Can you find area of the triangle? | (Without Trigonometry) |

Тәжірибелік нұсқаулар және стиль

Learn how to find the area of the scalene triangle whose side lengths are 10 and 12. Important Geometry skills are also explained: area of the triangle formula; Exterior angle theorem; Pythagorean theorem. 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 area of t...
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 area of the triangle? | (Without Trigonometry) | #math #maths #geometry
Olympiad Mathematical Question! | Learn Tips how to solve Olympiad Question without hassle and anxiety!
#TriangleArea #FindAreaOfTheTriangle #PythagoreanTheorem #LawOfSines #Heron'sFormula #Triangle #GeometryMath #AreaOfTriangle #CongruentTriangles #IsoscelesTriangles
#MathOlympiad #SimilarTriangles
#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 #ExteriorAngleTheorem #PythagoreanTheorem #IsoscelesTriangles #AreaOfTriangleFormula #AuxiliaryLines
#blackpenredpen #MathOlympiadTraining #Olympiad Question #GeometrySkills #GeometryFormulas #AreaOfTriangles #CompetitiveExams #CompetitiveExam
#MathematicalOlympiad #OlympiadMathematics #LinearFunction #TrigonometricRatios
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
Olympiad Question
Find Area of the Shaded Triangle
Geometry
Geometry math
Geometry skills
Right triangles
Exterior Angle Theorem
pythagorean theorem
Isosceles Triangles
Area of the Triangle Formula
Competitive exams
Competitive exam
Find Area of the Triangle without Trigonometry
Find Area of the Triangle
Auxiliary lines method
Pythagorean Theorem
Square
Diagonal
Congruent Triangles
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.

Пікірлер: 75

  • @kaziummeruman524
    @kaziummeruman5243 ай бұрын

    LOVE THIS❤️‍🔥🥰

  • @PreMath

    @PreMath

    3 ай бұрын

    Excellent! Glad to hear that! Thanks ❤️

  • @JSSTyger
    @JSSTyger3 ай бұрын

    You are one of the best geometry instructors on KZread.

  • @thinker821
    @thinker8213 ай бұрын

    Sine rule: 10/sin(3x) = 12/sin(2x), expanding sin(2x) and sin(3x) and solving a quadratic in cos(x) gives cos(x) = either 3/4 or -1/3, the latter being impossible since 2x needs to be less than 180 degrees. Hence cos(x) = 3/4 and sin(x) = √7/4. Area = 1/2 c a sin B = 1/2 * 10 * 12 * √7/4 = 15√7.

  • @valentinebestofthevoiceall9289

    @valentinebestofthevoiceall9289

    3 ай бұрын

    Yes, I've chosen the same way. It's easier, i think.

  • @rudychan8792

    @rudychan8792

    3 ай бұрын

    Hey, this is my idea too^^ Sinus Rules! More Effective! Cut this video from 14' to 7' 😉

  • @PreMath

    @PreMath

    3 ай бұрын

    Thanks ❤️

  • @DanielNeedham2500

    @DanielNeedham2500

    3 ай бұрын

    How do you deduce the remaining angle is 3x to use the Sine rule?

  • @valentinebestofthevoiceall9289

    @valentinebestofthevoiceall9289

    3 ай бұрын

    @@DanielNeedham2500 sin (180° - a) = sin a

  • @mariopopesco
    @mariopopesco3 ай бұрын

    Another way without trigonometry : Draw bisector of angle BAC, AD. In triangle ACD , angle DAC is x, angle ADC is 2x. So triangle ADC is similar with triangle BAC. Triangle ABD has angles BAD and ABD equal with x, so is isosceles, so AD is equal with BD. Lets say AC is y and BD is a. So AD is a also. BD + CD is 12, so CD= 12- a From similar triangles we have : AD/AB = AC/BC = CD/AC, so a/10 = y/12 = 12-a/y > a = 10y/12, and in second equation we have y/12 = (12- 10y/12)/y -> y^2 = 144 - 10y, so y = 8 the only positive solution. We have all the sides, 8,10,12, so with Heron formula s = (8+10+12)/2 = 15 Aria = sqrt (15*7*5*3) = 15 sqrt7

  • @PreMath

    @PreMath

    3 ай бұрын

    Excellent! Thanks ❤️

  • @gibbogle

    @gibbogle

    2 ай бұрын

    I did the same.

  • @Abby-hi4sf
    @Abby-hi4sf3 ай бұрын

    So great to see how to solve it without trig! Beautiful question and solution

  • @PreMath

    @PreMath

    3 ай бұрын

    Glad you enjoyed it! Thanks ❤️

  • @misterenter-iz7rz
    @misterenter-iz7rz3 ай бұрын

    You successfully solve it without any means of trigonometry. 😮

  • @PreMath

    @PreMath

    3 ай бұрын

    Yes! Thanks ❤️

  • @michaelkouzmin281
    @michaelkouzmin2813 ай бұрын

    Another solution using trigonometry: 1) Angle ACB = 180-3x; 2) sin(180-3x)=sin(3x); 3) 12/sin(2x)=10/sin(3x); 4) sin(3x)/sin(2x)=10/12=5/6; => cos(x)=3/4 => 5) sin(x) = =sqrt(1-(cos(x))^2)= sqrt(1-(5/6)^2) = sqrt(7)/4 6) AreaABC = 10*12/2*sin(x)= 60*sqrt(7)/4=15*sqrt(7) = approx 39.686 sq units.

  • @PreMath

    @PreMath

    3 ай бұрын

    Excellent! Thanks ❤️

  • @jimlocke9320
    @jimlocke93203 ай бұрын

    At 1:00, there is an implication that side AB is the base. Actually, side BC or side AC can be treated as the base. However, PreMath's solution is dependent on using AB as the base, so that is the reason why he chose AB as the base. There are probably other solutions that use the other sides as the base. Also, it is possible that using line segments to divide ΔABC into 2 or more triangles would produce solutions where AB is not used as the base. For example, once PreMath has values for h and a, he could have used CP as a common base of triangles ΔACP and ΔBCP and computed and added their areas to equal the area of ΔABC. As for the trigonometric solutions provided in the comments, these are great, but the title states that trigonometry is not used.

  • @PreMath

    @PreMath

    3 ай бұрын

    Thanks ❤️

  • @jamestalbott4499
    @jamestalbott44993 ай бұрын

    Thank you!

  • @PreMath

    @PreMath

    3 ай бұрын

    You are very welcome! Thanks ❤️

  • @bigm383
    @bigm3833 ай бұрын

    Quite convoluted but strangely satisfying!

  • @PreMath

    @PreMath

    3 ай бұрын

    Thanks ❤️

  • @alster724
    @alster7243 ай бұрын

    After seeing the quadratic equation, I did basic factoring instead of quadratic formula

  • @davidstaphnill9636
    @davidstaphnill96363 ай бұрын

    Could you explain please why you replaced the 2 in the numerator with 1/4😊

  • @misterenter-iz7rz
    @misterenter-iz7rz3 ай бұрын

    Using sin rule, trigonometric means yields a simpler solution, as 12/sin 2x=10/sin 3x, give cos x=3/4, so sin x=sqrt(7)/4, thus the area is 1/2×12×10×(sqrt(7)/4)=15sqrt(7).😊

  • @PreMath

    @PreMath

    3 ай бұрын

    Thanks ❤️

  • @user-cb8lx4ot4y

    @user-cb8lx4ot4y

    Ай бұрын

    能寫下具體的計算過程嗎?

  • @user-cb8lx4ot4y

    @user-cb8lx4ot4y

    Ай бұрын

    Can you write down the specific calculation process?

  • @misterenter-iz7rz

    @misterenter-iz7rz

    Ай бұрын

    @user-cb8lx4ot4y s×in 3x/sin 2x=10/12=5/6, sin 3x=sin x(3-4sin^2 x),sin 2x=2sin xcos x,....

  • @joeschmo622
    @joeschmo6223 ай бұрын

    Magic!

  • @PreMath

    @PreMath

    3 ай бұрын

    Glad to hear that! Thanks ❤️

  • @CloudBushyMath
    @CloudBushyMath3 ай бұрын

    Interesting

  • @PreMath

    @PreMath

    3 ай бұрын

    Glad to hear that! Thanks ❤️

  • @gibbogle
    @gibbogle2 ай бұрын

    I drew a different line, bisecting CAB, meeting CB at D. Then triangles ABC and DAC are similar. Let w = AC, y = AD = BD, z = DC. z/w = w/12, i.e. z = (w^2)/12, and y/w = 10/12, i.e. y = (5/6)w But BC = BD + DC = y + z = (w^2)/12 + 5w/6 = 12, which gives w = 8 With s = (triangle perimeter)/2 = (a+b+c)/2 where a,b,c are the side lengths area = sqrt(s(s-a)(s-b)(s-c)) We have s = (12+10+8)/2 = 15 and area = sqrt(15*3*5*7) = sqrt(1575) = 15sqrt(7).

  • @aura_beast2678
    @aura_beast2678Ай бұрын

    This is an application of inmo question 1998 . you can prove that a²= b(c+b) 144=10x + x² Just use quadratic formula to get x and appy herons formula Easier than your method

  • @user-cb8lx4ot4y

    @user-cb8lx4ot4y

    Ай бұрын

    Can you write down the specific calculation process?

  • @professorrogeriocesar
    @professorrogeriocesar3 ай бұрын

    Problema maravilhoso. Muito obrigado. Fiz de outra forma, traçando a bissetriz relativa ao ângulo A. Muito obrigado e parabéns!!

  • @PreMath

    @PreMath

    3 ай бұрын

    You are very welcome! Thanks ❤️

  • @prossvay8744
    @prossvay87443 ай бұрын

    Area of the triangle=1/2(12)(10)sin(41.41)=39.69 square units.❤❤❤ Thanks sir.

  • @PreMath

    @PreMath

    3 ай бұрын

    Great job! You are very welcome! Thanks ❤️

  • @user-cb8lx4ot4y

    @user-cb8lx4ot4y

    Ай бұрын

    Can you write down the specific calculation process?

  • @sorourhashemi3249
    @sorourhashemi32493 ай бұрын

    How do you know the angle CTA is 2x?

  • @johnbrennan3372

    @johnbrennan3372

    3 ай бұрын

    By construction he made it equal to 2x

  • @user-sk9oi9jl2g
    @user-sk9oi9jl2g3 ай бұрын

    Можно без дополнительных построений. Применяя теорему синусов и синус тройного угла, находит тригонометрическую функцию x. Удобнее cos( x)=0.75. Зная cos( x) находим sin( x), высоту, проведённую к стороне AB.

  • @PreMath

    @PreMath

    3 ай бұрын

    Thanks ❤️

  • @user-cb8lx4ot4y

    @user-cb8lx4ot4y

    Ай бұрын

    Can you write down the specific calculation process?

  • @user-sk9oi9jl2g

    @user-sk9oi9jl2g

    Ай бұрын

    @@user-cb8lx4ot4y BC/sin(2x)=AB/sin(180-3x), 12/sin(2x)=10/sin(3x), we use the reduction formula. sin(3x)=3sin(x)-4sin^3(x). Substitute 12/(2sin(x)*cos(x))=10/(3sin(x)-4sin^3(x)). We reduce everything that is possible and get 3/cos(x)=5/(3-4sin^2(x)). We replace the sine with the cosine on the right side of the equality using the basic trigonometric identity sin^1(x)+cos^2(x)=1. 3/cos(x)=5/(3-4+4cos^2(x)). 3/cos(x)=5/(4cos^2(x)-1). After the transformations, we obtain the quadratic equation 12cos^2(x)-5cos(x) -3=0. This equation has 2 roots, one of which is negative, which cannot be. cjs(x)=0.75. We find sin(x)=√1-0,75:2=√7/4 and the area according to the formula.

  • @user-cb8lx4ot4y

    @user-cb8lx4ot4y

    Ай бұрын

    ​@@user-sk9oi9jl2g Great, this is what I needed, thank you!

  • @user-sk9oi9jl2g

    @user-sk9oi9jl2g

    Ай бұрын

    @@user-cb8lx4ot4y According to the sine theorem CB/sin(2x)=AB/sin(180-3x), 12/sin(2x)=10/sin(3x), we use the reduction formula. sin(3x)=3sin(x)-4sin^3(3x). 12/(2sin(x)*cos(x))=10/(3sin(x)-4sin^3(x). We're cutting everything we can. 3/cos(x)=5/(3-4sin^2(x)). We use the basic trigonometric identity sin^2(x)+cos^2(x)=1. 3/cox(x)=5/(3-4sin^2(x)). After the transformations, we get a quadratic equation with respect to cos(x), which has two roots, one of which is negative, which is impossible. So cos(x)=0.75. sin(x)=√1-(0,75)^2=√7/4. S=0.5*12*10 * sin(x)=15√7.

  • @wackojacko3962
    @wackojacko39623 ай бұрын

    One needs to have a measured sense of reason to solve this problem. 🙂

  • @PreMath

    @PreMath

    3 ай бұрын

    Thanks ❤️

  • @giuseppemalaguti435
    @giuseppemalaguti4353 ай бұрын

    Teorema dei seni 10/sin3x=12/sin2x....risulta cosx=3/4...A=12*10*sinx/2=15√7

  • @PreMath

    @PreMath

    3 ай бұрын

    Excellent! Thanks ❤️

  • @user-cb8lx4ot4y

    @user-cb8lx4ot4y

    Ай бұрын

    Can you write down the specific calculation process?

  • @pralhadraochavan5179
    @pralhadraochavan51793 ай бұрын

    Good night sir

  • @PreMath

    @PreMath

    3 ай бұрын

    Same to you❤️

  • @marcgriselhubert3915
    @marcgriselhubert39153 ай бұрын

    The area of the triangle is (1/2).BA.BC.sin(x) = (1/2).10.12.sin(x) = 60.sin(x). Now let(s calculate sin(x) We have: sin(x)/AC = sin(2.x)/12 = sin(180° -3.x)/10 in triangle ABC, so 10.sin(2.x) = 12.sin(3.x), or 5.sin(2.x) = 6.sin(3.x) Then: 5.2.sin(x).cos(x) = 6.(3.sin(x) - 4.(sin(x))^3) We divide by sin(x) which is non equal to 0: 10.cos(x) =18 - 24.(sin(x))^2 We simplify by 2 and replace (cos(x))^2 by 1 - (sin(x))^2, we obtain: 5.cos(x) = 9 -12.(1 -(cos(x)^2) or: 12.(cos(x)^2 -5.cos(x) -3 =0 Delta = (-5)^2 -4.12.(-3) = 25 +144 = 169 = (13)^2, so cos(x) = (5 -13)/24 = -1/3 which is rejected as negative, or cos(x) = (5 +13)/24 = 18/24 = 3/4 Then (cos(x)) = 9/16 and (sin(x))^2 = 1 - (9/16) = 7/16 and sin(x) = sqrt(7)/4 (as sin(x)>0, 0

  • @PreMath

    @PreMath

    3 ай бұрын

    Thanks ❤️

  • @marcgriselhubert3915

    @marcgriselhubert3915

    3 ай бұрын

    At the last line, please naturaly read : .... Area = 60.sin(x) = 15.sqrt(7).

  • @unknownidentity2846
    @unknownidentity28463 ай бұрын

    Let's find the area: . .. ... .... ..... Let's introduce a point D on BC such that ∠BAD=x. Then we have two triangles ABD and ACD with: ∠BAD = x ∠ABD = x ∠ADB = 180° − ∠BAD − ∠ABD = 180° − 2x ∠CAD = x ∠ADC = 180° − ∠ADB = 180° − (180° − 2x) = 2x ∠ACD = 180° − ∠CAD − ∠ADC = 180° − x − 2x = 180° − 3x Therefore ABD is an isosceles triangle (AD=BD) and the triangles ABC and ACD are similar: AB/AD = AC/CD = BC/AC AC/CD = BC/AC ⇒ AC² = CD*BC AB/AD = AC/CD AB²/AD² = AC²/CD² AB²/AD² = CD*BC/CD² AB²/AD² = BC/CD AB²/AD² = BC/(BC − BD) AB²/AD² = BC/(BC − AD) 10²/AD² = 12/(12 − AD) 100/AD² = 12/(12 − AD) 25/AD² = 3/(12 − AD) 300 − 25*AD = 3*AD² 3*AD² + 25*AD − 300 = 0 AD = [−25 ± √(25² + 4*3*300)] / (2*3) = [−25 ± √(625 + 3600)] / 6 = (−25 ± √4225) / 6 = (−25 ± 65) / 6 Obviously only one of the two solutions is useful: AD = (−25 + 65)/6 = 40/6 = 20/3 Since ABD is an isosceles triangle, it can be divided into two congruent right triangles. So we can conclude: cos(x) = (AB/2)/AD = (10/2)/(20/3) = 5*3/20 = 3/4 sin(x) = √(1 − cos²(x)) = √(1 − 9/16) = √(7/16) = √7/4 Finally we are able to calculate the area of the triangle ABC: A(ABC) = (1/2)*AB*BC*sin(x) = (1/2)*10*12*√7/4 = 15√7 Best regards from Germany

  • @unknownidentity2846

    @unknownidentity2846

    3 ай бұрын

    I also made a comment with an alternative solution to the problem you published yesterday. I'm looking forward to your opinion.

  • @PreMath

    @PreMath

    3 ай бұрын

    Excellent! Thanks ❤️

  • @user-ri6rn7ti5h
    @user-ri6rn7ti5h3 ай бұрын

    (10)^2=100 (12)^2=144 2x(45°)=90°x. 3x(15°)=45°x 3(15°)=45°x {90°x+45°x+45°x}=180°x^3 b{100+144}=244 ,{180°x^3-244}=0√76°x^3 4^√19x^2 4^√19^1x^3 √4^√1^√1x^3√ 2^2x^3 √2^1^2x^3 √1^√1^2x^3 23 (x+2x-3)

  • @PreMath

    @PreMath

    3 ай бұрын

    Thanks ❤️

  • @rudychan8792
    @rudychan87923 ай бұрын

    Looks Complicated, Messy, Too Long. Sinus Rules, Please: sin(3x) ÷ 10 = sin2x ÷ 12 sin3x / sin2x = 10/12 ... do the rest, you will get 12p"-5p -3 = 0 ; p = cos x = 3/4 Turn into sin x = √7 ÷ 4 Area = (1/2)•10•12•(√7/4) = 15√7 = 39,7 sq units Much-much Easier\Practical. --- Anyway, i'll give you thumbsup, anyway^ "Longtime no see" Dec'23 today Mar'24 😉 👍

  • @PreMath

    @PreMath

    3 ай бұрын

    Good to see you again 😀 Thanks ❤️

  • @adept7474
    @adept74743 ай бұрын

    СК || АВ. АК - bisector ∠ВАС. СН ⟂ АВ. АСКВ - Isosceles trapezoid. СК = АС = ВК = х, АН = y. х² - y² = 12² - (10-y)², х = 10 - 2y. (10 - 2y)² - y² = 12² - (10 - y)². y² - 15y + 14 = 0. y = 1. СН = √(12² - 9²) = √63. S(АВС) = (10√63)/2 = 15√7.

  • @SkinnerRobot

    @SkinnerRobot

    3 ай бұрын

    I really like how you let AC = 10 - 2y , and the rest is so easy. Bravo! Thumbs up.

  • @adept7474

    @adept7474

    3 ай бұрын

    @@SkinnerRobot Thanks

  • @PreMath

    @PreMath

    3 ай бұрын

    Thanks ❤️

Келесі