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Area of a Trapezoid


Date: 04/27/97 at 11:50:55
From: John Browne
Subject: Area of an uneven rectangle

The measurements of an uneven rectangle are 80' x 80' x 80' x 120'.  
We would like to know the area and the the acreage of this region and 
how you get the answer.  Thanks for your help.

John


Date: 07/14/97 at 17:57:40
From: Doctor Terrel
Subject: Re: Area of an uneven rectangle

Dear John,

What an interesting problem you are proposing!  First, an "uneven"
rectangle is a new idea for me. By definition, true rectangles always 
have two pairs of sides equal in length; plus the pairs of "opposite 
sides" are always parallel.  So your 80' x 80' x 80' x 120' figure 
really isn't actually a rectangle.

However, I've been trying to think what it might be, and the only idea 
that comes to me right now is a TRAPEZOID.  Here's how I think your 
figure might be:


                A___________B
                /           \                   AB = BC = AD = 80
               /             \
              /               \
             /                 \                CD = 120
           D/___________________\C

(Angles ADC and BCD have the same measure.)

Now the area of a trapezoid is rather easy to find.  We use this
formula:

     h(b + b')
A = ----------       h = height, b = one base, and b' = other base
        2

In our problem, b = CD = 120 and b' = AB = 80.  We just need to find 
the height "h".  For that we need to add some more information to our
drawing.


                A___________B
                /|         |\                   
               / |         | \
              /  |         |  \
             /   |         |   \                
           D/____|_________|____\C
                 E         F

Since our trapezoid is an "isosceles" one [i.e. the non-parallel 
sides, AD and BC, are the same length], we can draw the segments AE 
and BF. Either one of these can be considered the height of our 
figure. Since AB = EF = 80, then DE = FC = 20 and we can use the 
Pythagorean theorem on either "right" triangle - the one on the right 
(BCF) or the one on the left (AED) - to find the length of those 
segments.

AE^2 + DE^2 = AD^2
AE^2 + 20^2 = 80^2
AE^2 + 400  = 6400
       AE^2 = 6000
         AE = sqrt(6000)

Now substituting our numbers into the formula, we have:

     sqrt(6000)(80 + 120)      sqrt(6000)(200)
A = ---------------------   = -----------------  =  100 sqrt(6000)
             2                       2

Using a calculator to simplify the result, we obtain 7746 sq ft
(approximately).

Finally, you asked for its acreage. There are 43,560 sq ft in one
acre. So divide 7746 by 43560 (another good time to use your
calculator!), yielding 0.1778 acre (again a rounded value).  

It is interesting to observe that this shape or field is a little bit
more than one-sixth of an acre (1/6 = 0.1667 (approx.)).

I hope this helps you.  Write again.

-Doctor Terrel,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/   
    
Associated Topics:
High School Geometry
High School Triangles and Other Polygons

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