Calculate side lengths of the right triangle | Area and Perimeter are known

Calculate side lengths of the right triangle | Area and Perimeter are known | Geometry Olympiad

Рет қаралды 62,245

Жыл бұрын

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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Пікірлер: 105

@garypaulson5202 Жыл бұрын

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!

@PreMath Жыл бұрын

Glad to hear that! Thanks for your feedback! Cheers! You are awesome, Gary. Keep smiling👍 Love and prayers from the USA! 😀

@himo3485 Жыл бұрын

@PreMath Жыл бұрын

Thanks for sharing! Cheers! You are awesome. Keep it up 👍 Love and prayers from the USA! 😀

@franklai4203 Жыл бұрын

❤

@johnplong3644 Жыл бұрын

I like the fact that you show all the steps involved .So one can see how you work the problem

@cameronspalding9792 Жыл бұрын

This triangle is like the 3-4-5 triangle but scaled up by a factor of 21

@aymanabdellatief1572 Жыл бұрын

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.

@PreMath Жыл бұрын

Quite sobering! Thanks for your feedback! Cheers! You are awesome, Ayman. Keep it up 👍 Stay blessed 😀

@mega_mango Жыл бұрын

Душнила

@nikoloijames-jarrett6613 4 ай бұрын

The world does it this way with the exception of the USA

@ASChambers Жыл бұрын

Great fun! So much work involved, but incredibly satisfying watching the answer unfold.

@PreMath Жыл бұрын

Excellent! Thanks for your feedback! Cheers! You are awesome. Keep smiling👍 Love and prayers from the USA! 😀

@theoyanto Жыл бұрын

Sublime, truly sublime !! This is currently beyond my reach, your mental gymnastics is an absolute joy to watch,

@PreMath Жыл бұрын

Excellent! Glad you think so! Thanks for your continued love and support! You are awesome, Ian. Keep smiling👍 Love and prayers from the USA! 😀

@stephencindrich6787 Жыл бұрын

A very convoluted proof for what is a relatively simple problem.

@davidfromstow Жыл бұрын

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!

@PreMath Жыл бұрын

Welcome back! Thanks for your continued love and support! You are awesome, David. Keep smiling👍 Love and prayers from the USA! 😀

@timk8079 Жыл бұрын

Very clear demonstration.

@HappyFamilyOnline Жыл бұрын

Very interesting👍 Thanks for sharing😊

@alexdelpiero8693 Жыл бұрын

Very nice Thank you

@KAvi_YA666 Жыл бұрын

Thanks for video.Good luck sir!!!!!!!!!

@dabbirusrinivasarao3758 Жыл бұрын

Good explanation

@maryrosepulma6813 Жыл бұрын

I like your calculations. Show me all your hypothesis.

@ybodoN Жыл бұрын

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

@marcovargasglobant7923 Жыл бұрын

Simplification: c = (P² - 4A) / 2P a = ½ [(P - c) + √((P-c)² - 8A)] b = ½ [(P - c) - √((P-c)² - 8A)]

@vidyadharjoshi5714 Жыл бұрын

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

@PreMath Жыл бұрын

Excellent! Thanks for sharing! Cheers! You are awesome. Keep it up 👍 Love and prayers from the USA! 😀

@richardslater677 Жыл бұрын

Most enjoyable. What made you think (at 3.15) to square a+b=252-c? I would never have guessed that step.

@allanflippin2453 Жыл бұрын

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.

@allanflippin2453 Жыл бұрын

Sorry, 2646 / 441 = 6, not 12.

@PreMath Жыл бұрын

Thanks for your feedback! Cheers! You are awesome. Keep it up 👍 Love and prayers from the USA! 😀

@ramanivenkata3161 Жыл бұрын

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.

@PreMath Жыл бұрын

Excellent! Glad you think so! Thanks for your continued love and support! You are awesome, Ramani dear. Keep smiling👍 Love and prayers from the USA! 😀

@ramanivenkata3161 Жыл бұрын

@@PreMath 🙏

@ajbonmg Жыл бұрын

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.

@murdock5537 Жыл бұрын

Many thanks, I walked the same way.

@PreMath Жыл бұрын

Thanks for your feedback! Cheers! You are awesome. Keep it up 👍 Love and prayers from the USA! 😀

@sudhirjoshi7782 Жыл бұрын

👍 In a lighter mood, we can calculate the radius of the inscribed circle and find out other dimensions. Great effort. Best wishes.

@PreMath Жыл бұрын

Thank you! Cheers!

@Ramkabharosa Жыл бұрын

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

@PreMath Жыл бұрын

Excellent! Thanks for sharing! Cheers! You are awesome. Keep it up 👍 Love and prayers from the USA! 😀

@Tornado.363 Жыл бұрын

ur channel is a class for itself

@PreMath Жыл бұрын

Thanks

@bertblankenstein3738 Жыл бұрын

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).

@asielelsaben3587 Жыл бұрын

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.

@PreMath Жыл бұрын

Thanks for your feedback! Cheers! You are awesome. Keep it up 👍 Love and prayers from the USA! 😀

@tomtke7351 Жыл бұрын

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

@QuestoesdeMatematica_ Жыл бұрын

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

@vara1499 Жыл бұрын

An exciting problem with equally exciting solution

@PreMath Жыл бұрын

Glad you think so! Thanks for your continued love and support! You are awesome, Varadarajan. Keep smiling👍 Love and prayers from the USA! 😀 Stay blessed 😀

@vara1499 Жыл бұрын

@@PreMath Thank you, Prof. Always looking forward to viewing your problem solving techniques.

@madmad5605 5 ай бұрын

Is it possible to solve for the sides of a triangle given perimeter, area, and one angle?

@MrPaulc222 Жыл бұрын

I get to the same place, but I do seem to over-rely on the quadratic formula to arrive there :)

@alster724 Жыл бұрын

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 🇵🇭

@misterenter-iz7rz Жыл бұрын

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.😊

@Copernicusfreud Жыл бұрын

Yay! I solved it.

@PreMath Жыл бұрын

Excellent! Thanks for sharing! Cheers! You are awesome. Keep smiling👍 Love and prayers from the USA! 😀

@wackojacko3962 Жыл бұрын

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.😇

@PreMath Жыл бұрын

Thanks for your feedback! Cheers! You are awesome. Keep smiling👍 Love and prayers from the USA! 😀

@Mycroft616 Жыл бұрын

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.

@PreMath Жыл бұрын

Thanks for your feedback! Cheers! You are awesome. Keep smiling👍 Love and prayers from the USA! 😀

@martincohen8991 11 ай бұрын

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).

@quigonkenny 3 ай бұрын

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)

@davidmontgomery5193 Жыл бұрын

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.

@richardl6751 Жыл бұрын

There is a bit of a shortcut. Once c was found, there is a limited number of values for a and b,

@mathbbn2676 4 ай бұрын

Explain well the Pythagorean theorem

@JSSTyger Жыл бұрын

My answer is a = 84, b = 63, and c = 105

@SrisailamNavuluri Жыл бұрын

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.

@alancs85 10 ай бұрын

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

@sharonmarshall3671 Жыл бұрын

Could find b by substituting for a in b = 5292/a instead of substituting for both a and c

@PreMath Жыл бұрын

Thanks for your feedback! Cheers! You are awesome. Keep smiling👍 Love and prayers from the USA! 😀

@Xyz-sd7ub Жыл бұрын

84,,,63,,,105 respectivally.

@HenriLaporte-kv6qq 3 күн бұрын

try to trick with simply factor the sizes by 21

@ikatras_who 4 ай бұрын

This question was asked in one of our periodic assessment😅

@santiagoarosam430 Жыл бұрын

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.

@PreMath Жыл бұрын

Excellent! Thanks for sharing! Cheers! You are awesome. Keep it up 👍 Love and prayers from the USA! 😀

@soli9mana-soli4953 Жыл бұрын

Once known a+b (sum) and ab (product) why don't apply x^2 - (sum)x + (product) = 0 ?

@karimalabi8639 Жыл бұрын

Exactly

@lnmukund6152 Жыл бұрын

The pythagorean triplet 3,4,5, is the best and easy method Mukundsir

@williamwingo4740 Жыл бұрын

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. 🤠

@calspace Жыл бұрын

It's actually a 3-4-5 triangle. Those lengths are those values multiplied by 21.

@williamwingo4740 Жыл бұрын

@@calspace Right. Didn't realize that until later.

@johnplong3644 Жыл бұрын

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 ..

@DB-lg5sq Жыл бұрын

شكرا لكم يمكن وضع a=4xوb=3xوc=5xمعx>0 نجدx=21 ثم نبين أن لا حل آخر للمسألة .....

@PreMath Жыл бұрын

Excellent! Thanks for sharing! Cheers! You are awesome. Keep it up 👍 Love and prayers from the USA! 😀 Shukran!

@raghavbatra3588 Жыл бұрын

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.

@mohanramachandran4550 Жыл бұрын

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

@PreMath Жыл бұрын

Excellent! Thanks for sharing! Cheers! You are awesome. Keep it up 👍 Love and prayers from the USA! 😀

@mohanramachandran4550 Жыл бұрын

Thanks for encourages

@mohanramachandran4550 11 ай бұрын

Simple method

@pa28cfi Жыл бұрын

Easier way: With both the area and perimeter being integers, we know the sides of the triangle must be pythagorean triples. Pythagorean triples also must have at least 1 even side. Using this information find the prime factors of 252 to find the smallest triple. In this case, 2 ,3, 7. Divide the area by the square of the factors since area squares in relation to side length. (A shape with sides 5 times longer will have and area 25 times larger.) If the result of this is still an integer, we can reduce the peremiter by the corresponding prime factors. In this case you can divide the area by 9 and 49, so you can divide 252 by 3 and 7 resulting in a pythagorean triple perimeter of 12. 3,4,5 is the only pythagorean triple that has a perimeter of 12. This is our base triple. Multiply the 3,4,5 by 3 and 7 and you get sides of 63, 84, and 105.

@PreMath Жыл бұрын

Thanks for your feedback! Cheers! You are awesome. Keep it up 👍 Love and prayers from the USA! 😀

@misterenter-iz7rz Жыл бұрын

ab=5292=4x27x49, a^2+b^2=c^2, a+b+c=252=4x9x7, ...🤔🤔🤔

@PreMath Жыл бұрын

Thanks for your feedback! Cheers! You are awesome. Keep it up 👍 Love and prayers from the USA! 😀

@mohamadtaufik5770 Жыл бұрын

a=84, b=63 and c=105

@PreMath Жыл бұрын

Thanks for sharing! Cheers! You are awesome. Keep it up 👍 Love and prayers from the USA! 😀

@johncampbell7868 11 ай бұрын

why not solve the quadratic of a for its roots, rather than guess!

@giuseppemalaguti435 Жыл бұрын

63,84,105...84,63,105

@PreMath Жыл бұрын

Thanks for sharing! Cheers! You are awesome. Keep it up 👍 Love and prayers from the USA! 😀

@antoniochadwick722 Жыл бұрын

I came to another solution : a= 113,106 b= 23,393 c= 115,5

@mohammadmoaz5748 Жыл бұрын

In india this question is 7th 8th level

@user-uu4eo4zt9c Жыл бұрын

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