# The area of a trapezoid

#### Albert

##### Well-known member
1,2,3,4 are four side length of a trapezoid,please find the area of

this trapezoid

#### Wilmer

##### In Memoriam
Re: the area of a trapezoid

Parallel sides must be 1 and 4.

Area = 5*SQRT(32/9) / 2 = ~4.714

#### Albert

##### Well-known member
Re: the area of a trapezoid

yes, your answer is correct =$\dfrac {10\sqrt 2}{3}$ #### Wilmer

##### In Memoriam
Re: the area of a trapezoid

We don't do homework here OK; of the 6 possibilities, only 1:4 is possible as parallels;
can be easily shown using triangle inequality. Good nuff?
Code:
         A     1     B

2                         3

D  3-e   F     1     E          e           C
Let height AF = BE = h, and EC = e; then:
triangleADF: h^2 = 2^2 - (3 - e)^2
triangleBCE: h^2 = 3^2 - e^2

SO:
2^2 - (3 - e)^2 = 3^2 - e^2
solve: e = 7/3
so DF = 2/3 and h = (4/3)SQRT(2)

Leads to area = 5*SQRT(32/9) / 2
which may be simplified to 10SQRT(2) / 3.

#### Wilmer

##### In Memoriam
Re: the area of a trapezoid

This is not my homework...
I know that, Albert; I was making a joke I see no difference between your solution and mine.