Can you calculate area of the triangle? | (Perimeter) | #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 calculate area of the triangle? | (Perimeter) | #math #maths | #geometry

Olympiad Mathematical Question! | Learn Tips how to solve Olympiad Question without hassle and anxiety!

FindArea #Area #TriangleArea #Perimeter #GeometryMath
#MathOlympiad #IntersectingChordsTheorem #RightTriangle #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.

[abeonthehill]

Today i learned some Grammar as well …..” conjugate !”
Increased my Maths knowledge and simultaneously my vocabulary !
Thanks for another clear example Professor ; i do so much enjoy your teaching modus operandi ! 😛🙏😉

[marioalb]

a+b+c= 48
c cos30° + c sin30° + c = 48
c (√3/2 + 1/2 + 1) = 48
c = 20, 287 cm

A = ½ a.b
A = ½ c cos30° c sin30° = √3/8 c²
A = 89, 11 cm² ( Solved √ )

[alexniklas]

Let us denote: AB= a; BC=c; AC= c.
c= 2a; b= √(4a^2-a^2)=a√3.
a+2a+a√3=48; a=48/(3+√3);
b= 48√3/(3+√3)
S= ab/2= 89.108
Thanks sir🤝

[quigonkenny]

Let AB = k. As ∆ABC is a 30-60-90 right triangle, CA = 2k and BC = √3k.

AB + BC + CA = P
k + √3k + 2k = 48
3k + √3k = 48
k = 48/(3+√3)
k = 48(3-√3)/(3+√3)(3-√3)
k = 48√3(√3-1)/(9-3)
k = 48√3(√3-1)/6
k = 8√3(√3-1) ≈ 10.14 units

A = √3k(k)/2
A = √3k²/2
A = √3(8√3(√3-1))²/2
A = (√3/2)(192(3-2√3+1))
A = 96√3(4-2√3)
A = 384√3 - 192(3)
A = 384√3 - 576 ≈ 89.11 sq units

[SkinnerRobot]

x + 2x + √3x = 48
x · (3 + √3) = 48
x = 48 / (3 + √3)

Area = ½ · x · √3x
= ½ · 48 / (3 + √3) · √3 · 48 / (3 + √3)
= 1152 · √3 / (3 + √3)²
= 1152 · √3 / (12 + 6√3)
= 192 · √3 / (2 + √3)
= 192 · √3 · (2 - √3) / (4 -3)
= 192 · (2√3 - 3)
≈ 89.11 square units

[himo]

AB=x BC=√3x CA=2x
x+√3x+2x=(3+√3)x=48
x=48(3-√3)/6=24-8√3

area of the triangle :

=288√3-288-288+96√3=384√3-576

[yalchingedikgedik]

Very nice and enjoyable
Thanks Sir for your efforts
Good luck with respects
❤❤❤❤

[alster]

P= 48

a+b+c= 48

c= 2b
a= b√3
b+b√3+2b= 48

3b+b√3= 48
b(3+√3)= 48
b= 48/(3+√3)*(3-√3)/(3-√3)

After rationalizing
48(3-√3)/6

8(3-√3)

b= 24-8√3
a= b√3
a= (24-8√3)(√3)
a= 24√3-24



A∆= ab/2

A∆= (24-8√3)(24√3-24)/2
A∆= 576√3-576-192(3)+192√3/2
A∆ = 768√3-2(576)/2
A∆= 384√3-576
A∆= 89.11 units²

[santiagoarosam]

AC=c---> AB=c/2---> BC=c√3/2---> (c/2)(3+√3)=48---> c=96/(3+√3)---> c²=256(12-6√3).
Área ABC =½(c/2)(c√3/2)= c²√3/8 =384√3-576 =89, 1075...ud².
Gracias y saludos

[cyruschang]

The ratio of the 3 sides of a 30° - 60° - 90° triangle = 1 : ✓3 : 2
48/(1 + ✓3 + 2) = 48/(3 + ✓3) = 48(3 - ✓3)/6 = 8(3 - ✓3)
area = 8(3 - ✓3) x 8✓3(3 - ✓3) ÷ 2 = 64✓3(12 - 6✓3)/2 = 192✓3(2 - ✓3) = 192(2✓3 - 3)

[murdock]

Nice! φ = 30° → sin⁡(3φ) = 1; ∆ ABC →
AB = a; AC = 2a; BC = a√3; BCA = φ
CAB = 2φ → ABC = 3φ
(a√3)(√3 + 1) = 48 → a = (8√3)(√3 - 1) →
area ∆ ABC = (1/2)sin⁡(2φ)2a^2 = (192√3)(2 - √3) ≈ 89, 11

[hemantdikshit]

How can forget to subscribe having listened simple explanation and a modest voice of yours!

[andryvokubadra]

We can use :
Sin 30° --> me
Cos 30°
Tan 30°
Sin 60°
Cos 60°
Tan 60°


Sin 30° = ½ --> a = 1, c = 2
b² = c² - a²
b = √(c² - a²)
b = √(2² - 1²)
b = √3 or -√3
b = √3

P = x(a+b+c)
48 = x(1+√3+2)
48 = x(3 + √3)
x = 48/(3+√3)
x = 48(3-√3)/6
x = 8(3-√3)

AB = ax
BC = bx
AC = cx

Area = ½*AB*BC
Area = ½*ax*bx
Area = ½x²*a*b
Area = ½(8(3-√3))²*1*√3
Area = ½*8²*(3-√3)²*√3
Area = ½*64*(12-6√3)*√3
Area = 32*(12√3-18)
Area = 192(2√3-3) 8:48
Area ≈ 89.1

[anestismoutafidis]

a+b+c=48 <=> a^2+b^2=c^2 <=>
[30°+60°+90°=180°] <=> [48/6+48/3+48/2=8+16+24=48]
9, 83^2+17, 83=20, 35^2 <=>
9, 83+17, 83+20, 35=48, 0
Area of the triangle = (9, 83×17, 83)/2=87, 63 area-units.

[PrithwirajSen-njqq]

Perpendicular ঃ Base ঃhypotenuse =1ঃ√3ঃ2
1+√3+2 =4.732
48/4.732= 10.143

Then perpendicular is 10.143
Base is 10.143*1.732
=17.567676
Area =(10.143 * 17.567676)/2
=89.0944 sq units approx

[unknownidentity]

Let's find the area:
.
..
...
....


Since ABC is a 30°-60°-90° triangle, we can conclude:

AB:BC:AC = sin(30°):sin(60°):sin(90°) = (1/2):(√3/2):1 = 1:√3:2

From the known perimeter P of the triangle we obtain:

AB + BC + AC = P
AB + √3*AB + 2*AB = 48
3*AB + √3*AB = 48
AB*(3 + √3) = 48
⇒ AB = 48/(3 + √3) = 48*(3 − √3)/[(3 + √3)(3 − √3)] = 48*(3 − √3)/(9 − 3) = 48*(3 − √3)/6 = 8*(3 − √3)

Now we are able to calculate the area of the triangle:

A = (1/2)*AB*BC = (1/2)*AB*(√3*AB) = (√3/2)*AB²
A = (√3/2)*8²*(3 − √3)² = (√3/2)*64*(9 − 6√3 + 3) = 32√3*(12 − 6√3) = 192*(2√3 − 3) ≈ 89.11

Best regards from Germany

[sorourhashemi]

Thanks. Easy. The side in front of 30° in right triangle is half of the chord and in front of 60° is rad.3/2 of chord so if we mark chord as C, we have: C/2+ rad.3/2 and C =48. C= 16(3-rad.3), BC= 8rad.3(3-rad.3) and AB = 8(3-rad.3). And the area 89.11

[michaeldoerr]

The area is 192[2sqrt(3)-3]. I shall use that for practice. I also hope that there is a special playlist that involve special triangles and really counter-intuitive problems.

[lasalleman]

A = 89. Just set up the triangle and use the sine, cosine and resulting line AC using basic Pythagorean formula. Line AC comes out to 1. Find perimeter which comes out to 2.366. Line AC is thus .4226 of total perimeter. Applied to total perimeter of 48 in the example, turns out that line AC = 20.2848. Sine AB is thus 10.1424, and cosine line BC is 17.5654. Using the formula for area of a right triangle, turns out that the area of this triangle is 89 units.

[marcgriselhubert]

For such a triangle, if P is the perimeter and A the area, then A = (P^2).((2.sqrt(3) -3)/12) or P = 2.sqrt((2.sqrt(3) +3).A).