- #1
Albert1
- 1,221
- 0
1,2,3,4 are four side length of a trapezoid,please find the area of
this trapezoid
this trapezoid
Can you find the area of a trapezoid with four given side lengths?
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Discussion Overview
The discussion revolves around finding the area of a trapezoid given four side lengths (1, 2, 3, and 4). Participants explore different methods and reasoning related to the geometry of trapezoids, including the identification of parallel sides and the application of the triangle inequality.
Discussion Character
- Exploratory
- Technical explanation
- Debate/contested
- Mathematical reasoning
Main Points Raised
- One participant suggests that the parallel sides of the trapezoid must be 1 and 4, leading to a proposed area calculation of approximately 4.714.
- Another participant agrees with the area calculation and requests a detailed solution.
- A different participant emphasizes that only one pair of sides (1 and 4) can be parallel, supported by the triangle inequality, and provides a geometric breakdown to derive the area.
- One participant presents their own solution involving equations for height and base lengths, arriving at the same area result of \( \frac{10\sqrt{2}}{3} \).
- Another participant humorously clarifies that their previous comment was not meant to imply homework assistance, while noting similarities between their solution and another's.
Areas of Agreement / Disagreement
Participants generally agree on the area calculation of \( \frac{10\sqrt{2}}{3} \), but there is no consensus on the approach to plotting the trapezoid or the implications of the triangle inequality. Multiple methods and interpretations are presented without resolution.
Contextual Notes
The discussion includes various assumptions about the trapezoid's configuration and relies on the triangle inequality, but these assumptions are not universally accepted or resolved among participants.
- #2
Wilmer
- 303
- 0
Re: the area of a trapezoid
Parallel sides must be 1 and 4.
Area = 5*SQRT(32/9) / 2 = ~4.714
Parallel sides must be 1 and 4.
Area = 5*SQRT(32/9) / 2 = ~4.714
- #3
Albert1
- 1,221
- 0
Re: the area of a trapezoid
yes, your answer is correct =$\dfrac {10\sqrt 2}{3}$
can you show your detailed solution please ?
yes, your answer is correct =$\dfrac {10\sqrt 2}{3}$
can you show your detailed solution please ?
- #4
Wilmer
- 303
- 0
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?
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.
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 CtriangleADF: 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.
- #5
Albert1
- 1,221
- 0
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 $x=\dfrac {7}{3},\,\, and ,\,\, h=\dfrac {4\sqrt 2}{3}$
$\therefore area=2.5h=\dfrac {10\sqrt 2}{3}$
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 $x=\dfrac {7}{3},\,\, and ,\,\, h=\dfrac {4\sqrt 2}{3}$
$\therefore area=2.5h=\dfrac {10\sqrt 2}{3}$
Attachments
- #6
Wilmer
- 303
- 0
Re: the area of a trapezoid
I see no difference between your solution and mine.
I know that, Albert; I was making a jokeAlbert said:
I see no difference between your solution and mine.