Area congruence in two quadrilaterals

  • Context: MHB 
  • Thread starter Thread starter My Name is Earl
  • Start date Start date
  • Tags Tags
    Area
Click For Summary

Discussion Overview

The discussion revolves around the area congruence of two quadrilaterals, specifically focusing on the relationship between triangle AED and quadrilaterals ABCD and EFGA. Participants explore the implications of assuming EFGA is a parallelogram and the challenges in proving its properties based on the given information.

Discussion Character

  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • Some participants reference area formulas for triangles, parallelograms, and trapezoids as foundational to solving the problem.
  • It is suggested that triangle AED contains half the area of both given parallelograms, contingent on the assumption that EFGA is a parallelogram.
  • Others argue that without proving EFGA is a parallelogram, the area cannot be determined, as EFGA could represent any type of quadrilateral.
  • Concerns are raised about the lack of information regarding the position of point D and its implications for the problem.
  • A participant expresses frustration over the inability to prove the necessary properties of EFGA, indicating that the problem may have been misprinted or miscommunicated.
  • Another participant shares that the author of the book confirmed the problem is unsolvable without the assumption that EFGA is a parallelogram, suggesting a printing error in the problem statement.

Areas of Agreement / Disagreement

Participants generally disagree on the solvability of the problem without additional information about EFGA. There is no consensus on how to approach the problem without assuming EFGA is a parallelogram.

Contextual Notes

Limitations include the unclear definition of quadrilateral EFGA and the implications of point D's position, which affect the ability to apply area formulas confidently.

My Name is Earl
Messages
12
Reaction score
0
Using the theorems:
Area of a triangle = (1/2)bh
Area of a parallelogram = bh
Area of a trapezoid = (1/2)h(b1+b2)
Solve the following:

View attachment 8517
 

Attachments

  • parallelogram.jpg
    parallelogram.jpg
    67.7 KB · Views: 144
Mathematics news on Phys.org
Hint ...

triangle AED contains half the area of both given parallelograms
 
skeeter said:
Hint ...

triangle AED contains half the area of both given parallelograms

That is true if you assume quadrilateral EFGA is a parallelogram. Only quadrilateral ABCD is defined as a parallelogram. You would have to prove that EFGA is a parallelogram, then the problem becomes easy using area formulas. How do you prove that EFGA is a parallelogram with the given information?
 
My Name is Earl said:
That is true if you assume quadrilateral EFGA is a parallelogram. Only quadrilateral ABCD is defined as a parallelogram. You would have to prove that EFGA is a parallelogram, then the problem becomes easy using area formulas. How do you prove that EFGA is a parallelogram with the given information?

We can't prove that EFGA is a parallelogram.
It could be any type of quadrilateral, and we can't say anything about its area either.
Note that we can pick F anywhere we want, and after picking F we can pick G anywhere on the line DF.

In other words, it seems that the problem statement should have said that EFGA is a parallelogram, and that it was accidentally left out.
 
I find it strange that there is no reference to the position of point $D$.
 
Klaas van Aarsen said:
We can't prove that EFGA is a parallelogram.
It could be any type of quadrilateral, and we can't say anything about its area either.
Note that we can pick F anywhere we want, and after picking F we can pick G anywhere on the line DF.

In other words, it seems that the problem statement should have said that EFGA is a parallelogram, and that it was accidentally left out.

I spent days on this one to no avail. I can pick an E such that AE is parallel to GF, but that does not allow me to show that AE = GF. I really believe the author meant EFGA not to be initially defined as a parallelogram. I sure do wish I could see their solution to this one. Thanks for the reply!
 
My Name is Earl said:
I spent days on this one to no avail. I can pick an E such that AE is parallel to GF, but that does not allow me to show that AE = GF. I really believe the author meant EFGA not to be initially defined as a parallelogram. I sure do wish I could see their solution to this one. Thanks for the reply!

I contacted the author of the book. To my surprise I got a quick response. He stated that the problem is not solvable without saying that EFGA is a parallelogram. There was an error during the printing process and the problem was supposed to state this fact. Problem solved!
 

Similar threads

  • · Replies 4 ·
Replies
4
Views
2K
Replies
21
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
4
Views
2K
Replies
4
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 4 ·
Replies
4
Views
3K