The length of either MN or BC may be omitted and the problem can still be solved. If BC is omitted, we note that ΔAMN and ΔABC are similar. Therefore, AM/AB = MN/BC and, computing AM, AB and MN from x = 1, we can compute BC. Note, also, that PreMath really should check that the computed value of x satisfies AM/AB = MN/BC. It does, but if it did not, the problem would not be valid. Another way to solve the problem is to solve AM/AB = MN/BC. We get (8x - 7)/(13x - 10) = (2x - 1)/(2x + 1). There are 2 positive solutions, x = 1 and x = 1.7. However, there are other equations which must be satisfied. Let's take PreMath's equation: AM/MB = AN/NC. We find that x = 1 satisfies this equation but x = 1.7 does not. So, x = 1.7 is invalid and x = 1 is used to compute the perimeter. If either MN or BC were omitted, we would still get one solution to the problem. If AM, MB, AN, or NC were omitted (any one of the 4 lengths), there would be two valid solutions.
@ 1:17 one way to prove the ratio is to look at the ratio of congruent triangles ADE and BDE. DE . Line segment DE Is common too both triangles (Bases) and because parallel lines are equidistant everywhere both triangles have the same height and so they have equal area. Similarly the area ratio of triangles CED and CEA have same Base (CE) and same Height. Is also true for triangles DEC and DEB .. same Base (DE) and same Height. ... How much fun can a person have in one day? I don't know, I'm goin back to bed. 😊
Even faster method: AM = AN so: 8X - 7 = 4X - 3, therefore X = 1. Now that you know the value of X, you can have the whole perimeter just by substitution.
∠A ≅ ∠A by the Reflexive Property of Congruence. ∠AMN ≅ ∠B by the Corresponding Angles Theorem. So, △ABC ~ △AMN by AA. Use the Triangle Proportionality Theorem. AM/BM = AN/CN (8x - 7)/(5x - 3) = (4x - 3)/(3x - 1) (8x - 7)(3x - 1) = (5x - 3)(4x - 3) 24x² - 8x - 21x + 7 = 20x² - 15x - 12x + 9 24x² - 29x + 7 = 20x² - 27x + 9 4x² - 2x - 2 = 0 2x² - x - 1 = 0 (ac = -2, [-2, 1]) 2x² - 2x + x - 1 = 0 2x(x - 1) + (x - 1) = 0 (2x + 1)(x - 1) = 0 2x + 1 = 0 or x - 1 = 0 2x = -1 x = 1 x = -1/2 But when x = -1/2, the side lengths are either negative or zero. So, x = 1. Substitute this value in the equations. We can use the Segment Addition Postulate to solve quicker. AB = AM + BM = (8x - 7) + (5x - 3) = 13x - 10 AC = AN + CN = (4x - 3) + (3x - 1) = 7x - 4 AB = 13x - 10 = 13(1) - 10 = 13 - 10 = 3 AC = 7x - 4 = 7(1) - 4 = 7 - 4 = 3 BC = 2x + 1 = 2(1) + 1 = 2 + 1 = 3 P = AB + AC + BC = 3 + 3 + 3 = 9 So, the perimeter of the large triangle is 9 units. (By the way, solving proves that the triangles are equilateral. This doesn't help much in solving, but I thought I'd point it out.)
This task brings us to face one more question which at least I cannot answer: 1. Triangles ABC and AMN are similar, so (at least it should be) BC/MN = AB/AM = AC/AN; 2. some preliminary estimations: AB= 8*x-7+5x-3 = 13x-10; AC=4x-3+3x-1=7x-4; 3. Let us solve for x equations listed in p2: (2x+1)/(2x-1)=(13x-10)/(8x-7) we get 2 real positive roots: x1=1; x2 = 1.7 - both are reasonable - I see no reason to exclude one ((( (2x+1)/(2x-1)=(7x-4)/(4x-3) we get 2 real positive roots: x3=1; x4 = 7/6 - both are reasonable - I see no reason to exclude one ((( 3rd attempt (13*x-10)/(8*x-7)=(7*x-4)/(4*x-3) we get 1 real positive root: x5=1; x6 = -1/2 - Now I know how to exclude x6 ))))) x1=x3=x5=1 is ok but how should we treat x2 and x4 ????
Ok from what i checked x=7/6 is not a solution of the AM/AB=MN/BC therefore x=1 is the only solution.Same thing happens with x=1,7 which is not valid too.
First time I've been interested math in a while. Some of the steps don't make full sense to me anymore after a decade out of school, but I could at least follow along.
No. The triangles ABC and AMN are obviously similar, so we have: BC/MN = AB/AM = AC/AN If only the equation BC/MN = AB/AM is considered, two solutions are obtained: x = 1 and x = 1.7 For x=1 the ratio AC/AN is equal to the ratios BC/MN and AB/AM, therefore x=1 is a valid (and the only valid) solution. For x=1.7 the ratio AC/AN is not equal to the ratios BC/MN and AB/AM anymore, therefore x=1.7 is not a valid solution. Best regards from Germany
Thank you!
You are very welcome! ❤️
Thanks for the feedback ❤️
The length of either MN or BC may be omitted and the problem can still be solved. If BC is omitted, we note that ΔAMN and ΔABC are similar. Therefore, AM/AB = MN/BC and, computing AM, AB and MN from x = 1, we can compute BC. Note, also, that PreMath really should check that the computed value of x satisfies AM/AB = MN/BC. It does, but if it did not, the problem would not be valid.
Another way to solve the problem is to solve AM/AB = MN/BC. We get (8x - 7)/(13x - 10) = (2x - 1)/(2x + 1). There are 2 positive solutions, x = 1 and x = 1.7. However, there are other equations which must be satisfied. Let's take PreMath's equation: AM/MB = AN/NC. We find that x = 1 satisfies this equation but x = 1.7 does not. So, x = 1.7 is invalid and x = 1 is used to compute the perimeter.
If either MN or BC were omitted, we would still get one solution to the problem. If AM, MB, AN, or NC were omitted (any one of the 4 lengths), there would be two valid solutions.
Let's find the perimeter:
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The triangles ABC and AMN are obviously similar, so we can conclude:
BC/MN = AB/AM = AC/AN
BC/MN = (AM + BM)/AM = (AN + CN)/AN
BC/MN = (AN + CN)/AN
(2x + 1)/(2x − 1) = [(4x − 3) + (3x − 1)]/(4x − 3)
(2x + 1)/(2x − 1) = (7x − 4)/(4x − 3)
(2x + 1)*(4x − 3) = (7x − 4)*(2x − 1)
8x² − 6x + 4x − 3 = 14x² − 7x − 8x + 4
0 = 6x² − 13x + 7
0 = 6x² − 6x − 7x + 7
0 = 6x*(x − 1) − 7*(x − 1)
0 = (6x − 7)*(x − 1)
First solution:
6x − 7 = 0 ⇒ x = 7/6
MN = 2x − 1 = 2*(7/6) − 1 = 7/3 − 1 = 7/3 − 3/3 = 4/3
BC = 2x + 1 = 2*(7/6) + 1 = 7/3 + 1 = 7/3 + 3/3 = 10/3
BC/MN = (10/3)/(4/3) = 5/2
AN = 4x − 3 = 4*(7/6) − 3 = 14/3 − 3 = 14/3 − 9/3 = 5/3
AC = 7x − 4 = 7*(7/6) − 4 = 49/6 − 4 = 49/6 − 24/6 = 25/6
AC/AN = (25/6)/(5/3) = 5/2 ✅
AM = 8x − 7 = 8*(7/6) − 7 = 28/3 − 7 = 28/3 − 21/3 = 7/3
AB = 13x − 10 = 13*(7/6) − 10 = 91/6 − 10 = 91/6 − 60/6 = 31/6
AB/AM = (31/6)/(7/3) = 31/14 ⚠
This is not a valid solution.
Second solution:
x − 1 = 0 ⇒ x = 1
MN = 2x − 1 = 2*1 − 1 = 2 − 1 = 1
BC = 2x + 1 = 2*1 + 1 = 2 + 1 = 3
BC/MN = 3
AN = 4x − 3 = 4*1 − 3 = 4 − 3 = 1
AC = 7x − 4 = 7*1 − 4 = 7 − 4 = 3
AC/AN = 3 ✅
AM = 8x − 7 = 8*1 − 7 = 8 − 7 = 1
AB = 13x − 10 = 13*1 − 10 = 13 − 10 = 3
AB/AM = 3 ✅
This is a valid solution. Therefore the perimeter of the triangle ABC turns out to be:
P = AB + AC + BC = 3 + 3 + 3 = 9
Best regards from Germany
Solved it by similarity of the triangles.
@ 1:17 one way to prove the ratio is to look at the ratio of congruent triangles ADE and BDE. DE . Line segment DE Is common too both triangles (Bases) and because parallel lines are equidistant everywhere both triangles have the same height and so they have equal area. Similarly the area ratio of triangles CED and CEA have same Base (CE) and same Height. Is also true for triangles DEC and DEB .. same Base (DE) and same Height. ... How much fun can a person have in one day? I don't know, I'm goin back to bed. 😊
(8x-7)(3x-1) = (5x-3)(4x-3)
24x^2 - 29x + 7 = 20x^2 - 27x + 9
4x^2 - 2x - 2 = 0
2x^2 - x - 1 = 0
(1+or-sqrt(1 - 4*2*-1))/4 = x
(1+sqrt(9))/4
or
(1-sqrt(9))/4
x = -2 gives side lengths
My way of solution ▶
[MN] // [BC]
∠AMN= ∠ABC
∠MAN= ∠BAC
∠ANM= ∠ACB
⇒
ΔAMN ~ ΔABC
Both triangles are similar !
If we write the proportion of the side lengths to each other :
[AM]/[AB]= [MN]/[MC]= [AN]/[AC]
⇒
(8x-7)/(13x-10)= (2x-1)/(2x+1)= (4x-3)/(7x-4)
By taking the first two equations above, we can write:
(8x-7)*(2x+1)= (13x-10)*(2x-1)
16x²+8x-14x-7= 26x²-13x-20x+10
10x²-27x+17=0
Δ= 27²-4*10*17
Δ= 49
√Δ= 7
x₁= (27-7)/2*10
x₁= 20/20
x₁= 1
x₂= (27+7)/20
x₂= 34/20
x₂= 17/10
for x₁= 1
[AM]= 1
[AB]= 3
[MN]= 1
[BC]= 3
[AN]= 1
[AC]= 3
⇒
1/3= 1/3= 1/3 ✅
x₂= 17/10
[AM]= 33/5
[AB]= 121/10
[MN]= 12/5
[BC]= 22/5
[AN]= 19/5
[AC]= 79/10
⇒
(33/5)/(121/10)= 66/121= 6/11
(12/5)/(22/5)= 6/11
(19/5)/(79/10)= 38/79
⇒
6/11 = 6/11 ≠ 38/79 ❌
⇒
x= 1
P(ΔABC)= [AB] + [BC] + [CA]
[AB]= 3
[BC]= 3
[CA]= 3
⇒
P(ΔABC)= 3+3+3
P(ΔABC)= 9 length units
ABC perimeter=3+3+3=9 units.❤
Even faster method: AM = AN so: 8X - 7 = 4X - 3, therefore X = 1. Now that you know the value of X, you can have the whole perimeter just by substitution.
No.
@@pedroteran5885 Yes!
Very nice & concise!
It's was easy
∠A ≅ ∠A by the Reflexive Property of Congruence.
∠AMN ≅ ∠B by the Corresponding Angles Theorem.
So, △ABC ~ △AMN by AA. Use the Triangle Proportionality Theorem.
AM/BM = AN/CN
(8x - 7)/(5x - 3) = (4x - 3)/(3x - 1)
(8x - 7)(3x - 1) = (5x - 3)(4x - 3)
24x² - 8x - 21x + 7 = 20x² - 15x - 12x + 9
24x² - 29x + 7 = 20x² - 27x + 9
4x² - 2x - 2 = 0
2x² - x - 1 = 0 (ac = -2, [-2, 1])
2x² - 2x + x - 1 = 0
2x(x - 1) + (x - 1) = 0
(2x + 1)(x - 1) = 0
2x + 1 = 0 or x - 1 = 0
2x = -1 x = 1
x = -1/2
But when x = -1/2, the side lengths are either negative or zero. So, x = 1. Substitute this value in the equations. We can use the Segment Addition Postulate to solve quicker.
AB = AM + BM = (8x - 7) + (5x - 3) = 13x - 10
AC = AN + CN = (4x - 3) + (3x - 1) = 7x - 4
AB = 13x - 10 = 13(1) - 10 = 13 - 10 = 3
AC = 7x - 4 = 7(1) - 4 = 7 - 4 = 3
BC = 2x + 1 = 2(1) + 1 = 2 + 1 = 3
P = AB + AC + BC
= 3 + 3 + 3
= 9
So, the perimeter of the large triangle is 9 units.
(By the way, solving proves that the triangles are equilateral. This doesn't help much in solving, but I thought I'd point it out.)
Bom dia Mestre
Grato pela aula
perimeter=1+1+2+2+3=9 unit.
Calculating side lengths reveals that the triangles are equilateral.
(8x-7)(3x-1)=(4x-3)(5x-3)
24x^2-29x+7=20x^2-27x+9.
4x^2-2x-2=0 2x^2-x-1=0
x=2
(8x-7)/5x-3=(4x-3)/(3x-1)=>24x^2-29x+7=20x^2-27x+9=0=>4x^2-2x-2=0=>2x^2-x-1=0=>2x^2-2x+x-1=0=>(x-1)(2x+1)=0=>x=1.perimeter=9 unit.ans
P=22x-13=44-13=31
This task brings us to face one more question which at least I cannot answer:
1. Triangles ABC and AMN are similar, so (at least it should be) BC/MN = AB/AM = AC/AN;
2. some preliminary estimations: AB= 8*x-7+5x-3 = 13x-10; AC=4x-3+3x-1=7x-4;
3. Let us solve for x equations listed in p2:
(2x+1)/(2x-1)=(13x-10)/(8x-7)
we get 2 real positive roots:
x1=1; x2 = 1.7 - both are reasonable - I see no reason to exclude one (((
(2x+1)/(2x-1)=(7x-4)/(4x-3)
we get 2 real positive roots:
x3=1; x4 = 7/6 - both are reasonable - I see no reason to exclude one (((
3rd attempt
(13*x-10)/(8*x-7)=(7*x-4)/(4*x-3)
we get 1 real positive root:
x5=1; x6 = -1/2 - Now I know how to exclude x6 )))))
x1=x3=x5=1 is ok but how should we treat x2 and x4 ????
; x4 = 7/6 when you substitute it back to check proportionalty, it fails
x = 7/6
MN = 2x − 1 = 2*(7/6) − 1 = 7/3 − 1 = 7/3 − 3/3 = 4/3
BC = 2x + 1 = 2*(7/6) + 1 = 7/3 + 1 = 7/3 + 3/3 = 10/3
BC/MN = (10/3)/(4/3) = 5/2
AN = 4x − 3 = 4*(7/6) − 3 = 14/3 − 3 = 14/3 − 9/3 = 5/3
AC = 7x − 4 = 7*(7/6) − 4 = 49/6 − 4 = 49/6 − 24/6 = 25/6
AC/AN = (25/6)/(5/3) = 5/2
AM = 8x − 7 = 8*(7/6) − 7 = 28/3 − 7 = 28/3 − 21/3 = 7/3
AB = 13x − 10 = 13*(7/6) − 10 = 91/6 − 10 = 91/6 − 60/6 = 31/6
AB/AM = (31/6)/(7/3) = 31/14
@@Abby-hi4sf Thanx!
Shouldn't x=7/6 be a solution too?
Ok from what i checked x=7/6 is not a solution of the AM/AB=MN/BC therefore x=1 is the only solution.Same thing happens with x=1,7 which is not valid too.
It could be also proved that the triangle is equilateral
Класс
First time I've been interested math in a while. Some of the steps don't make full sense to me anymore after a decade out of school, but I could at least follow along.
by Thales theorem.
I used another method in half the time! Right answer. 9
Too easy!
Very noninteresting 😂😅😢
Я вот не понял, в чем сложность? Мой ребенок 9 лет тоже не оценил сложности
X can't be 1.7 ???????
No. The triangles ABC and AMN are obviously similar, so we have:
BC/MN = AB/AM = AC/AN
If only the equation BC/MN = AB/AM is considered, two solutions are obtained:
x = 1 and x = 1.7
For x=1 the ratio AC/AN is equal to the ratios BC/MN and AB/AM, therefore x=1 is a valid (and the only valid) solution. For x=1.7 the ratio AC/AN is not equal to the ratios BC/MN and AB/AM anymore, therefore x=1.7 is not a valid solution.
Best regards from Germany
Thanks I love your videos