Thanks for the interesting video. I spent several time to examine pythagorean triples, making up a table systematically, until I noticed a scheme that you can take the 53, subtract 1, divide by 2 to find n=26, add 1 to find m=27, then generate the missing values by calculating 2mn and m²+n². But your solution is obviously faster and better, just one has to get the right idea 💡 Thanks for sharing 👍
Here perpendicular = 53 units (an odd integer) so base = 53^2/2 - 0.5 = 1404 and (use n^2 / 2 - 0.5) hypotenuse = 53^2 / 2 + 0.5 = 1405 (use n^2 / 2 + 0.5) Perimeter = p + b + h = 2862 units Area = 1/2 p x b = 37206 sq units.
Let a=53 ; BC=b ; AC=C So c^2=53^2+b^2 c^2-b^2=2809 (c-b)(c+b)=2809(1) So c-b=1 c+b=2809 So 2c=2810 c=1405 b=2809-1405=1404 Perimeter=1405+1405+53=2862 units Area=1/2((35)(1404)=37206 square units.❤❤❤
This task is undefined, it has infinite solutions. The solution you accepted 2809*1 can also be taken as 1404.5*2, or 702.25*4 and so on. Your solution can be considered correct, but it is not the only solution.
A questao é clara; PERTENCE a Z+( inteiro nao negativo) , e, tb sera' Z*+( inteiro positivo) por nao existir segmento de reta valor zero para lado de POLI'GONO ,ou seja ; e' N *( natural diferente de zero ),e, nunca existirá dois números naturais ,cuja soma ou diferença resulte decimal como 1404,5 : que pertencente ao conjunto dos RACIONAIS Q. todo número formado pelo produto de números primos como 53x53 :os divisores serao (1, ele mesmo e os primos).assim nao precisará arrancar todo o cabelo do subaco procurando outros produtos...
^=read as to the power h^2 - b^2=(53)^2 (h+p)(h-p)=2809 (h+p)(h-p)=2809×1 So, h+p=2809....eqn1 h-p=1........ Eqn2 Adding both the equations, we shall get the following h+p+h-p=2809+1=2810 2h=2810 h=1405 P=2810-1405=1404 Perimeter = 1405+1404+53=2862 Area =1/2(1404×53) =702×53 =37206 Numerous answers are there
This is very easy. If the smallest side of a Right Angle Traingle is given and is an odd number and other sides are integers, there is a standard way to find them. 1. Square the smallest side. 2. Subtract 1 from that number. 3. Divide it by 2. It wil be the other side. 4. Add one to that. It will be the hypotuneus. So for above 53 sqr = 2809 So other side will be 1404 And Hypotuneous will be 1405 All Pythagorus Triplet in sequence follow this 3 4 5 5 12 13 7 24 25 9 40 41 And so on
Hello, how are you I need your help. I have a problem. Triangle ABC, AM is a median, BH divide AC in proportion 1 to 2 (AH = 1, HC = 2). P is the intersection to AM with BH, what is the area APH?, thank you.
From point M draw a parallel to BH. Its point of intersection with AC is denoted by D. Since BM = MC, it follows that HD = DC (and also AH = HD = DC). The area of the triangle APH is four times smaller than that of the triangle AMD (its base and height are twice smaller). The area of triangle AMC is half the area of triangle ABC. If the area of the triangle DMC is subtracted from the area of the triangle AMC, we get the area of the triangle AMD. This area, divided by 4, will give us the area of the triangle APH. But what is the height of the triangle DMC ? It can easily seen that it is one third of the height of the triangle ABC, and this because AH = HD = DC, as I have shown. Thus: Area DMC = 1/6 area ABC. Area AMC = 1/2 area ABC. Area AMD = 1/3 area ABC. Area APH = 1/12 area ABC.
Let's do some math: . .. ... .... ..... The triangle ABC is a right triangle. Therefore we can apply the Pythagorean theorem: AB² + BC² = AC² 53² + BC² = AC² 53² = AC² − BC² 53² = (AC + BC)(AC − BC) The unknown side lengths AC and BC should be positive integers. Since 53 is a prime number, we obtain only one solution: AC + BC = 53² = 2809 AC − BC = 1 ⇒ AC = (2809 + 1)/2 = 1405 ∧ BC = (2809 − 1)/2 = 1404 Now we are able to calculate the perimeter P and the area A of this triangle: P = AB + BC + AC = 53 + 1404 + 1405 = 2862 A = (1/2)*AB*BC = (1/2)*53*1404 = 37206 Best regards from Germany
Use number theory 😉 As the side lengths are known to be positive integers, this implies a Pythagorean triple. So we can use the parametrization: b = 53 = m² - n² a = 2mn c = m² + n² Using difference of squares: (m - n)(m + n) = 53. As 53 is a prime, the only factorization in positive integers is 53 ∙ 1, so we can solve the system: m - n = 1 m + n = 53 Adding the equations: 2m = 54 and finally m = 27 and (by substitution into one of the equations above) n = 26. So we get: a = 1404 and c = 1405. Finally, Area = 37206 and Perimeter = 2862.
Using the difference of 2 squares formula, the only way for 53^2 to equal ((c-b)(c+b) is for ((c-b) = 1, and (c+b)^2 =53^2 or 2809. Therefore, c-b=1, c=b+1. Substitute back in: ((b+1)^2 =b^2 + 2809. Solving that, b=1404. C=b+1=1405. The rest is simple.
53^2+b^2=c^2, 53=(c-b)(c+b), c=b +1, 2b+1=53^2, b=(53^2-1)/2=1404, c=1405, so 53^3+1404^2=1405^2, the area is 1/2×53×1404=37206, the perimeter is 53+1404+1405=2862.🎉
I'm having trouble making sense of this. Given your solution, can't you just multiply the length BC (to 2808 or higher) and let the hypotenuse reach that new target C at a shallower angle? Or is it the fact that all lengths must be integers that constrains the solutions of a^2 + b^2 = c^2?
From school, I Have learnd that the opposite line from the triangel the opposite letter must be. I am from the Netherlands. So excuse me for mij bad English. Is it possible that you give some more simple problems? I am 75 years old.
The way you describe helps teach the concept and memorize the equation of the Pythagorean Theorum. It's how I still draw and teach the math, but it isn't really considered a standard or rule.
Let m be the smallest side in a right angle triangle,by triplets rule the ratio are ,m: ,m²+1/2:m²-1/2 so by that 53²+1 /2:53²-1 /2= 2809+1/2:2809-1/2= 1405:1404
There is another short-cut method. In a right angle triangle, if one of the sides of right angle is an odd number say, a, the diagonal and the other side are respectively (a²+1)/2 and (a²-1)/2. Because [(a²+1)/2]² - [(a²-1)/2]² = a². In this case, the sides are 1405 and 1404. The rest follows.
Very interesting problem! Since the triangle represents a Pythagorean triple, it may be possible to use the Pythagorean triple formula: When c is the hypotenuse of a right triangle, c=p^2+q^2, b=p^2-q^2 and a=2pq (p, q ∈ ℤ, p>q). Since 53 is not even, 53=p^2-q^2. In this case p=27, q=26.
This is the crisis I face... the distinction of numbers like primes integers rational numbers ect. on the one hand and lengths and proportions whether rational or not on the other hand. Up here in the Appalachians we call that "Some Strange A$$ Schiff"! 🙂
The set Z is the set of integers, so rational numbers (the set R) are excluded. However, he does tend to painstakingly explain the easy bits and skip over the less obvious ones, so you're probably not the only person wondering about this.
@@jhenshaw102 Yes - the thumbnail that you see before selecting the video misses this vital piece of information. But once the video starts, it shows an extra line: side length ∈ ℤ⁺ People who are unfamiliar with Set Theory, won't know what this means. People who are familiar with Set Theory will find the rest of the problem too simple to bother with. Even with this vital piece of information, there are still an infinite number of solutions because "side length" is singular, but let's just ignore that. There are much better maths problem presenters on RUclips - have a look at those by Presh Talwalkar.
Yes, that was my immediate thought too when I saw the thumbnail, which is why I clicked on it. But once the video starts, an extra vital line appears: side length ∈ ℤ⁺ Anyone unfamiliar with Set Theory won't realise that this means that the sides are positive whole numbers. Anyone who is familiar with Set Theory, will find the rest of the problem too simple to bother with. (Strictly speaking, that's not correct, because "side length" needs to be plural, otherwise there are still an infinite number of solutions, but we'll ignore that error.) There are better maths problem presenters on RUclips - have a look, for example, at some of those by Presh Talwalkar.
Thanks for the interesting video.
I spent several time to examine pythagorean triples, making up a table systematically, until I noticed a scheme that you can take the 53, subtract 1, divide by 2 to find n=26, add 1 to find m=27, then generate the missing values by calculating 2mn and m²+n².
But your solution is obviously faster and better, just one has to get the right idea 💡
Thanks for sharing 👍
Excellent!
You are very welcome!
Thanks for the feedback ❤️
Here perpendicular = 53 units (an odd integer)
so base = 53^2/2 - 0.5 = 1404 and (use n^2 / 2 - 0.5)
hypotenuse = 53^2 / 2 + 0.5 = 1405 (use n^2 / 2 + 0.5)
Perimeter = p + b + h = 2862 units
Area = 1/2 p x b = 37206 sq units.
Excellent! Keep rocking 😀
Thanks for sharing ❤️
@@PreMath Thank you so much.
Let a=53 ; BC=b ; AC=C
So c^2=53^2+b^2
c^2-b^2=2809
(c-b)(c+b)=2809(1)
So c-b=1
c+b=2809
So 2c=2810
c=1405
b=2809-1405=1404
Perimeter=1405+1405+53=2862 units
Area=1/2((35)(1404)=37206 square units.❤❤❤
Excellent!
Thanks for sharing ❤️
BC=x AC=y
53²+x²=y² y²-x²=2809 (y+x)(y-x)=2809
y+x=2809 y-x=1 2x=2808 x=1404 y=1405
Perimeter = 53+1404+1405 = 2862 Area = 53*1404*1/2 = 37206
Excellent!
Thanks for sharing ❤️
P= a+b+c
P= a+c+53
c²= a²+b²
c²=a²+(53)²
c²-a²= 2809
(c+a)(c-a)= (2809)(1)
c+a= 2809
c-a= 1
=========
2c= 2810
c= 1405
a= 1404
P= 1405+1403+53
P= 2861 units
A= (702)(53)
A= 37206 units²
This task is undefined, it has infinite solutions. The solution you accepted 2809*1 can also be taken as 1404.5*2, or 702.25*4 and so on.
Your solution can be considered correct, but it is not the only solution.
A questao é clara; PERTENCE a Z+( inteiro nao negativo) , e, tb sera' Z*+( inteiro positivo) por nao existir segmento de reta valor zero para lado de POLI'GONO ,ou seja ; e' N *( natural diferente de zero ),e, nunca existirá dois números naturais ,cuja soma ou diferença resulte decimal como 1404,5 : que pertencente ao conjunto dos RACIONAIS Q. todo número formado pelo produto de números primos como 53x53 :os divisores serao (1, ele mesmo e os primos).assim nao precisará arrancar todo o cabelo do subaco procurando outros produtos...
Rethink it. The problem set up specifies the sides must be positive integers. Only one solution to the problem with the constraints given.
^=read as to the power
h^2 - b^2=(53)^2
(h+p)(h-p)=2809
(h+p)(h-p)=2809×1
So,
h+p=2809....eqn1
h-p=1........ Eqn2
Adding both the equations, we shall get the following
h+p+h-p=2809+1=2810
2h=2810
h=1405
P=2810-1405=1404
Perimeter = 1405+1404+53=2862
Area =1/2(1404×53)
=702×53
=37206
Numerous answers are there
This is very easy. If the smallest side of a Right Angle Traingle is given and is an odd number and other sides are integers, there is a standard way to find them.
1. Square the smallest side.
2. Subtract 1 from that number.
3. Divide it by 2. It wil be the other side.
4. Add one to that. It will be the hypotuneus.
So for above
53 sqr = 2809
So other side will be 1404
And Hypotuneous will be 1405
All Pythagorus Triplet in sequence follow this
3 4 5
5 12 13
7 24 25
9 40 41
And so on
Hello, how are you I need your help. I have a problem. Triangle ABC, AM is a median, BH divide AC in proportion 1 to 2 (AH = 1, HC = 2). P is the intersection to AM with BH, what is the area APH?, thank you.
From point M draw a parallel to BH. Its point of intersection with AC is denoted by D. Since BM = MC, it follows that HD = DC (and also AH = HD = DC).
The area of the triangle APH is four times smaller than that of the triangle AMD (its base and height are twice smaller). The area of triangle AMC is half the area of triangle ABC. If the area of the triangle DMC is subtracted from the area of the triangle AMC, we get the area of the triangle AMD. This area, divided by 4, will give us the area of the triangle APH.
But what is the height of the triangle DMC ? It can easily seen that it is one third of the height of the triangle ABC, and this because AH = HD = DC, as I have shown. Thus:
Area DMC = 1/6 area ABC.
Area AMC = 1/2 area ABC.
Area AMD = 1/3 area ABC.
Area APH = 1/12 area ABC.
Inconceivable !
We can prove 3-4-5 right angle triangle in this way. Thanks a lot.
Very nice and useful
Thanks Sir for understanding .
You are very very good
with my respects
❤❤❤❤❤
Let's do some math:
.
..
...
....
.....
The triangle ABC is a right triangle. Therefore we can apply the Pythagorean theorem:
AB² + BC² = AC²
53² + BC² = AC²
53² = AC² − BC²
53² = (AC + BC)(AC − BC)
The unknown side lengths AC and BC should be positive integers. Since 53 is a prime number, we obtain only one solution:
AC + BC = 53² = 2809
AC − BC = 1
⇒ AC = (2809 + 1)/2 = 1405
∧ BC = (2809 − 1)/2 = 1404
Now we are able to calculate the perimeter P and the area A of this triangle:
P = AB + BC + AC = 53 + 1404 + 1405 = 2862
A = (1/2)*AB*BC = (1/2)*53*1404 = 37206
Best regards from Germany
Excellent!
Thanks for sharing ❤️
Use number theory 😉 As the side lengths are known to be positive integers, this implies a Pythagorean triple. So we can use the parametrization:
b = 53 = m² - n²
a = 2mn
c = m² + n²
Using difference of squares:
(m - n)(m + n) = 53.
As 53 is a prime, the only factorization in positive integers is 53 ∙ 1, so we can solve the system:
m - n = 1
m + n = 53
Adding the equations:
2m = 54
and finally
m = 27 and (by substitution into one of the equations above) n = 26.
So we get:
a = 1404 and
c = 1405.
Finally, Area = 37206 and Perimeter = 2862.
Thanks for sharing ❤️
φ = 30° → sin(3φ) = 1; ∆ ABC → sin(BCA) = 1; BC = 53 = k → AB = (k^2 + 1)/2 = 1405 → AC = (k^2 - 1)/2 = 1404
Excellent!
Thanks for sharing ❤️
Nice!
Using the difference of 2 squares formula, the only way for 53^2 to equal ((c-b)(c+b) is for ((c-b) = 1, and (c+b)^2 =53^2 or 2809. Therefore, c-b=1, c=b+1. Substitute back in: ((b+1)^2 =b^2 + 2809. Solving that, b=1404. C=b+1=1405. The rest is simple.
53^2+b^2=c^2, 53=(c-b)(c+b), c=b +1, 2b+1=53^2, b=(53^2-1)/2=1404, c=1405, so 53^3+1404^2=1405^2, the area is 1/2×53×1404=37206, the perimeter is 53+1404+1405=2862.🎉
✨Magic!✨
Thanks for the feedback ❤️
I'm having trouble making sense of this. Given your solution, can't you just multiply the length BC (to 2808 or higher) and let the hypotenuse reach that new target C at a shallower angle? Or is it the fact that all lengths must be integers that constrains the solutions of a^2 + b^2 = c^2?
Very nice. Did not thought about primes, giving the way to go.
From school, I Have learnd that the opposite line from the triangel the opposite letter must be.
I am from the Netherlands. So excuse me for mij bad English.
Is it possible that you give some more simple problems? I am 75 years old.
The way you describe helps teach the concept and memorize the equation of the Pythagorean Theorum. It's how I still draw and teach the math, but it isn't really considered a standard or rule.
The value of c should have been substituted in c - a = 1 to get a = 1404 rather than in the second equation.
Good very nice sharing sir❤❤
Let m be the smallest side in a right angle triangle,by triplets rule the ratio are ,m: ,m²+1/2:m²-1/2 so by that 53²+1 /2:53²-1 /2= 2809+1/2:2809-1/2= 1405:1404
There is another short-cut method. In a right angle triangle, if one of the sides of right angle is an odd number say, a, the diagonal and the other side are respectively (a²+1)/2 and (a²-1)/2. Because [(a²+1)/2]² - [(a²-1)/2]² = a². In this case, the sides are 1405 and 1404. The rest follows.
Very smart! I would just use a=c-1 to find a faster…
... it is same for any b prime number>2.
There is always only 1 solution.
STEP-BY-STEP RESOLUTION PROPOSAL :
01) AB^2 = b^2 = 53^2 = 2.809
02) BC = a ; AC = c
03) c^2 = a^2 + b^2 ; b^2 = c^2 - a^2 ; 2.809 = (c - a) * (c + a)
04) 2.809 / 2 = 1.404.5
05) c = 1.405
06) a = 1.404
07) 1.405^2 = 53^2 + 1.404^2 ; 1.974.025 = 2.809 + 1.971.216 ; 1.974.025 = 1.974.025 ; Check!
08) Perimeter = 1.404 + 1.404 + 53 = 2.862 lin un
09) Area = (53 * 1.404) / 2 = 37.206 sq un
Thus,
OUR ANSWER IS :
Perimeter equal to 2.862 Linear Units
Area equal to 37.206 Square Units
It is only one solution because we can extend BC to any length and accordingly AC
Reread the constraints given. We're looking for the case where a and c are positive integers. Only one solution.
Simple.
This seems to show that any odd integer could be part of a Pythagorean triple.
Side length applies to all sides
You don’t ask “Which side was he talking about?🤦♂️”
A silly question I think 😂
This is more of a number theory problem than geometry
Thanks for the feedback ❤️
Very interesting problem! Since the triangle represents a Pythagorean triple, it may be possible to use the Pythagorean triple formula:
When c is the hypotenuse of a right triangle, c=p^2+q^2, b=p^2-q^2 and a=2pq (p, q ∈ ℤ, p>q). Since 53 is not even, 53=p^2-q^2. In this case p=27, q=26.
This is the crisis I face... the distinction of numbers like primes integers rational numbers ect. on the one hand and lengths and proportions whether rational or not on the other hand. Up here in the Appalachians we call that "Some Strange A$$ Schiff"! 🙂
😀
Thanks for the feedback ❤️
The solution is unreal. Why not 28, 45, 53. That makes sense.
53 is not the hypotenuse
Это уже не треугольник, а какая-то иголка получается...
इस प्रश्न के बहुत से उत्तर हो सकते हैं। अतः यह प्रश्न ही ग़लत है।
This is impossible the triangles absolutely need minimum of three given parameters. This is the way it is.
It's not 3, but 2. In this case the restriction on the sides to be integers counts for the second.
I added the his when comm nts were 53😂
the sides of the triangle are irrational - ( 1405 , 1404 , 53 ) there is a big gap between ( 53 ) and (1405 or 1404). this unacceptable.
Thanks for the feedback ❤️
Not soluble. BC can be any length.
nope, this is not valid. there are an infinite number of values for a and c. these are the only whole number of values which is not in the problem.
The set Z is the set of integers, so rational numbers (the set R) are excluded. However, he does tend to painstakingly explain the easy bits and skip over the less obvious ones, so you're probably not the only person wondering about this.
That was my first thought looking at this, we need one more piece of information - an angle, a side, or "assuming all side lengths are integers"
Would be very different if all angles are integers.
@@jhenshaw102 Yes - the thumbnail that you see before selecting the video misses this vital piece of information. But once the video starts, it shows an extra line:
side length ∈ ℤ⁺
People who are unfamiliar with Set Theory, won't know what this means. People who are familiar with Set Theory will find the rest of the problem too simple to bother with.
Even with this vital piece of information, there are still an infinite number of solutions because "side length" is singular, but let's just ignore that. There are much better maths problem presenters on RUclips - have a look at those by Presh Talwalkar.
Yes, that was my immediate thought too when I saw the thumbnail, which is why I clicked on it. But once the video starts, an extra vital line appears:
side length ∈ ℤ⁺
Anyone unfamiliar with Set Theory won't realise that this means that the sides are positive whole numbers. Anyone who is familiar with Set Theory, will find the rest of the problem too simple to bother with. (Strictly speaking, that's not correct, because "side length" needs to be plural, otherwise there are still an infinite number of solutions, but we'll ignore that error.)
There are better maths problem presenters on RUclips - have a look, for example, at some of those by Presh Talwalkar.
ไม่ใช่คำตอบนี้ครับเพราะตัวเลขไม่สัมพันธ์กันครับ