The area of a trapezoid

[Albert]

1,2,3,4 are four side length of a trapezoid,please find the area of

this trapezoid

 

[Wilmer]

Re: the area of a trapezoid

Parallel sides must be 1 and 4.

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

 

[Albert]

Re: the area of a trapezoid

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

can you show your detailed solution please ?

 

[Wilmer]

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?

         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.

 

[Albert]

Re: the area of a trapezoid

This is not my homework ,at first you must decide how to plot this trapezoid(this part may be a challenge)
here is my solution :

$h^2=9-x^2---(1)$
$h^2=4-(3-x)^2---(2)$
from (1)(2) we get $x=\dfrac {7}{3},\,\, and ,\,\, h=\dfrac {4\sqrt 2}{3}$
$\therefore area=2.5h=\dfrac {10\sqrt 2}{3}$

 

[Wilmer]

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.