MHB Can you find the area of a trapezoid with four given side lengths?

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The discussion revolves around calculating the area of a trapezoid with side lengths 1, 2, 3, and 4. The parallel sides are identified as 1 and 4, leading to an area calculation of approximately 4.714 or simplified to 10√2/3. Participants share their methods for deriving the height and base lengths, confirming the validity of their solutions. The conversation also touches on the challenge of plotting the trapezoid accurately. Ultimately, the area calculation is established through geometric principles and algebraic manipulation.
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1,2,3,4 are four side length of a trapezoid,please find the area of

this trapezoid
 
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Re: the area of a trapezoid

Parallel sides must be 1 and 4.

Area = 5*SQRT(32/9) / 2 = ~4.714
 
Re: the area of a trapezoid

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

can you show your detailed solution please ?
 
Re: the area of a trapezoid

We don't do homework here (Sun)

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.
 
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 :
View attachment 1118
$h^2=9-x^2---(1)$
$h^2=4-(3-x)^2---(2)$
from (1)(2) we get $x=\dfrac {7}{3},\,\, and ,\,\, h=\dfrac {4\sqrt 2}{3}$
$\therefore area=2.5h=\dfrac {10\sqrt 2}{3}$
 

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Re: the area of a trapezoid

Albert said:
This is not my homework...
I know that, Albert; I was making a joke :cool:

I see no difference between your solution and mine.
 
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