5 and 1/3. However, the triangles can not exist. A good follow up question would be, "Assume that X=5. What should the minimum value of Y be so the triangles are valid triangles ?"
I reached the solution given in the video by using proportions. The math works, but the triangles do not work. The length of 6 is > (2 + 3) which cannot happen in a triangle. The length of 10 is > than (2 + 4/3 + 2 + 3) which also cannot happen in a triangle. I thought I did the problem wrong with my solutions, but the problem is actually wrong.
Buenos días señores: PreMath, reciban un cordial saludo. Tengo una pregunta: ¿ El triángulo que se determina será que No existe...? Gracias por la atención que se me preste. Éxitos.
DE/BC = AD/AB, so (x+1)/(2.x) = (2.x+7)/(3.x-10), so (x+1).(3.x-10) = (2.x).(2.x-7), so 3.(x^2) +3.x-10.x-10 = 4.(x^2)-14.x, so (x^2)-7.x+10 = 0 Delta = 49-40 = 9 = 3^2, so x = (7-3)/2 = 2 (rejected because all lengthes must be positive), or x = (7+3)/2 = 5. Finally x = 5. Now AD/AB = AE/AC, so 3/5 = (3.y+1)/(7.y+1), so (3).(7.y+1) = (5).(3.y+1), so 21.y+3 = 15.y+5, so 6.y = 2, and finally y = 1/3. We now have AD = 3, AE = 2 and DE = 6, it is not possible as DE must be inferior to AD+AE, so the triangle ADE cannot be constructed (and the triangle ABC as well.)
You are absolutely right: this combination of side lengths does it make impossible to construct the triangles ABC and ADE. I can't believe I missed this even though it was so obvious.😠 Best regards to France
He resuelto el ejercicio por el mismo proceso y llego a la misma solución. Pero es imposible trazar un triángulo con esas medidas porque, si BC=10 y AB+AC
MY RESOLUTION PROPOSAL : 01) Thales Theorem Statement : " If a line is drawn parallel to one side of a triangle intersecting the other two sides in distinct points, then the other two sides are divided in the same ratio. " 02) AD / AB = DE / DC 03) (2X - 7) / (3X - 10) = (X + 1) / (2X) 04) X = 5 05) Proportionality Ratio (Linear Ratio) = 3 / 5 06) (3Y + 1) / (7Y + 1) = 3 / 5 07) Y = 1 / 3 08) Checking dor Y : 09) (3Y + 1) / (7Y + 1) = 3 / 5 ; 15Y + 1 = 21Y + 3 ; 6Y = 2 ; Y = 2 / 6 ; Y = 1 / 3 So, MY BEST ANSWER : X = 5 and Y = 1 / 3
As already mentioned in other comments: this combination of side lengths does it make impossible to construct the triangles ABC and ADE. I missed this even though it was so obvious.😠
(2x - 7)/(3x - 10) = (3y + 1)/(7y + 1) = (x + 1)/(2x) (2x)(2x - 7) = (x + 1)(3x - 10) x^2 - 7x + 10 = 0 (x - 2)(x - 5) = 0 x = 2 is not possible (left side of the green triangle 2x - 7 cannot be negative) x = 5 (3y + 1)/(7y + 1) = (x + 1)/(2x) = 3/5 5(3y + 1) = 3(7y + 1) => 6y = 2 (x, y) = (5, 1/3) The three sides of the green triangle: 2x - 7 = 3 x + 1 = 6 3y + 1 = 2 This is not possible because 3 + 2 < 6 ( the sum of two sides cannot be smaller than the third side) There are no solutions to the problem.
Bom dia Mestre
Vou tentar resolver por semelhança de 🔼 utilizando o Teorema de Tales
R) x = 5, y = 1/3
Grato pela aula
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x=5 and y=1/3 BUT it is impossible for the lengths of the sides to form a triangle, either for 3, 6, 2 or 5, 10, 10/3 !
Yes! Everything else is cool, but the triangle is impossible :D oopsie!
Take it down. Put it up with realistic dimensions and don't represent a scalene triangle to look like an equilateral.triangle
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5 and 1/3.
However, the triangles can not exist. A good follow up question would be, "Assume that X=5. What should the minimum value of Y be so the triangles are valid triangles ?"
Excellent!
Thanks for the feedback ❤️
Soo good 👍
Excellent!
I reached the solution given in the video by using proportions. The math works, but the triangles do not work. The length of 6 is > (2 + 3) which cannot happen in a triangle. The length of 10 is > than (2 + 4/3 + 2 + 3) which also cannot happen in a triangle. I thought I did the problem wrong with my solutions, but the problem is actually wrong.
Thanks for the feedback ❤️
Great question professor 🙂
Glad to hear that!
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Thank you!
DE>DA+AE ! Le triangle n’est pas constructible.
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Buenos días señores: PreMath, reciban un cordial saludo. Tengo una pregunta: ¿ El triángulo que se determina será que No existe...? Gracias por la atención que se me preste. Éxitos.
Excellent!
Thanks for the feedback ❤️
That's one messed up triangle, but since I've become a a scientologist it makes perfect sense. 😊
😀😀
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DE/BC = AD/AB, so (x+1)/(2.x) = (2.x+7)/(3.x-10), so (x+1).(3.x-10) = (2.x).(2.x-7), so 3.(x^2) +3.x-10.x-10 = 4.(x^2)-14.x, so (x^2)-7.x+10 = 0
Delta = 49-40 = 9 = 3^2, so x = (7-3)/2 = 2 (rejected because all lengthes must be positive), or x = (7+3)/2 = 5. Finally x = 5.
Now AD/AB = AE/AC, so 3/5 = (3.y+1)/(7.y+1), so (3).(7.y+1) = (5).(3.y+1), so 21.y+3 = 15.y+5, so 6.y = 2, and finally y = 1/3.
We now have AD = 3, AE = 2 and DE = 6, it is not possible as DE must be inferior to AD+AE, so the triangle ADE cannot be constructed (and the triangle ABC as well.)
You are absolutely right: this combination of side lengths does it make impossible to construct the triangles ABC and ADE. I can't believe I missed this even though it was so obvious.😠
Best regards to France
@@unknownidentity2846 Thanks. Best regards.
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(5; 1/3)
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Thales' theorem
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(x+1)/2x=(2x-7)/(3x-10) (x+1)(3x-10)=2x(2x-7)
3x²-7x-10=4x²-14x x²-7x+10=0 (x-2)(x-5)=0
x=2 is rejected , x=5 is correct
(3y+1)/(7y+1)=6/10 10(3y+1)=6(7y+1) 30y+10=42y+6 12y=4 y=1/3
x = 5 , y = 1/3
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He resuelto el ejercicio por el mismo proceso y llego a la misma solución.
Pero es imposible trazar un triángulo con esas medidas porque, si BC=10 y AB+AC
Segment AE 2y+1 (oral) or 3y+1 (in the diagram)?
My apologies for any mistake🙏
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MY RESOLUTION PROPOSAL :
01) Thales Theorem Statement :
" If a line is drawn parallel to one side of a triangle intersecting the other two sides in distinct points, then the other two sides are divided in the same ratio. "
02) AD / AB = DE / DC
03) (2X - 7) / (3X - 10) = (X + 1) / (2X)
04) X = 5
05) Proportionality Ratio (Linear Ratio) = 3 / 5
06) (3Y + 1) / (7Y + 1) = 3 / 5
07) Y = 1 / 3
08) Checking dor Y :
09) (3Y + 1) / (7Y + 1) = 3 / 5 ; 15Y + 1 = 21Y + 3 ; 6Y = 2 ; Y = 2 / 6 ; Y = 1 / 3
So,
MY BEST ANSWER :
X = 5 and Y = 1 / 3
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Thanks for sharing ❤️
My answer, as well. But these values do not form a valid triangle.
Let's find x and y:
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..
...
....
.....
The triangles ABC and ADE are obviously similar, so we can conclude:
AD/AB = AE/AC = DE/BC
AD/AB = DE/BC
AD*BC = DE*AB
AD*BC = DE*(AD + BD)
(2x − 7)*(2x) = (x + 1)*[(2x − 7) + (x − 3)]
4x² − 14x = (x + 1)*(3x − 10)
4x² − 14x = 3x² − 10x + 3x − 10
x² − 7x + 10 = 0
(x − 2)*(x − 5) = 0
⇒ x = 2 ∨ x = 5
The first solution x=2 is not useful since AD=2x−7=2*2−7=−3
As already mentioned in other comments: this combination of side lengths does it make impossible to construct the triangles ABC and ADE. I missed this even though it was so obvious.😠
But there is no triangle with these criteria.
@@AzouzNacir So it is: the problem has a problem.
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(2x - 7)/(3x - 10) = (3y + 1)/(7y + 1) = (x + 1)/(2x)
(2x)(2x - 7) = (x + 1)(3x - 10)
x^2 - 7x + 10 = 0
(x - 2)(x - 5) = 0
x = 2 is not possible (left side of the green triangle 2x - 7 cannot be negative)
x = 5
(3y + 1)/(7y + 1) = (x + 1)/(2x) = 3/5
5(3y + 1) = 3(7y + 1) => 6y = 2
(x, y) = (5, 1/3)
The three sides of the green triangle:
2x - 7 = 3
x + 1 = 6
3y + 1 = 2
This is not possible because 3 + 2 < 6 ( the sum of two sides cannot be smaller than the third side)
There are no solutions to the problem.
Thanks for the feedback ❤️
@ Thank YOU
(2x-7)/(3x-10)= (x+1)/(2x)
(3x-10)(x+1)=(2x)(2x-7)
3x²-7x-10= 4x²-14x
x²-7x+10= 0
(x-5)(x-2)
x= 2,5
x= 5
3/2= (3y+1)/(4y)
12y= 6y+2
6y= 2
y= ⅓
5,10,10/3
15>10/3 ✓
10+10/3>5
30+10>15
40> 15 ✓
10/3+5> 10
10+15> 30
25> 30 ❌
Triangle does not exist