Can you solve for X? | Green Trapezoid | Trapezium | Important Geometry and Algebra skills explained

[@rishudubey1533]

i think you are the best teacher of the world ❤😊

[@PreMath]

So nice of you, dear Thank you! Cheers! 😀

[@bigm383]

😂👍

[@francismoles9852]

we can applicate Thales theorem: CE/EA equal of DE/EB which gave 1/x+4 equal to x-3/13x-44 in cross multiply it give x^2-12x+32 equal to 0 when x is 4 we have 2 isoceles triangles CE equal of DE is 1 and AE equal of BE is 8

[@nineko]

It's interesting and unusual that both solutions actually are acceptable, I was totally expecting that one had to be rejected like it's often the case, but not today 😁

[@PreMath]

Excellent!. Thank you! Cheers! 😀

[@miguelgnievesl6882]

Although these diagrams are not to scale, it is obvious that X cannot equal 8. Triangles ABE and CDE are isosceles.

[@tspis]

Nothing in here OTHER THAN THE DIAGRAM suggests that ABE & CDE are isosceles. Both solutions are acceptable, but with x=8, the ratio of the non-parallel sides in each triangle is 1:5 (vs. 1:1), the scale ratio of the triangles themselves is 1:12 (vs. 1:8), and both triangles become VERY obtuse.

[@RobG1729]

Your careful, step-by-step explanations make the solutions so clear. However, only one value of x fits the geometry of that trapezoid, but this video did not clarify that. Thank you for your time and effort.

[@PreMath]

Thanks for your feedback! Cheers! 😀 You are awesome. Keep it up 👍

[@HappyFamilyOnline]

Great video👍 Thanks for sharing😊

[@PreMath]

Thanks for visiting

[@khalidayubi01]

Sir your explanation is very good...

[@PreMath]

Thanks Keep watching

[@misterenter-iz7rz]

x+4=13x-44/x-3, (x+4)(x-3)=13x-44, x^2+x-12=13x-44, x^2-12x+32, x=4 or 8.😊

[@PreMath]

Excellent! Thank you! Cheers! 😀

[@DannyClassics]

It cannot be two answers, the fundamental theorem of algebra states "every polynomial equation of degree n with complex number coefficients has n roots, or solutions, in the complex numbers" and there is only x to the 1st power in the problem. So therefore an extraneous solution was introduced to the equation when solving and must be removed from the solution.

[@tspis]

I think the best way to interpret this is that only one solution is valid for a specific shape. If we were given at least one angle measurement, we would have been locked to a specific shape. However, we were not - therefore, from the information given (which clearly specifies that the diagram is not to scale, which I'm assuming to also mean that the scale factors between dimensions aren't locked), both solutions are valid.

[@DannyClassics]

@@tspis this is a good argument. I see how they are both valid, but I feel as though the introduction of a second solution violates the FTA (fundamental theorem of algebra). Does it not violate it?

[@tspis]

@@DannyClassics While I'll prefix that I'm no expert, I don't believe it violates the FTA. Remember that just because none of the initial terms seem to contain the variable in the 2nd degree, that doesn't necessarily mean that it's not a quadratic trinomial. If it WAS such a case, then merely factoring a quadratic would seem to change its degree. To evaluate roots, like terms have to be collected, simplified, and the whole expression arranged such that it is equal to zero. So although the initial equation that's set up from the ratios of sides between similar triangles looks like it only contains 1st degree terms [i.e. (x+4)/1 = (13x-44)/(x-3)], once terms are cross-multiplied and collected, we get x^2-12x+32=0, showing that the algebraic relationship is indeed a quadratic one. But again, I'm not an expert, so if you or anyone has other thoughts (or knows better), do chime in please.

[@LENAKSOY]

Awesome question🎉 Best regards

[@PreMath]

Thank you! Cheers!

[@michaelgarrow3239]

You’re a math calculating wizard.

[@PreMath]

Thanks for your feedback! Cheers! 😀

[@mohammedkhettab9965]

❤❤❤

[@PreMath]

Thanks for your feedback! Cheers! 😀

[@vidyadharjoshi5714]

EA/EC = EB/ED gives xsq - 12x + 32 = 0 so x = 4 or x = 8

[@PreMath]

Excellent! Thank you! Cheers! 😀

[@raffaeleguerrieri5482]

X2-12x+32=0 sono 2 soluzioni: 2 e 4. 2 si rigetta e valida solo 4.

[@ΓΕΩΡΓΙΟΣΛΕΚΚΑΣ-μ9μ]

Σήμερα είδα και ασχολήθηκα με την άσκηση. Όντως αυτές φαίνεται να είναι οι λύσεις της. Αν x=4, το τραπέζιο έχει ίσες διαγωνίους άρα είναι ισοσκελές. Στην 2η περίπτωση αν κάποιος μπορέσει να κατασκευάσει το σχήμα νομίζω ότι θα κερδίσει NOBEL😂. Έχω τη γνώμη λοιπόν ότι πρέπει να αναδιατυπωθεί η άσκηση και - πρόταση - στο τέλος να μπει 2ο ερώτημα: "για ποιά τιμή του x το τραπέζιο είναι ισοσκελές" οπότε νομίζω ξεκαθαρίζουν τα πράγματα. Ευχαριστώ.

[@crazywarxyz911]

Mathematics is the best subject

[@PreMath]

Yes!

[@is7728]

I do also think so

[@Mathskylive]

Tính toán hay quá❤❤

[@PreMath]

Thanks for your feedback! Cheers! 😀

[@bigm383]

👍❤️😀🥂

[@PreMath]

Super! Thank you! Cheers! 😀

[@wackojacko3962]

You know I it! 🙂

[@PreMath]

Thanks for your feedback! Cheers! 😀

[@tonytillman3260]

4 looked like the answer from the thumbnail. I can’t see 8 being an answer

[@derekcoyle1147]

4 is an obvious answer, but 8 makes no sense. With 4, DE and EC are both 1. AE and EB are both 8. If 8 is a solution, DE=1, EC=5, AE=60, and EB=12. Based on the diagram, it’s impossible.

[@tspis]

Correct, based on the diagram, it's impossible. But when you're told the diagram is not to scale, that is true in its entirety. A skewed and stretched version of the diagram with the side lengths you provided is still a valid trapezoid - just one where the triangles are VERY obtuse, with the ratio of the non-parallel sides in each triangle being 1:5 (vs 1:1), and the scale ratio of the triangles themselves being 1:12 (vs. 1:8).

[@shrikrishnagokhale3557]

X=8