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The Area of a Trapezoid

Date: 1/27/96 at 14:18:9
From: Anonymous
Subject: Area of a Trapezoid

My 7th grade daughter has been assigned a project to derive the 
area of a trapezoid, i.e. not simply look up the formula in CRC.  
My math was in prehistoric times and I have banged my head for a 
couple of days on this.  

Any direction on how to derive this from basic geometric principles 
would be appreciated i.e. not just 1/12 . l3 . sqrt(2).  


Date: 2/5/96 at 9:38:43
From: Doctor Elise
Subject: Re: Area of a Trapezoid


Okay.  Draw yourself a trapezoid - any old trapezoid.  It has two 
parallel sides of different lengths, and two non-parallel sides of 
different lengths (unless it's an isosceles trapezoid, and then 
these two are the same length).  Just so we can talk about this 
trapezoid, draw the two parallel lines horizontally with the 
longer side on the bottom.

Now label the upper left corner "A", the upper right corner "B", 
the lower right "C", and the lower left "D".  Yay!  Now we have a 
trapezoid we can talk about.

You don't do trigonometry yet in 7th grade, do you?  I'll assume 

In order to find the area of this trapezoid, we have to know a few 

The length of the line segment AB (how long is the side between A and B)
The length of the line segment DC
The length of AD, and the length of BC,
The height of the trapezoid, OR the angles of the corners.

Okay. Look at your trapezoid again, and draw a line from "A" 
perpendicular to DC.  Call the point where the new line hits DC 
point E. Then draw another line from B perpendicular to DC, and 
call that point F.  Still with me?  You should have a rectangle 
and two triangles, like this:

        A _____________________________________________ B
         /|                                            |\
        / |                                            | \
       /  |                                            |  \
      /   |                                            |   \
     /    |                                            |    \
    /     |                                            |     \
   D      E                                            F       C

Since we know that AB and DC are parallel line segments, and we 
drew AE and BF perpendicular to DC, we know a bunch of things:

AB = EF (because ABEF is a rectangle)

DAE is a right triangle (because angle AED is a right angle)
BCF is a right triangle (because angle BFC is a right angle).

So, the area of the trapezoid is equal to the area of the
rectangle plus the area of each of the triangles.

Let's take a trapezoid where AB = 5, DC = 9, the
height (which is AE AND BF) = 3, and lets say
it's an isosceles trapezoid where AD = BC.

You might want to label your trapezoid picture so you can follow

Okay.  We already know that AB = EF, because it's a rectangle, so
they're both 5. And we know that the height is 3, because they 
told us. So the area of the rectangle is the base times the 
height, 5 * 3 = 15.

Since this is an isosceles trapezoid, we know that the two 
triangles are exactly the same size and proportions, which we call 
"congruent" triangles.

(If you cut two congruent triangles right out of the paper, turn
them around the right way, and put them on top of each other, they 
are exactly the same.)

So DE = FC.   Since we know EF = 5, (because AB = 5), and DC = 9.
DE has to be half of DC - AB so we know that DE and FC each have 
to be 2.

The area of any right triangle is 1/2 base times height.  We know
the base is 2, and the height is 3 (because it's our old friend 
AE), so the area of one of these triangles is 1/2 * 2 * 3 = 3.  
And we have two of these triangles, so the total area of this 
trapezoid would be 15 + 3 + 3 = 21.

If I were going to make a generic formula for an isoceles 
trapezoid, I'd do it this way:

AREA = base * height + 1/2 base1 * height + 1/2 base2 * height

AREA = AB * AE   + 1/2 (1/2(DC - AB))* AE + 1/2 (1/2(DC - AB)) * AE

AREA = AB * AE + 1/2(DC - AB) * AE

AREA = AE * (AB + 1/2(DC - AB))

AREA = AE * (AB + 1/2 * DC - 1/2 * AB)

AREA = AE * (1/2 * AB + 1/2 * DC)

AREA = 1/2 * AE * (AB + DC)

If you check my work here, and plug in AE = 3, AB = 5, and DC = 9,
do you still get 21?

Hope this helps!

-Doctor Elise,  The Math Forum

Associated Topics:
Middle School Geometry
Middle School Triangles and Other Polygons

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