Can you find area of the Green shaded triangle? | (Semicircle) | #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 Green shaded triangle? | (Semicircle) | #math #maths #geometry

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

#GreenTriangleArea #Rectangle #Semicircle #GeometryMath #PythagoreanTheorem #SimilarTriangles
#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.

[parthtomar]

Very nice and easy solution sir with no mistake thx🎉

[jimlocke]

Very well done! A minor shortcut. At 8:20, we can treat CD as the base of ΔBCD, then the length of CF or DP becomes the height. Applying the area of a triangle formula, A = (1/2)bh, (1/2)(2√(34))(√(30)) = (√(34))(√(30)) = √(1020) = 2√(255), as PreMath also found.

When taking the area of a right triangle, we are used to taking the product of the sides and dividing by 2. When the hypotenuse and the distance from the right angle's vertex are known, we may overlook the fact that we can treat them as base and height in the area formula!

[tombufford]

Taking a second glance, A very fast technique for this one, draw a rectangle DCGF, where G and F are the perpendicular intersections from D and C to AE. Reflect the rectangle about axis AE to form a square, noting DC is parallel with AE with D and C on the circumference of the semicircle radius, 16/2. The green shaded area is half the area of the rectangle = 1/4 the area of the square. Then the green triangle area = 0.25 * (Diagonal^2) /( sqrt(2)^2 )= 0.25 * 16^2/2 = 32 square units

[hookahsupplier.]

Your solution escorted with unmatched clarity and honed problem solving skills are amazing and pretty inspirational!

[limfilms]

AP = FE = x.
(a) Δ DBC ~ Δ BPD ~ Δ BFC. Take ratios y^2= (10-x)(6-x) (i)
(b) Draw right triangle Δ ACE. Δ ACE ~ Δ AFC ~ Δ EFC. Take ratios y^2= x(16-x) (ii).
(c) thus (i) and (ii): (10-x)(6-x)=x(16-x), so x= 2.07 approx.
(d) thus y=5.36 approx.
(e) DC= 16-2x = 11.86 approx.
(f) A = (1/2)(11.86)(5.36)= 31.8 aprox.

[aryankushwaha]

We can easily do it with symmetry and chrod intersection property

[solimana-soli]

Being AP=FE=x and applying Euclid theorem on ADE we have:
DP²=x*(16-x) [1]
On DBC we have:
DP² =(6-x)*(10-x) [2]
Comparing [1] with [2] you find x
16 - 2x is the hypotenuse of the green triangle
√ x*(16-x) is its height

[DB-lgsq]

شكرا لكم على المجهودات
يمكن استعمال a=COE
180-a=COB
180-2a=COD
x=BC
y=BD
x^2=8^2+2^2-2×2×cos(180-a)
x^2=67+32cosa
y^2=8^2+2^2-2×2×8cosa
y^2=68-32cosa
x^2+y^2=136
DC^2=136
DC^2=8^2+8^2-2×8×8cos(180-2a)
cos2a=1/16
cosa=(جذر34)/8
x^2=68+4(34جذر)
y^2=68-4(34جذر)
x^2 × y^2=4080
xy=4(34جذر)
S=1/2 xy

S=2(255جذر)

[LuisdeBritoCamacho]

I solved this problem geometrically. Let the coordinates of point A be the Origin (0 ; 0)
Let the coordinates of point B equal (0 ; 6)
The Equation of the given Circle is: (x - 8)^2 + y^2 = 64
The coordinates of point D are the interception of (x - 8) ^2 + y^2 = 64 and x^2 +y^2 = 36
Solution: x = 9/4 and y = (3*sqrt55))/4
Drawing a Straight Line, y = 3*sqrt(55)/4, I can calculate the Distance between points D and C; wich is 11, 5 lin un (13, 75 - 2, 25)
Now, knowing the Base of triangle [BCD] and the height = (3*sqrt(55))/4 I can calculate the Area of triangle [BCD]
Area = [11, 5 * (3*sqrt(55))/4] / 2 = (34, 5 * sqrt(55)) / 8 ~ 255, 859 / 8 ~ 31, 982 sq un
Answer:
The Green Triangle Area is equal to approx. 31, 982 Square Units.

[johnwilson]

I don't understand where the fact that dc is parallel to ae comes from?

[ybodoN]

Draw the chord GBH perpendicular to AE. Then GB = BH = √(6 · 10) = √60 and DP = CF = √(60 / 2) = √30.
Since the radius of the semicircle is (10 + 6) / 2 = 8 then OP = OF = √(64 − 30) = √34 and PF = 2√34.
Therefore, the area of the green triangle is ½ 2√34 √30 = √1020 = 2√255 ≈ 31.937 square units.

[arthurschwieger]

Is there a rule that says two parallel chords of a circle are equally divided by a perpendicular line that runs through the center of the circle?

[tznephm]

Why does angle FCB equals angle PDB, and angle CBF equals angle BDP?

[marcgriselhubert]

Let's use an adapted orthonormal as often, center O, first axis (AB), the radius of the semi circle is 8, its equation is x^2 + y^2 = 64.
Let's nama a the distance from D or C to (AB), then C(sqrt(64 -a^2); a) and D(-sqrt(64 -a^2), and B(-2; 0)
Then VectorBC(sqrt(64 -a^2) +2; a) and VectorBD(-sqrt(64 -a^2) +2; a). These vectors are orthogonal, so: (sqrt(64 - a^2) +2).(-sqrt(64 -a^2) +2) + a.a = 0
or 4 - (64 -a^2) + a^2 = 0, so 2.(a^2) = 60 and a= sqrt(30)
Then the length of the green triangle is DC = 2.sqrt(64 -a^2) = 2.sqrt(34) and its height is a = sqrt(30) and finally its area is (1/2).2.sqrt(34).sqrt(30) = sqrt(1020) = 2.sqrt(255)

[ybodoN]

In any similar construction where a rectangle is divided into three similar right triangles, DP = CF = √AB × √BE / √2 🤩

[misterenter-izrz]

x^2+y^2=8^2=64, ((x-2)^2+y^2)+((x+2)^2+y^2)=4x^2, x^2-y^2=4, so x^2=34, y^2=30, therefore the area is approximately 😅

[tombufford]

At a quick glance, Rotate Triangle BDC about the line AE to form a square in the Circle with diagonals 2 * radius = 10+ 6 = 16. Then dc = 16 /SQRT(2) and height = 8/sqrt(2) Then area of Green shaded triangle = 0.5 / 2 * 16 * 8 = 32 square units.

[dirksteele]

Why are DP and CF equal in length? Was it stated that CD and AE are parallel?

[Atharva-jfcq]

Thanks for giving us knowledge by tricky question which were asked in competitive exams

[misterenter-izrz]

It is very clumsy and difficult puzzle. 😅😅😅