Calculate side lengths of the right triangle | Area and Perimeter are known | Geometry Olympiad
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- เผยแพร่เมื่อ 4 ต.ค. 2024
- Learn how to find the side lengths of the right triangle. Area and Perimeter are given as 2646 and 252 respectively. Important Geometry and algebra skills are also explained. Step-by-step tutorial by PreMath.com
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Calculate side lengths of the right triangle | Area and Perimeter are known | Geometry Olympiad
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When I was a kid in school I never would have imagined enjoying math so much. I look forward to these videos, thank you professor!
Glad to hear that!
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You are awesome, Gary. Keep smiling👍
Love and prayers from the USA! 😀
b
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❤
This is high school level math but in the American public education system we learn these subjects like algebra, geometry, trigonometry etc. as separate subjects. We don’t learn to combine these subjects to solve more complex problems like you do in these videos. This puts you at a disadvantage when studying calculus, differential equations, and more advanced mathematics at the university level.
Quite sobering!
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You are awesome, Ayman. Keep it up 👍
Stay blessed 😀
Душнила
The world does it this way with the exception of the USA
I like the fact that you show all the steps involved .So one can see how you work the problem
Mr. Premath,
Thanks for the excellent solution. One thing I notice which would simplify the solution process is to factor out common multipliers. In this example, we have a perimeter and an area. Any factor that is in the perimeter with a corresponding square of that factor in the area can be divided out. In this example after factoring to primes, I find a factor if 21 in the perimeter with 21^2 (441) in the area. You can replace perimeter with 12 and 2646 with 12 and solve it from there. The answers you'll get (b - 3, a - 4, c = 5) can be scaled back up by 21 to get 63, 84 and 105.
The same methods are used otherwise. My point is I find there's an advantage in eliminating large numbers when possible.
Sorry, 2646 / 441 = 6, not 12.
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This triangle is like the 3-4-5 triangle but scaled up by a factor of 21
asq + bsq = csq, a+b+c = 252, ab/2 = 2646, put c = 252 - (a+b) in asq eqn. c = 105. a = 5292/b, put this a + b = 147, Solve for a quadratic
a = 84 or 63. since b less so a = 84, b - 63, & c = 105
Excellent!
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Excellent explanation.
One amazing part i notice in your working is that you are writing down every step.
Whereas many Mathematics Masters skip some steps.
Your teaching method helps students who are dull in Mathematics.
Excellent!
Glad you think so!
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You are awesome, Ramani dear. Keep smiling👍
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@@PreMath 🙏
Sublime, truly sublime !! This is currently beyond my reach, your mental gymnastics is an absolute joy to watch,
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Glad you think so!
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You are awesome, Ian. Keep smiling👍
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Great fun! So much work involved, but incredibly satisfying watching the answer unfold.
Excellent!
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Generalization where A is the area and P is the perimeter:
a = ½ (P - (P² - 4A) / 2P) + √((P - (P² - 4A) / 2P)² - 8A)
b = ½ (P - (P² - 4A) / 2P) - √((P - (P² - 4A) / 2P)² - 8A)
c = (P² - 4A) / 2P
Simplification:
c = (P² - 4A) / 2P
a = ½ [(P - c) + √((P-c)² - 8A)]
b = ½ [(P - c) - √((P-c)² - 8A)]
After getting the side c(105), apply the value in eq.2. We will get a+b = 147 --> eq.4
From eq.3, we can find a-b by subtracting 2ab on both sides.
Then we'll get a-b = 21 --> eq. 5
By adding eq.4 & eq.5
2a = 168
a = 84 units
b = 63 units
Anyway good explanation. Waiting for more videos.
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Legal. Eu respondi em 1 minuto assim:
Como os valores estão com aparência de ser um triângulo retângulo perfeito de lados proporcionais a 3, 4 e 5, eu fiz a seguinte equação para descobrir a proporção:
3x + 4x + 5x = 252
12x = 252
X = 21
Significa que cada lado foi aumentado em 21 vezes, portanto, é um triângulo de lados 63, 84 e 105.
Para confirmar, utilizei a fórmula da área
63 . 84 / 2 = 2646
Até mais
An exciting problem with equally exciting solution
Glad you think so!
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@@PreMath Thank you, Prof. Always looking forward to viewing your problem solving techniques.
Very interesting👍
Thanks for sharing😊
👍 In a lighter mood, we can calculate the radius of the inscribed circle and find out other dimensions.
Great effort. Best wishes.
Thank you! Cheers!
A very convoluted proof for what is a relatively simple problem.
Perimeter P = a+b+c and area A = ab/2, so 2A=ab. But c² = a²+b², so c²+4A = c² + 2ab = a²+b² + 2ab = (a+b)².
So c² + 4A = (a+b+c - c)² = (P-c)² = c² - 2cP + P². So 4A = - 2cP + P². Thus 2cP = P² - 4A & so c = (P² - 4A)/2P.
Now ab = 2A, so b = 2A/a. Thus a = P - c - b = (P-c) - 2A/a. Hence a² = (P-c).a - 2A, so a² - (P-c).a + 2A =0.
Thus a = {(P-c) + √[(P-c)² - 8A]}/2 & b = 2A/b = {(P-c) - √[(P-c)² - 8A]}/2, since b
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Not been looking at your channel for a while - great to be back. Got as far as your equation 3 then ground to a halt... Superb question and solution - thank you!
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Very nice
Thank you
Seeing as the perimeter is a whole number, a, b and c must be in the ratio of a pythagorean triple. 3:4:5 is the obvious first guess (if that hadn't worked, I would have tried 5:12:13).
3+4+5=12. P=252=12*21.
So try a=3*21=63; b=4*21=84, c=5*21=105.
Area = ½*a*b = ½*63*84 = 2646, as required.
There are occasions when just trying something to see if it works is the best, and mathematically accurate method.
Many thanks, I walked the same way.
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I solved this by noting the perimeter is an integer, suggesting it is a scaled up pythagorean triple. Kicking around a few things, noting that both perimeter and area are divisible by 126, and looking up factors of 252, I did some trial and error, 3+4+5 = 12, 21×12=252, then tried the area and yes it works, 21 times 3,4,5. I know you can equate the area against the perimeter and that would be a more solid approach (and I'm sure I'll see you do that as I watch the rest of the video).
I like your calculations.
Show me all your hypothesis.
ur channel is a class for itself
Thanks
Thanks for video.Good luck sir!!!!!!!!!
three unknowns requires three equations:
eq.1 permineter:
a + b + c = 252
eq.2 area:
(1/2)(a×b) = 2646
a×b = 5292
a = 5292/b
eq.3 hypotenuse:
a^2 + b^2 = c^2
substitute --
eq.2 into both eq.1 and eq.3
eq.1:
a + b + c = 252
5292/b + b + c = 252
5292 + b^2 + cb = 252b
b^2 - 252b + cb = -5292 eq.4
eq.3:
a^2 + b^2 = c^2
(5292/b)^2 + b^2 = c^2
Very clear demonstration.
Not going to lie: when I saw Area and Perimeter, I thought for sure this was a Heron's Formula moment. I got stonewalled at:
55,566 = (126 - a)(126 - b)(126 - c)
I knew that was going to create more problems. So I went back and hit up the Pythagorean Theorem and found myself in lock-step with you. Literally the only difference is I did not assume a > b out of the gate and solved it both ways; I am so accustomed to your diagrams not being to scale I concluded any such assumption could be erroneous.
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Most enjoyable. What made you think (at 3.15) to square a+b=252-c? I would never have guessed that step.
Explain well the Pythagorean theorem
I remember replacing my brain cells with top fuel eliminator math last February 10, 2022 when you posted similar perimeter problem and solved quadratic by factoring . Side length B > Side length A is is a little more trickier too solve.😇
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I remembered seeing this refreshing problem in a previous video.
(Disclaimer: Solved on my own after seeing the tnumbnail)
Here's what I did
P∆= 252
A∆= 2646
Find the sides
a+b= 252-c
ab= 5292
(a+b)²= a²+2ab+b²
and c²= a²+b²
By substitution...
(252-c)²= c²+10584
63504-504c+c²=c²+10584
63504-504c=10584
52920= 504c
105= c
a+b+105= 252
a+b= 147
ab= 5292
I used Vieta's Formula and basic monomial factoring for a and b
Vieta's Formula
x²-(a+b)x+(ab)= 0
x²-147x+5292= 0
(x-84)(x-63)
x= 84, 63 (the legs of the right triangle)
Final answer
a= 84, b= 63, c= 105
(Based on the figure, a>b)
Same drill as usual, fast forwarded to the end to double check if it matches which turned out to be.
Cheers from The Philippines 🇵🇭
ab=5292, a^2+b^2=c^2, then a+b=sqroot of c^2+2x5292=c^2+10584, so sqroot of (c^2+10584)+c=252, c^2+10584=(252-c)^2=63504-504c+c^2, 504c=63504-10584=52920, c=52920/504=105, therefore ab=5292, a^2+b^2=11025, a+b=sqroot of 11025+10584=sqroot of 21609=147, a, b are given by (147+sqroot(147^2-4x5292))/2=(147+21)/2=84 and (147-21)/2=63, a,b,c are 84, 63, 105.😊
Could find b by substituting for a in b = 5292/a instead of substituting for both a and c
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Good explanation
The numbers in these problems can get kind of unwieldy. By observation, 252 = 2(126) = 12(21) and 2646 = 2520+126 = 21(126) so let's have u = 21.
a + b + c = 252 = 12u
a + b = 12u - c ---- [1]
ab/2 = 2646 = 126u = 6u²
ab = 12u² ---- [2]
a² + b² = c² ---- [3]
(a+b)² = a² + b² + 2ab
(12u-c)² = c² + 2(12u²)
Is it possible to solve for the sides of a triangle given perimeter, area, and one angle?
The formula for the a and b in terms of A and P is (P^2+4A\pm\sqrt{P^4-24AP^2+16A^2})/(4P).
Perimeter is one dimensions.Area two dimensions.
252=7×3×12=21(12)
2646=7×7×3×3×6=21^2(6)😮
If area is 6 and perimeter=12 then pythogarian triplet is 3,4,5.
So the sides are 21×3,21×4,21×5=63,84,105.
There is a bit of a shortcut. Once c was found, there is a limited number of values for a and b,
I did solve the problem but since it's been so long since I was in high school I had forgotten all but the formulas for area and the Pythagorean theorem. So I factored 5292 into 7x7x3x3x3x2x2. The theorem told me a and b would each have to be smaller than c so easy to spot the only 2 combinations that meet this test 98x 54 and 84 x 63 and obviously only the latter solves the theorem. The problem tells that a>b so there you have it.
Once known a+b (sum) and ab (product) why don't apply
x^2 - (sum)x + (product) = 0 ?
Exactly
The pythagorean triplet 3,4,5, is the best and easy method
Mukundsir
I get to the same place, but I do seem to over-rely on the quadratic formula to arrive there :)
Antes de meterse en cálculos farragosos es conveniente verificar si el triángulo propuesto es de tipo (3-4-5) →→→ 2646/(3x4/2)=441=21² → 21(3+4+5)=252 → Sí, el triángulo propuesto es de tipo 3-4-5 y la razón de semejanza entre ambos es =21 → a=21x4=84 ; b=21x3=63 ; c=21x5=105
Bonito problema. Gracias y un saludo cordial.
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Yay! I solved it.
Excellent!
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Here again is a slightly different approach:
Area = ab/2 = 2646; ab = (2)(2646) = 5292; b = 5292/a. (equation 1)
Perimeter = a + b + √(a^2 + b^2) = 252; rearrange to get the square root term by itself on the right:
a + b - 252 = √(a^2 + b^2); square both sides, multiplying out manually on the left:
a^2 + ab - 252a + ab + b^2 + 252b - 252a - 252b + (252)^2 = a^2 + b^2; collect terms and simplify on the left:
a^2 + b^2 + 2ab - 504(a + b) + (252)^2 = a^2 + b^2; subtract a^2 + b^2 from both sides:
2ab - 504(a + b) + (252)^2 = 0; substitute b = 5292/a from equation (1) above:
2a(5292/a) - 504(a + 5292/a) + 63504 = 0; cancel out a in the first term:
2(5292) - 504a - (504)(5292)/a + 63504 = 0; multiply through by a:
10,584 a - 504a^2 - (504)(5292) + 63504a = 0; rearrange and collect terms:
-504a^2 +74,688a - (504)(5292) = 0; divide everything by -504:
a^2 - (74,688/504)a + 5292 = 0; and finally:
a^2 - 147a + 5292 = 0, the same quadratic equation that you got.
From here I used the quadratic formula. I'll skip the tedious details, but everything fell into place and I came out with the same answers. The two solutions to the quadratic are a and b: either a or b is 63 and the other is 84.The triangle turns out to be an integer Pythagorean, 63-84-105, but I didn't learn them that high.
And now it's time for breakfast.
Cheers. 🤠
It's actually a 3-4-5 triangle. Those lengths are those values multiplied by 21.
@@calspace Right. Didn't realize that until later.
My answer is a = 84, b = 63, and c = 105
My solution:
- Insert a circle into the triangle
- As this triangle's area is equal to the semiperimeter times the inserted circles's radius (r), r turns out to be 21
- Then I drew three lines starting from the center of the circle to the extreme points of the triangle, which divided the triangle into 1 square of side r + two pairs of identical triangles, whose heights are also r and bases a-r and b-r
- As b=5.292/a, the sum of the areas is equal to r² + r(a-r) + r(5.292/a-r). Note that I did not divide the triangles' areas by 2 because they are doubled and identical
- Equating this sum of areas to the given area of 2.646, I arrived at the quadratic equation a² - 147a + 5292 = 0, whose solutions for a are 84 and 63. Accepted solution is 84 to satisfy the condition b
84,,,63,,,105 respectivally.
شكرا لكم
يمكن وضع a=4xوb=3xوc=5xمعx>0
نجدx=21 ثم نبين أن لا حل آخر للمسألة .....
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Shukran!
Is has been a long time since I did Trig Once you have c the hypotenuse of the triangle. I am looking at this as a 60 , 30 , 90 Special right triangle I am seeing a different way of doing this problem Any one care to weigh in on this As I said it has been a while about 45 years Since I did this ..
try to trick with simply factor the sizes by 21
After finding a+b=147 & ab=5292,we should calculate a-b which comes out 21.So without solving for quadratic equation my solution is simpler than yours.
This question was asked in one of our periodic assessment😅
ab=5292=4x27x49, a^2+b^2=c^2, a+b+c=252=4x9x7, ...🤔🤔🤔
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ab. = 5292
c. = 105
a. + b = ( 252 - 105 ). = 147
( a + b ) ^² -- 4ab = ( a -- b ) ^²
( 147 * 147 ) - ( 4 * 5292 ) = 441
( a - b ) ^² = 441
a - b = 21
(A + B ) - ( A - B ) = ( 147 + 21 ) = 168
2A. = 168
A = 84
B = ( 147. - 84 ). = 63
A = 84. B = 63. C = 105
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Thanks for encourages
Simple method
a=84, b=63 and c=105
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why not solve the quadratic of a for its roots, rather than guess!
63,84,105...84,63,105
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I came to another solution : a= 113,106 b= 23,393 c= 115,5
In india
this question is 7th 8th level
А значения 252 и 2646 ты из головы взял. Сама задача не трудная, систему уравнений можно записать зная определение периметра , площадь прямоугольного треугольника и теорему Пифагора, но решение делают громоздким эти значения, пусть я сделаю всё правильно, буду выражать одно через другое, дойду до дискриминанта , а тут возводи трёхзначные числа в квадрат вычитай что-то , а потом ещё гадай можно ли из этого извлечь корень. Такие задачи должны проверять знание геометрии, а не действия над числами
First go to a school and lean how to pronounce English Language. Your Math skills are good but pronunciation is to be improved
252/12 = 21 → a = 4(21) → b = 3(21) → c = 5(21) → ab/2 = 2464
btw: a + b + c = 252 ↔ ab/2 = 2464 → ∆ABC = pyth. triple = 21(3 - 4 - 5)