's left is to set up the ratio and solve for x. Then we 
will know where M is on the line BC. I'll set up the ratio, and you 
can solve for x. Since the ratio of the area of ABM to the area of the 
ADCM is 1/3, we have:

     2x       1
   ------- = ---
   16 - 2x    3

You can finish by cross-multiplying and solving the algebraic equation. 
Write back if you have further questions. Good luck!

- Doctor Teeple, The Math Forum
  http://mathforum.org/dr.math/   
    
Associated Topics:
Middle School Algebra
Middle School Geometry
Middle School Ratio and Proportion
Middle School Triangles and Other Polygons

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