Can you find area of the triangle? | (Without Trigonometry) | #math #maths #geometry

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!

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

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
#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.

[mariopopesco]

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

[JSSTyger]

You are one of the best geometry instructors on YouTube.

[thinker]

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.

[Abby-hisf]

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

[michaelkouzmin]

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.

[jimlocke]

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.

[aura_beast]

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

[bigm]

Quite convoluted but strangely satisfying!

[phungcanhngo]

Thank you professor for amazing geometry.

[gibbogle]

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).

[misterenter-izrz]

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

[prossvay]

Area of the square units.❤❤❤ Thanks sir.

[misterenter-izrz]

You successfully solve it without any means of trigonometry. 😮

[marcgriselhubert]

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<x<90°)
Finally the area of the triangle ABC is 60.sin(x) = 15.sin(x) (No need to think outside the box).

[alster]

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

[shrikrishnagokhale]

I was wrong in taking a=18instead of 8.Corrected myself.Thanks.

[giuseppemalaguti]

Teorema dei seni 10/sin3x=12/sin2x....risulta

[shrikrishnagokhale]

Using Heron's formula my answer is 40√2sq.unit

[professorrogeriocesar]

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

[davidstaphnill]

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