You shouldn't reject the value of x = 7 - sqrt(2). It doesn't look like the picture, since you've drawn it so x > 7. But as we say in math, "the picture may not be drawn to scale." There is a second solution with x < 7, in which the blue triangle is skinnier and taller. More generally, "it's not realistic" isn't a mathematical explanation. You didn't start with a problem in which x > 7 was a stated constraint.
@@PreMath You said the drawing was not to scale, and have used that statement in past problems to your advantage. Rejecting the value of x = 7 - sqrt(2) should not be accepted, plus it is an interesting valid result that produces interesting triangle shapes.
I solved it independently and the solution for X matched with yours except that my solution yielded two values for X. One is the same as yours and the other is obtained with - operator ie 14-7√2. This value can't be rejected as it is not an imaginary number (√-value). In fact the sketch conceives X as greater than 7. With X
@@PreMath Sure You are right. In that eventuality, the quadratic equation solution transforms to: X = ( -b+(b^2 -4a.c)^0.5)/2. I like your effort very much. I always try to solve your puzzles and these give my mind a twist.
I find it interesting that the other value for X works as well. The resulting x value is a little over 4 (4.1005050633883346583881789305321) which gives a height of about 17.66 (17.656854249492380195206754896839). This yields a trapezoid much taller than shown with bottom base narrower than the top. But the area calculation still works out to 98.
100% impossible X has to be longer than 7, as drawing shows. One thing is that drawing could be slightly out of scale, other completely different thing is that drawing is wrong..And wrong is not.
@@PreMath No problem Sir. I just thought it was interesting that both work in this case as the alternate stayed positive. So often only one case works.
@@marioalb9726 Not saying the Professor is wrong. Just interesting that the alternate also works mathematically. As he stated above, the alternate solution does not fit the preconditions he set for the problem.
If you wanted to produce a scale drawing, how would you deduce the x coordinates of C, D and E, assuming that there is a unique solution given the knowns. Any thoughts?
This is a great problem to solve...thank you for posting! 🙂 Your Trapezoid has 2 parallel sides of lengths 7 and X. You don't know this but I compressed your Trapezoid till it became a single line with points ADCB. It's still a Trapezoid,...only flattened with Area = 0. For some reason I couldn't determine X. 🙃
Clever, but I feel like I have been sleighted 😉 I usually do these in CAD which has all these logics in the background, but this time it failed...miserably 🙂
Elegant solution, thanks Professor!❤🥂👍
Glad you like it!
Thanks for your feedback! Cheers! 😀
You are awesome. Stay blessed 👍
You shouldn't reject the value of x = 7 - sqrt(2). It doesn't look like the picture, since you've drawn it so x > 7. But as we say in math, "the picture may not be drawn to scale." There is a second solution with x < 7, in which the blue triangle is skinnier and taller.
More generally, "it's not realistic" isn't a mathematical explanation. You didn't start with a problem in which x > 7 was a stated constraint.
My sincere apologies for not making it clear that the side length CD is smaller than AB!
@@PreMathyour drawings can be a bit off but that is definitely clear
@@PreMath You said the drawing was not to scale, and have used that statement in past problems to your advantage. Rejecting the value of x = 7 - sqrt(2) should not be accepted, plus it is an interesting valid result that produces interesting triangle shapes.
@@johncampbell7868 I agree.
Easiest solution thanks sir
I solved it independently and the solution for X matched with yours except that my solution yielded two values for X. One is the same as yours and the other is obtained with - operator ie 14-7√2. This value can't be rejected as it is not an imaginary number (√-value). In fact the sketch conceives X as greater than 7. With X
My sincere apologies for not making it clear that the side length CD is smaller than AB!
@@PreMath Sure You are right. In that eventuality, the quadratic equation solution transforms to: X = ( -b+(b^2 -4a.c)^0.5)/2. I like your effort very much. I always try to solve your puzzles and these give my mind a twist.
I find it interesting that the other value for X works as well. The resulting x value is a little over 4 (4.1005050633883346583881789305321) which gives a height of about 17.66 (17.656854249492380195206754896839). This yields a trapezoid much taller than shown with bottom base narrower than the top. But the area calculation still works out to 98.
I think so too...
100% impossible
X has to be longer than 7, as drawing shows.
One thing is that drawing could be slightly out of scale, other completely different thing is that drawing is wrong..And wrong is not.
My sincere apologies for not making it clear that the side length CD is smaller than AB!
@@PreMath No problem Sir. I just thought it was interesting that both work in this case as the alternate stayed positive. So often only one case works.
@@marioalb9726 Not saying the Professor is wrong. Just interesting that the alternate also works mathematically. As he stated above, the alternate solution does not fit the preconditions he set for the problem.
Great explanation 👍
Thanks for sharing 😊
My pleasure 😊
Thanks for video.Good luck sir!!!!!!!!!!!!
14-7*square root(2) = 14 - 7*1.41 = 4.1 and is not negative
Nice question. Thank you, Sir.
Most welcome😀
nice. very nice!
The second solution is < 7 while x must be > or = 7; for this reason is rejected. Thanks for your videos
good video!
Thanks!
If you wanted to produce a scale drawing, how would you deduce the x coordinates of C, D and E, assuming that there is a unique solution given the knowns. Any thoughts?
Area of yellow triangle:
A₁= ½ b.h = 14 cm²
h₁ = 2.A / b = 2 .14 / 7
h₁ = 4 cm
Area of blue triangle:
A₂ = ½ b.h = 28 cm²
h₂ = 2.A / x = 56/x
h = h₁+h₂
h = 4 + 56/x
Area of paralel trapezoid:
A = ½ (a+b).h = 14+21+35+28
½(7+x).h = 98
h = 196 / (7+x)
Equalling:
4 + 56/x = 196 / (7+x)
(x+7).(56/x+4) = 196
56+4x+392/x+28 = 196
4x+392/x = 112
4x² - 122x +392 = 0
x² - 28x +98 = 0
x = 23,9 cm ( Solved √ )
This is a great problem to solve...thank you for posting! 🙂
Your Trapezoid has 2 parallel sides of lengths 7 and X. You don't know this but I compressed your Trapezoid till it became a single line with points ADCB. It's still a Trapezoid,...only flattened with Area = 0. For some reason I couldn't determine X. 🙃
Excellent!
Thanks for your feedback! Cheers! 😀
You are awesome. Keep it up 👍
Very yhanks
You are very welcome!
Thank you! Cheers! 😀
You are awesome. Keep it up 👍
I very glad
Clever, but I feel like I have been sleighted 😉
I usually do these in CAD which has all these logics in the background, but this time it failed...miserably 🙂
38..scusa 28..A=(7+x)h/2...h=h1+h2=28*2/x+14*2/7...
con A=14+35+28+21=105
Thanks for sharing! Cheers!
You are awesome. Keep it up 👍
Yay! I solved it, but with the 2 solutions.
Bravo👍
I have a problem
14*28 =21 *35 ?
Because ABCD trapezius
14
Fantastic solution 👍, thank you teacher 🙏.
Very difficult for I can't think out of box.😅
No worries!
Thanks for your feedback! Cheers! 😀
You are awesome. Keep it up 👍
made it
Super!
You are awesome. Keep it up 👍
My answer is 14+7sqrt(2)
I got this by using a height of 4+56/x. The area of th trapezoid is then ((7+x)/2)(4+56/x) = 98. I ended up with a quadratic. x^2-28x+98=0.
x=23.9 if x>7
x=4.1 if x
Area of yellow triangle:
A₁= ½ b.h = 14 cm²
h₁ = 2.A / b = 2 .14 / 7
h₁ = 4 cm
Area of blue triangle:
A₂ = ½ b.h = 28 cm²
h₂ = 2.A / x = 56/x
Area of paralel trapezoid:
A = ½ (a+b).h = 14+21+35+28
½(7+x).h = 98
½(7+x).(h₁+h₂) =98
(x+7).(56/x+4) = 196
56+4x+392/x+28 = 196
4x+392/x = 112
4x² - 122x +392 = 0
x² - 28x +98 = 0
x = 23,9 cm ( Solved √ )
asnwer=23cm o.k