Can you find the missing side lengths of the triangle? | (Perimeter) |
HTML-код
- Опубликовано: 21 июн 2024
- Learn how to find the missing side lengths of the triangle. Important Geometry and Algebra skills are also explained: Perimeter; area of a triangle formula. Step-by-step tutorial by PreMath.com
Today I will teach you tips and tricks to solve the given olympiad math question in a simple and easy way. Learn how to prepare for Math Olympiad fast!
Step-by-step tutorial by PreMath.com
• Can you find the missi...
Need help with solving this Math Olympiad Question? You're in the right place!
I have over 20 years of experience teaching Mathematics at American schools, colleges, and universities. Learn more about me at
/ premath
Can you find the missing side lengths of the triangle? | (Perimeter) |#math #maths | #geometry
Olympiad Mathematical Question! | Learn Tips how to solve Olympiad Question without hassle and anxiety!
#FindSideLengths #Perimeter #TriangleArea #GeometryMath
#MathOlympiad #IntersectingChordsTheorem #RightTriangle #RightTriangles
#PreMath #PreMath.com #MathOlympics #HowToThinkOutsideTheBox #ThinkOutsideTheBox #HowToThinkOutsideTheBox? #FillInTheBoxes #GeometryMath #Geometry #RightTriangles
#OlympiadMathematicalQuestion #HowToSolveOlympiadQuestion #MathOlympiadQuestion #MathOlympiadQuestions #OlympiadQuestion #Olympiad #AlgebraReview #Algebra #Mathematics #Math #Maths #MathOlympiad #HarvardAdmissionQuestion
#MathOlympiadPreparation #LearntipstosolveOlympiadMathQuestionfast #OlympiadMathematicsCompetition #MathOlympics #CollegeEntranceExam
#blackpenredpen #MathOlympiadTraining #Olympiad Question #GeometrySkills #GeometryFormulas #Angles #Height #ComplementaryAngles
#MathematicalOlympiad #OlympiadMathematics #CompetitiveExams #CompetitiveExam
How to solve Olympiad Mathematical Question
How to prepare for Math Olympiad
How to Solve Olympiad Question
How to Solve international math olympiad questions
international math olympiad questions and solutions
international math olympiad questions and answers
olympiad mathematics competition
blackpenredpen
math olympics
olympiad exam
olympiad exam sample papers
math olympiad sample questions
math olympiada
British Math Olympiad
olympics math
olympics mathematics
olympics math activities
olympics math competition
Math Olympiad Training
How to win the International Math Olympiad | Po-Shen Loh and Lex Fridman
Po-Shen Loh and Lex Fridman
Number Theory
There is a ridiculously easy way to solve this Olympiad qualifier problem
This U.S. Olympiad Coach Has a Unique Approach to Math
The Map of Mathematics
mathcounts
math at work
Pre Math
Olympiad Mathematics
Two Methods to Solve System of Exponential of Equations
Olympiad Question
Find Area of the Shaded Triangle in a Rectangle
Geometry
Geometry math
Geometry skills
Right triangles
imo
Competitive Exams
Competitive Exam
Calculate the length AB
Pythagorean Theorem
Right triangles
Intersecting Chords Theorem
coolmath
my maths
mathpapa
mymaths
cymath
sumdog
multiplication
ixl math
deltamath
reflex math
math genie
math way
math for fun
Subscribe Now as the ultimate shots of Math doses are on their way to fill your minds with the knowledge and wisdom once again.
Well explained
Glad to hear that!
Thanks dear for the feedback ❤️
Note that a and b do not need to be integers but it is a good educated guess that the problem was created by choosing a Pythagorean triple, or multiple of a Pythagorean triple, with hypotenuse 277 and perimeter 644. Because 277 is prime, there is no multiple of a Pythagorean triple with hypotenuse of 277. The wikipedia article on Pythagorean triples lists 115, 252, 277 and the perimeter is 644, so a = 115 and b = 252 is a solution, as well as a = 252 and b = 115.
It can also be shown that these are the only solutions. The sum (a + b) must be 367 for the perimeter to be 644. If the absolute value of (a - b) is less than 137 (252 - 115), the perimeter will be shorter. If greater than 137, the perimeter will be longer.
Excellent!
Thanks for the feedback ❤️
Thank you so much!
STEP-BY-STEP RESOLUTION PROPOSAL :
01) Perimeter = 644
02) AC = Hypothenuse = 277
03) AB = a
04) BC = b
05) 644 - 277 = 367
06) a + b = 367 ; a = 367 - b or b = 367 - a
07) a^2 + b^2 = 277^2
08) a^2 + b^2 = 76.729
09) a^2 = (367 - b)^2
10) b^2 = (367 - a)^2
11) a^2 + (367 - a)^2 = 76.729
12) a^2 + 134.689 - 734a + a^2 = 76.729
13) 2a^2 - 734a + 57.960 = 0
14) a^2 - 367a + 28.980 = 0
15) Roots : a = 115 and : a = 252
16) AB = 252 and BC = 115
17) Checking Solutions :
18) 644 = 277 + 252 + 115 ; 644 = 644 ; True
19) 115^2 + 252^2 = 277^2
20) 13.225 + 63.504 = 76.729 ; 76.729 = 76.729 ; True
21) ANSWER : Missing Side Lengths are AB = 252 Linear Units and BC = 115 Linear Units
Our Answer from The International Islamic Institute for the Study of Ancient Mathematical Thinking, Knowledge and Wisdom - Cordoba Caliphate.
Super!👍
Thanks for sharing ❤️
You are the best!❤️
Let x=height of triangle,
a+b+c=644=perimeter.
a+b=644-c=644-277=367.
a=(367-x),b=x and c=277.
Use the famous Pythagoras Theorem
c^2=a^2+b^2
277^2=(367-x)^2+x^2
277^2=367^2+x^2-734x+x^2
2x^2-734x+57960=0
x^2-367x+28980=0
(x-115)(x-252)=0
x is 115 or 252 units.
Note these answers are the same thing and they check out the original equations.
a=(367-115),b=115,c=277
a=(367-252),b=252 and c=277
Therefore the three sides of the triangle are 252,115 and 277 units.
These check out with the perimeter and the Pythagoras Theorem on the triangle.
Thanks for the puzzle prof.
Alternately, on the hunch of a Pythagorean triangle and leveraging Euclid's parametrization of Pythagorean triples, look for two squares adding to 277:
277 = 256 + 21 - no good
= 225 + 52 - no good
= 196 + 81 - Bingo!
That suggests
(14^2 - 9^2, 2*14*9, 14^2 + 9^2) = (115, 252, 277)
Verify that
277 + 252 + 115 = 644 - Done; and we have our solution.
The exercise of searching for sum of two squares is so rewarding when successful, and such a small expense with these small numbers, that it really should be the first test.
Euclid (in modern terminology) showed that all Pythagorean triples are uniquely parametrized by natural numbers
k, m, n where m, n are coprime and not both even. Then by expansion it's clear that:
(2kmn)^2 + (k(m^2 - n^2))^2 = (k(m^2 + n^2))^2.
Since 277 is prime, k must be 1 in this problem.
The problem is to find a and b knowing that a + b = 644 - 277 and a^2 + b ^2 = 277^2.
So a + b = 367 and a^2 + b^2 = 76729 = (a + b)^2 - 2.a.b,
so a + b = 367 and a.b = (134689 - 76729)/2 = 28980.
a and b are solutions of the equation x^2 - 367.x + 28980 = 0. Delta = 18789 = 137^2.
So a = (367 + 137)/2 = 252 and b = (367 - 137)/2 = 115 (if a>b)
Excellent!🌹
Thanks for sharing ❤️
I made a typo early on in mine but my method was sound. I somehow got 377 instead of the correct 367. That's what I get for doing that part in my big head LOL.
I did it with your equations 1 and 2. Substituted a^2 in equation 1 with (367-b)^2. Then I used the quadratic equation to solve for X (our b), using the terms 2, 734, and 57960 (for a b and c in the quadratic equation). This resulted in our b = 115 (ignoring the negative result) and a= 252 after substituting 115 for b in your equation 2.
After I find a*b=28980, I find out 28980=2²*3²*5*7*23. After some calculations I found a=252, b=115.
But your method is much better.
Excellent!
Thanks for sharing ❤️
Thank you!
We can write 277 as
81+196=9^2+14^2
So one side is 2×9×14=252
Square of other side is (277^2-252^2)
=529×25=(23×5)^2
Third side is 115.
Third side is 196-81=115.
Very easy
Although I used Vieta's Formula.
Since I know that
a+b= 367 and ab= 28980
Vieta's Formula is
x²-(a+b)x+(ab)= 0
x²-367x+28980= 0
And after many trials and errors, I found the factors and did basic factoring
(x-252)(x-115)
x= 252, 115
Since these are both positive solutions, these are the measures of the two missing sides/legs.
Cheers from The Philippines 🇵🇭
1/ sum (a+b)=367
product (axb) =28980
--> by using Vieta theorem, we have the equation:
Sq x -367x+28980 = 0
--> delta= sq 137
We have
--> x=252 or x=115
--> a=115 and b=252
Or a=252 and b=115😊
Excellent!
Thanks for sharing ❤️
It is too simple to solve😅.....required side-lengths are 252 units and 115 units....love from INDIA🇮🇳
Excellent!
Thanks for the feedback ❤️
Love and prayers from the USA! 😀
Alot of people cant figure pythegorean therom. Some dont care but when you can solve for fabricating its awsome. Meaning also should say trig.
@@mikeandcolleenk9831 Thank you🙂
There's no way you did that in your head, though.
I know that a+b+277=644
a+b=367
a=367-b
Then
b²+(367-b)²=277²
b²+367²-734b+b2=277²
2b²-734b=277²-367²
And that's all I can get without a calculator.
a m=14; n=9
b=2mn=2•14•9=252
a=m²-n²=196-81=115
a+b+c=115+252+277=644 !!! 😁
How to find square root of large number like you did with √18769 in video?
c=277, a+b=644-277=367, b=367-a, c^2=a^2+b^2, c^2=a^2+(367-a)^2, a^2-367a+28980=0, a=(367+\/18769)/2=252,
b=(367-\/18769)/2=115.
Excellent!
Thanks for sharing ❤️
Let AB=a ; BC=b
Perimetter=a+b+277=644
So a+b=367 (1)
and a^2+b^2=(277)^2=76729(2)
(1) b=367-a
(2) a^2+(367-a)^2=76729
a=252 ; b=115
So a=252 ; b=115 .❤❤❤
Excellent!👍
Thanks for sharing ❤️
The Answer can be
a=115 and b=252 also.
I got the equation w^2 -367w + 28980 =0 which factored to (w-252)(w-115) =0 which got the same result after lots of squaring and factoring. Your method was easier. Are Lists of Pythagorean Triples allowed in tests?
115; 252
Excellent!
Thanks for sharing ❤️
252 and 115
277^2 = a^2 + (644-277-a)^2
Let's find the side lengths:
.
..
...
....
.....
Since ABC is a right triangle, we can apply the Pythagorean theorem. With P being the perimeter of the triangle we obtain:
AC² = AB² + BC²
AC² = AB² + (P − AC − AB)²
277² = AB² + (644 − 277 − AB)²
277² = AB² + (367 − AB)²
76729 = AB² + 134689 − 734*AB + AB²
0 = 2*AB² − 734*AB² + 57960
0 = AB² − 367*AB² + 28980
AB
= 367/2 ± √[(367/2)² − 28980]
= 367/2 ± √(134689/4 − 28980)
= 367/2 ± √(134689/4 − 115920/4)
= 367/2 ± √(18769/4)
= 367/2 ± 137/2
Finally we find the two solutions:
AB = 367/2 + 137/2 = 504/2 = 252 ⇒ BC = P − AC − AB = 644 − 277 − 252 = 115
AB = 367/2 − 137/2 = 230/2 = 115 ⇒ BC = P − AC − AB = 644 − 277 − 115 = 252
Best regards from Germany
Excellent!👍
Thanks for sharing ❤️
Not my favorite problem of yours, I'm sorry to say. Conceptually, it's very simple, but the numbers you selected turn the calculations into a tedious grind, assuming we are meant to solve it all without a calculator.
thx teacher for teach me i am student cambodia i
You are welcome, dear 🌹
Keep watching...
easy
If 2 sides are missing then it cannot be a triangle. My teacher taught me that a triangle has 3 sides. 😂
It's possible to solve the problem also without the perimeter, using old mathematc, from Italy
Thanks for the feedback ❤️
Уже решали, без Италии, так как более древние геометры, были в Греции, а алгебру изобрели арабы, когда Италии ещё не было!
@@sergeyvinns931Sud Italy wss Magnagrecia pythagora lived here....