Justify your answer | Can you find area of the Green shaded region? | #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

Justify your answer | Can you find area of the Green shaded region? | #math #maths | #geometry

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

#FindGreenArea #SimilarTriangles #Geometry #CongruentTriangles #IsoscelesTriangles #RightTriangles #Triangle #AreaOfSquare #AreaOfTriangle #CircleTheorem #GeometryMath #EquilateralTriangle #PerpendicularBisectorTheorem
#MathOlympiad #ThalesTheorem #RightTriangle #RightTriangles #PythagoreanTheorem
#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.

[mohabatkhanmalak]

Wonderfull geometry, thanks for posting.🌵

[AdemolaAderibigbe-js]

We could also use coordinate geometry as follows. Let D(0, 0) be our origin. Then the equation of DB is Y = (1/2)X and the equation of line EC is Y = (-5/6)X +50 and they intersect at P(37.5, 18.75). Thus the triangle EBP has a base EB of 36 and a height of (30 - 18.75 = 11.25) so its area is (1/2)*36*11.25 which is 202.5. So the area of the quadrilateral AEPD is that of triangle ABD which is (1/2)*30*60 minus that of triangle EPB or 900 - 202.5 = 697.5 square units.

[aliturkseven]

Hi professor . i believe the shorter way is calculate High of Triangle PEB and calculate area of triangle PEB directly !

Assume PK is high of triangle PDC and PH is high of PEB then we have
PDC~ PEB => (EB/DC)=(PH/PK) => PH/PK=(36/60)=(3/5) => PK=(5/3)*PH and PH+PK=30
PH+(5/3)*PH=30 => PH=(90/8)=(45/4)=11.25
Area of PEB = (11.25*36)/2=202.5

Green Area =900-202.5=697.5

[thepuzzlesolver.gaming]

I used the coordinate geometry here. Consider the bottom left corner as the origin and solve.

[montynorth]

Triangles DPC & EPB are similar.
Let height of triangle EPB be h from its base EB to point P.
Then height of triangle DPC = (30 - h) from its base DC to point P.
So h / 36 = (30 - h) / 60.
Cross multiplying.
60 h = 36 ( 30 - h ).
60 h = 1080 - 36 h.
96 h = 1080.
h = 11.25.
Thus area of triangle EPB = 1/2 x 36 x 11.25 = 202.5.
Area of triangle ADB = 1/2 x 30 x 60 = 900.
Therefore green area = 900 - 202.5.
697.5.

[ashutoshkumardalei]

Great thinking 👏🏻👏🏻👏🏻please do share more videos like this 🙏🏻🙏🏻🙏🏻

[nandisaand]

I also used a coordinate system: I assugbed C (0, 0) and created formulas to describe lines
DB (Y=X/2+30)
and
EC (Y=-5/6X)
By setting these eqns equal to each other, I found coordinates for
P (-22.5, 18.75)
This gave me the height for triangle EBP (11.25).


=697.5

[marcgriselhubert]

We use an orthonormal center D, first axis (DC)
Equation of (DB):y = x/2 (evident)
Equation of (CE): 5.x + 6.y -300 = 0 (easy)
Intersection: P(75/2; 75/4)
Triangle EPB: basis EP = 36, height 30 - ordinate of P = 45/4
Area of EPB: (1/2).(36).(45/4 = 405/2
Green area: area of ABD - 405/2 = 900 - 405/2 = 1395/2.
(Very easy)

[GabrieleIris-isbg]

On triangle EBP relative height to EB is equal to: EB/(EB+EB+AE)xAD = 36/(36+36+24)x30 = 36/96x30 = 45/4 So EBP Area is 36x45/4x1/2=202, 5. Area of the green shaded region is area of ABD= 60x30x1/2 minus area of EBP. 900-202, 5=697, 5

[himo]

AB=DC=60 36 : 60 = 3 : 5
EP : PC = 3 : 5
EBP=36*30/2*3/8=405/2

Area of Green region :
30*60/2 - 405/2 = 1395/2

[batavuskoga]

Like in some other cases, I solved it in a completely different way
BEP and CDP are similar triangles, till so far I have the same steps.

draw a perpendicular from P to the line CD, let's call this line h2
also draw a perpendicular from P to the line BE, let's call this line h1

h2/CD=h1/BE --> h2*BE=h1*CD
h2=60/36*h1, h1=36/60*h2
h2=5/3*h1, h1=3/5*h2
h1+h2=30, 3/5*h2+h2=30, h2=75/4 (replacing h1 by 3/5*h2)
area rectangle=60*30=1800
area BCE=36*30/2=540
area
green area = area rectangle - area BCE - area CDP = 1800 - 540 - 562.5
green area = 697.5

[ChuzzleFriends]

Because ABCD is a rectangle, ∠A & ∠ADC are right angles.
So, ADCE is a right trapezoid. This will come in handy later.
By the Segment Addition Postulate, AE + BE = AB.
So, AB = 24 + 36 = 60.
Also, because ABCD is, again, a rectangle, CD = 60 by the Parallelogram Opposite Sides Theorem.
By the Vertical Angles Congruence Theorem, ∠BPE ≅ ∠CPD.
Segments AB & CD are parallel by definition of a parallelogram.
Therefore, ∠ABD ≅ ∠BDC by the Alternate Interior Angles Theorem.
So, △BPE ~ △DPC by AA.
Label two certain points on segments AB & CD as F (AB) & G (CD), such that the segment they form, FG, is perpendicular to both segments and P is between F & G.
Segment FG is the width of the rectangle, and segments FP & GP are altitudes of their respective triangles. FG = 30.
Because △BPE ~ △DPC, the triangles have scale factors. The scale factor from △BPE to △DPC is 60/36 = 5/3. Use knowledge of corresponding segment lengths in similar triangles. Label FP = x.
x + (5/3)x = 30
(3/3)x + (5/3)x = 30
(8/3)x = 30
x = 30 * (3/8)
= 11.25
So, FP = 11.25 & GP = 30 - 11.25 = 18.75.
Now, the final formula.
Quadrilateral ADPE Area = Trapezoid ADCE Area - △DPC Area
A = 1/2(a + b)h
= 1/2 * (24 + 60) * 30
= 15 * 84
= 1260
A = (bh)/2
= 1/2 * 60 * 18.75
= 30 * 18.75
= 562.5
Quadrilateral ADPE Area = 1260 - 562.5
= 697.5
So, the area of the green shaded region is 697.5 square units.

[adept]

АЕ/EB = 2/3.EB/DC = BP/PD = EP/PC = 3/5. S(ADE) = 1/5, S(DEC) = 1/2, S(DEP) = 3/8 × 1/2 = 3/16.
S(green) = (1/5 + 3/16) × 30 × 60 = 697, 5.

[aljawad]

Nice! I started out by finding the (X, Y) coordinates of point P from the intersection of the two lines: CE & BD, then proceeded to calculate the area of BEP from Heron’s formula and deducting it from the area of ABD to reach the same result.

[phungpham]

1/Label the height PH= h to EB.
By similarity, PH/BH=AD/AB= 30/60=1/2
—> BH= 2h
2/ Focus on the triangle BEC and HEP
HP/ BC=EH/EB—> h/30=(36-2h)/36
—> h=45/4
3/ Area of the green region= Area of the triangle DAB- Area of the triangle EPB = 1/2 x 30x60 - 1/2 x36x 45/4 =(1800-405)/2 =697.5 sq units😢😊

[DB-lgsq]

شكرا على المجهودات.
S(AEPD)=S(AECD)-S(PDC)

S(AEPD)=697, 5

[h.g.buddne]

I believe my solution is easier. The hight of EBP and DCP relate like 36:60 giving me the hight for EBP at 11.25 and an area of 36*11.25/2=202.5
Now all I have to do is subtract that from the 900 for the big triangle.
900-202.5=697.5 square units

[jarikosonen]

It looks good job done.
6:24 But can it be assigned also as EP=3a, PC=5a, Areas EPB=3ah, CPB=5ah, EBC=8ah=540 => ah=540/8.
EPB=3ah=3*540/8=405/2.

[santiagoarosam]

R es la proyección ortogonal de P sobre DC ---> DCE=DAB=ABCD/2---> DPE=BPC ---> Pendiente de DB=1/2. Pendiente de CE =30/36=5/6----> Si RC=b: (60-b)/2=30-(b/2)---> b=45/2---> Verde = ADE+BPC =(24*30/2)+(30b/2) =1395/2=697, 50 ud².
Gracias y saludos .
--->

[josedavis]

Another method(pardon me if it feels a bit complicated😂!)

1) Draw a line parallel to BC through point P so that it touches both EB and DC.
Let these meeting points be F and G respectively .

2) draw a line perpendicular to BC from point P, and let the meeting point on BC be H.

3) Rectangle FBHP and rectangle ABCD are similar rectangles.

4) hence FP/PH = AD/DC = 30/60 =1/2.
Let FP = x, so PH = 2x and PG = 30-x
Also let EF = y

5) since triangles EFP~ PGC and PGD~PFB are similar,
EF/ GC = PF/ PG= FB/GD
i.e,
y/2x = x/(30-x) = 2x/ (60-2x) respectively
Solving this;
y = 2x^2 / (30-x)

6) EB = EF+FB = 36
ie y+2x = 36
Substituting for y, in terms of x from previous step, we get solution for
x = 11.25

7)area of triangle EPB = 1/2* FP * EB
ie, 1/2* x * 36 = 202.5 sq units

8) Hence shaded area = 900-202.5
= 697.5 sq units