what is the formula to find the area of a trapazoid

what is the formula to find the area of a trapazoid and please give an example

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I always relate this to the same "formula" as the area of a triangle.......(1/2)hb.

But, instead of one "base," we have to add the two parallel bases. Thus we have (1/2)h(b₁ + b₂).

Have a look at the following illustration.........

Let the height (CE) = 3

Let (AB) = base one = b₁  = 6

Let (CD) = base two = b₂ = 4

So the area = (1/2)h(b₁ + b₂) = (1/2)(3)(6 + 4) = (1/2)(3)(10) = (1/2)(30) = 15 sq. units

To see how this "formula" could be developed, look at this

Notice that we have a rectangle (CDFE) which has an area of (EF)*(CE)

And notice that we have two additional triangles with bases (AE) and (BF) and height (CE).

And the area of these two triangles = (1/2)(AE)(CE) + (1/2)(BF)(CE) = (1/2)(CE)(AE + BF)

So the total area is given by (EF)*(CE) + (1/2)(CE)(AE + BF) = (1/2)(CE)(2EF + AE + BF)

Now, note that 2EF = (EF + CD), since EF = CD     .... So we really have

(1/2)(CE)(2EF + AE + BF) = (1/2)(CE)(CD + EF + AE + BF)

But AE + EF + BF = AB   ...  So we have

(1/2)(CE)(CD + EF + AE + BF) = (1/2)(CE)(CD + AB) = (1/2)(CE)(AB + CD ) = (1/2)h(b₁ + b₂)  !!!

I hope this helps you to see how the "formula" "works".....

BTW.....we're glad to have you on the forum, bioschip !!

Hi bioschip,

I think a trapezoid is another name for a trapezium and the area of a trapezium is

A = the average of the two parralel sides x perpendicular height

Does that make sense to you? 

To get the average of 2 numbers, you add them together and divide by 2.  

top and base * hight

I'll second that.

It IS great to have you on the forum.