Can you find area of the Pink shaded triangle? | #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 Pink shaded triangle? | #math #maths | #geometry

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

#FindPinkTriangleArea #SquareArea #Squares #Square #RightTriangles #Triangle #AreaOfSquare #SimilarTriangles #AreaOfTriangle #CircleTheorem #GeometryMath #EquilateralTriangle #PerpendicularBisectorTheorem
#MathOlympiad #ThalesTheorem #RightTriangle #RightTriangles #CongruentTriangles
#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
#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 Radius
Equilateral Triangle
Pythagorean Theorem
Area of a circle
Area of the sector
Right triangles
Radius
Circle
Quarter circle
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.

[necdetguney]

I love how this man improves my math culture

[Mediterranean]

First find the side which is 30
Then focus on ABE and EDC
Both have sides 30 and 15 and a 90 degree angle so they are congruent
Then their hypothenuses are equal so BEC is an isoceles triangle
The hypothenuse value is 15root5 (by the Pythagorean theorem)
Let's focus on BEF
It's a right triangle in <BFE since it is 180-90=90
Let's call the angle EBF as @
Then EF = 15 sqrt5*sin @ and BF = 15sqrt5 *cos@
Let's focus on BFC
We have BEC is an isoceles triangle
So BE=EC=15*sqrt5
And EC=EF+FC
So FC=EC-EF=15sqrt5(1-sin@)
Let's find the sine of @ using the Pythagorean theorem
BC^2=FC^2+BF^2


225*4=225*5(1-2sin@+1)
4=5(2-2sin@)
4=10(1-sin@)
2=5-5sin@
-3=-5sin@
Then sin@ = 3/5
Let's focus on the pink triangle
We have sin@=3/5=EF/15sqrt5
So cross multiply to find EF =15sqrt5*3/5=9sqrt5
Since BEF is a 3;4;5 triangle then BF = 12sqrt5
Then calculate the area
The area is 270 like how you found it

[misterenter-izrz]

Not difficult. But need lengthy calculation😮, FC=30/sqrt(5), EC=15xsqrt(5), FC/EC=2/5, therefore the answers is 450×3/5=270.😅

[povijarrro]

1)get EC and BE from pythagorean theorem
2)get BF from identity BF*EC=AB*BC
3)get EF from pythagorean theorem
4)get BF*EF/2 as wanted area

[jimlocke]

Finding area ΔEFB almost directly: From Pythagoras, length CE = √(DE² +CD²) = √(15² + 30²) = 15√5. ΔBCF and ΔCDE are similar. CF/BC = DE/CE, CF = (BC)(DE)/CF = (30)(15)/(15√5) = 30/√5 = 6√5. BF/BC = CD/CE, BF = (BC)(CD)/CF = (30)(30)/(15√5) = 60/√5 = 12√5. EF = EC - CF = 15√5 - 6√5 = 9√5. Let b = EF and h = BF for ΔEFB. Then, area = (1/2)(9√5)(12√5) = (54)(5) = 270 square units, as PreMath also found.

[LuisdeBritoCamacho]

1) Total Area of Square [ABCD] = 900
2) Side of Square [ABCD] = 30 = sqrt(900)
3) Half Side of Square = 15
4) Area of Triangle [ABE] + Area of Triangle [CDE] = 450, because : 15 * 30 / 2 = 450 / 2 = 225
5) So, the Area of Triangle [BCE] = 450
6) Base of Triangle [EBC] = EC = sqrt(30^2 + 15^2) = sqrt(900 + 225) = sqrt(1.125) = 15*sqrt(5) ~ 33, 54
Now,
7) Base * h = 2*Area
8) 15*sqrt(5) * h = 900 ; h = 900 / 15*sqrt(5) ; h = 60 / sqrt(5) ; h = 60*sqrt(5) / 5 ; h = 12*sqrt(5) ; h ~ 26, 83
9) sqrt(FC) = sqrt(900 - 720) = sqrt(180) = 6*sqrt(5) ~ 13, 42
10) Area of Triangle [BCF] = 12*sqrt(5) * 6*sqrt(5) / 2 = 72 * 5 / 2 = 180
11) Area of Triangle [BEF] = 450 - 180 = 270
12 Answer : The Pink Shaded Region Area is equal to 270 Square Units.

[phungpham]

1/ Side of the square= 30
2/ All the white right triangles are similar so FC/FB= 1/2
Let FC = x so FB = 2x ——> sqr + 4sqx= 900—> sqx= 900/5 —> x=6sqrt5 and FB=12 sqrt5
Area of the smallest white right triangle = 36x 5=180
Area of the red triangle= 450-180= 270 sq units

[allanflippin]

Sorry, if I'm repeating somebody. I didn't see anybody use the method I'll propose. Start at the point in the video (about 5:00) where you decide to find the area of triangle BFC. Instead of drawing an extra line, simply compare triangle EDC with BFC and notice they are similar. Find the length of EC by pythagorean theorem: EC^2 = ED^2 + DC^2. EC^2 = 1125. EC = 15 * sqr(5). BC corresponds to EC. Let's find the ratio: BC/EC = 30 / (15*sqr(5)). This simplifies to 2/(sqr(5)). All sides of BFC are proportional to the sides of EDC by the same ratio. Based on this, the area of BFC is proportional to the area of EDC by that ratio squared. The square of 2/(sqr(5)) is 4/5. The area of EDC is 225. 225 * 4 / 5 = 180 which is the area of BFC. Thus the area of BFE is 450 - 180 = 270.

[marcgriselhubert]

In this problem using an orthonormal is very simple. We choose center D and first axis (DC). We then have D(0;0) A(0;30) b(30;30) C(30;0) E(0;15)
VectorCE(30;-15) is colinear to VectorU(-2;1). The equation of (CE) is: (x-30).(1) -(y).(-2) = 0 or x +2.y -30 = 0
VectorU is orthogonal to (BF), the equation of (BF) is (x-30).(-2) + (y -30).(1) = 0 or 2.x -y -30 = 0
By solving the system x +2.y -30 = 0 and 2.x -y -30 = 0 we find easily the coordinates of F: F(18;6)
Now VectorEF(18; -9) and EF^2 = 18^2 + (-9)^2 = 405 and EF = sqrt(405) = 9.sqrt(5); VectorBF(-12;-24) and BF^2 = (-12)^2 + (-24)^2 = 720 and BF = sqrt(720) = 12.sqrt(5). Finally the area of triangle EFB is (1/2).EF.BT = = 270.

[giuseppemalaguti]

AB=30..EF=a, FB=b...a^+b^2=900+225=1125...((√1125-a)^2+b^2=900..1350=2a√1125..a=9√5...b^2=1125-81*5=720..b=12√5...Apink=9√5*12√5/2=54*5=270

[francismoles]

CD is esual of 30, DE at 15, we have CE is equal of 15*sqrt5
With the area of BCE we can know BF which is equal of 900/CE it means 12*sqrt5
With the triplet 3, 4, 5 we can know EF which is equal of 9*sqrt5
For finishing, we calculate BF*EF/2 which is equal at 270

[Waldlaeufer]

s² = 900
Due to similarity of triangles CDE and BFC:
h² + (h/2)² = s² (h = BF, h/2 = CF)
h² + h²/4 = 900
5/4 h² = 900
h² = 900 * 4/5 = 180 * 4 = 720
h = 4 * 3 √5 = 12 * √5
h/2 = 6 * √5
A = square - triangle CDE - triangle ABE - triangle BCF
A = 900 - 2 * 225 - 1/2 * h/2 * h
A = 450 - 1/2 * 6 * √5 * 12 * √5
A = 450 - 36 * 5
A = 450 - 180 = 270 square units

[quigonkenny]

Square ABCD:
A = s²
900 = s²
s = √900 = 30

Triangle ∆EDC:
ED² + DC² = CE²
15² + 30² = CE²
225 + 900 = CE²
CE = √1125 = 15√5

A = bh/2 = 30(15)/2 = 225

BA = DC and AE = ED, so by symmetry, ∆BAE and ∆EDC are congruent.

If ∠CED = α and ∠DCE = β, where β = 90°- α, then ∠FCB = α and ∠CBF = β. Thus ∆EDC and ∆BFC are similar.

Triangle ∆BFC:
FC/CB =ED/CE
FC/30 = 15/15√5 = 1/√5
FC = 30/√5 = 6√5

BF/FC = DC/ED
BF/6√5 = 30/15 = 2
BF = 2(6√5) = 12√5

EF = EC - FC = 15√5 - 6√5 = 9√5

Pink Triangle ∆EFB:
A = bh/2 = 9√5(12√5)/2
A = 9(6)5 = 270 sq units

[wackojacko]

The coolest part of this problem is @ 3:05 and that the areas of triangles EAB & EDC are equivalent to area of EBC. A very cool property! One can do a similar problem with a rectangle inscribed in a triangle where one edge of the rectangle lies on any line of the triangle, then the final result of equivalent triangles is the same but It's journey to determine equivalence is different depending on choice of variables to use. 😉

[unknownidentity]

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


First of all we calculate the side length s of the square:

A(square) = s² = 900 ⇒ s = √900 = 30

The area of the triangle BCE is obviously half of the area of the square. Now we need to know the area of the right triangle BCF. Since ∠CDE=∠BFC=90°, ∠DCE=∠CBF=α and ∠CED=∠BCF=β, the triangles CDE and BCF are similar. So we can conclude:

BF/CF = CD/DE = s/(s/2) = 2

Since BCF is a right triangle, we can apply the Pythagorean theorem:

BF² + CF² = BC²
(2*CF)² + CF² = s²
4*CF² + CF² = s²
5*CF² = s²
⇒ CF² = s²/5

Now we can calculate the size of the pink area:

A(BCF) = (1/2)*BF*CF = (1/2)*(2*CF)*CF = CF² = s²/5

A(pink) = A(BEF) = A(BCE) − A(BCF) = s²/2 − s²/5 = 900/2 − 900/5 = 450 − 180 = 270

Best regards from Germany

[murdock]

φ = 30°; ∎ABCD → AB = BC = CD = AD = AE + DE = a/2 + a/2 = 30 → CE = BE = 15√5 = n
CE = CF + EF → sin⁡(EFB) = sin⁡(3φ) = 1 → DCE = CBF = δ; CF = k → EF = n - k
BF = m → sin⁡(δ) = √5/5 = k/a → k = 6√5 → EF = 9√5 → m = 12√5 → FBE = θ → sin⁡(θ) = 3/5 →
area ∆ BEF = (1/2)sin⁡(θ)nm = 270

[Heathenheart]

No need to use trigonometry, angles or costants. Just look at EBF and BFC triangles, they have the longest leg in common.

[michaelstahl]

Great Video again. I used your solution finding propotions over simular triangles with success. Now I want to propose another solution :
I calculated EC with the theorem of Pythagoras and got EC = 33, 45 LE . I know the area of the triangle EDC ( = ABE ) 225 squareunits.
The area of triangle EBC must be 450 squareunits. I know EC and now I can find BF the hight of triangle EBC . BF is 900 / EC = 26, 83 LE.
Using Pythagoras theorem you can find the lenght of EF . EF = 20, 13 LE. so I get for the pink area 20, 13 * 26, 83 /2 = 270, 02 squareunits.

[johnbrennan]

By pythag |EB|= sqroot 1125. So also is |EC|=sqroot 1125. The triangles bfc and edc are similar so |FB|\ 30= 30/ sqroot 1125. Therefore|FB|= 900/ sqroot 1125. In triangle ebf 1125= |EF|^2 plus ( 900/sqroot 1125)^2. So 1125= |EF|^2+ (810000/ 1125) that is |EF|^2 = 405. So |EF| = sqroot 405. Pink area = half |FB| by |EF| = 1/2 ( 900/ sqroot1125)( sqroot 405)= 1/2 by (900 by 9 sqroot5)/ 15 sqroot 5= 270.

[jamestalbott]

Thank you, this was a good way to start the day!