Can you find the area of the Green semicircle? | (Important Math skills explained) | #math #maths

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 the area of the Green semicircle? | (Important Math skills explained) | #math #maths

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

#SemicircleArea #FindArea #TriangleArea #Semicircle #Circle #Radius #Square #GeometryMath #PythagoreanTheorem
#MathOlympiad #SimilarTriangles #TwoTangentsTheorem
#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.

[jimlocke]

Solution by tangent double angle formula. Let radius of semicircle = r. Construct OC and OP. Note that <OCD and <OCP are equal and let them have measure Θ, then <PCD = 2Θ. From ΔOCD, tan(Θ) = r/2r = 1/2. From tangent double angle formula, tan(2Θ) = 4/3. By alternate interior angle theorem, <CAB = <PCD, which = 2Θ. From ΔABC, tan(2Θ) = BC/AB = 4/3. Length BC = 2r. Therefore, doing the algebra, length AB = (3/4)(2r) = 3r/2. Yellow Δ area = (1/2)bh = (1/2)(3r/2)(2r) = 3r²/2. But yellow Δ area is given as 96, so 3r²/2 = 96, r² = 64 and r = 8. The area of the half circle is  (1/2)πr² = (1/2)π(64) = 32π, as PreMath also found.

[jameshogge]

Construct the similar kites DOPC and EOPA. Recognise the second is half the size of the former thus EA is 1/4 the side length of the square.

Calculate the side length of the square using 96 = 0.5 * a * 0.75a then calculate the area of the semicircle

[Stuv]

Side of square is 2x
Circle rad is x
DC = CP = 2x
L(DCA) + L(EAC) = 180
L(DCA) + L(DOP) =180
L(DOP) = L(EAP)
DC/DO = OE/EA = 2
EA = PA = x/2
Hyp of triangle is 2x + x/2
(2.5x)^2 = (2x)^2 + (1.5x)^2
6.25x^2 = 4x^2 +2.25x^2

96 = (2x)(1.5x)/2
3x^2 = 96x2
X^2 = 64
X = 8

Area semi- circle is
A = pi x^2 /2 = 32pi
A= 100.531 sq units

[georgerobartes]

I just recognised the area as the result of a multiple of 12 in a 3, 4, 5 triangle giving me 16 for the side . Therefore the radius is 8cm and obtained the result using pi r 2 ÷ 2 . Quick and simple

[kimchee]

The math is simple but logic is head scratching until it's explained in this video as how to apply the two tangent theorem. Thank you, another clear solution.

[prossvay]

Let r is radius of semicircle
So site of the square= 2r
AE=AP=x
BC=2r; AB=2r-x
AC=2r+x
Area of the right triangle ABC
A=1/2(2r)(2r-x)=96
r(2r-x)=96
(2r)^2+(2r-x)^2=(2r+x)^2

4r^2-8rx=0
4r(r-2x)=0
x=r/2
r(2r-r/2)=96
r(3r)=192
r^2=64
So: r=8cm
Area of the green ❤❤❤ thanks

[brianscott]

Thinking out loud, the relationship of the semi-circle area is always pi / 3 of the triangle area and the side of the square is always 1/3 of the triangle area. Never noticed that before. Very well laid out and explained, as your videos always are, thank you.

[allanflippin]

Again, I have a stupid approach, but it works! I'll use your names: square side = a, triangle base is b. Also, ab = 192. Next I prove that triangle OEA is similar to CDO. If DC/DO = 2, then OE/EA = 2. DC = a, DO = a/2. OE is also a/2 so EA = a/4. b = a - EA so b = 3a/4. Then I can solve for a = 16, r = 8 and circle area = 32pi.
To prove OEA is similar to CDO:
1) Let alpha = the angle of DCP
2) The quadrilateral ODCP angles must total 360. alpha + 90 + 90 + DOP = 360. DOP = (180 - alpha).
3) Per straight angle theorem, POE = 180 - DOP. Simplifying, POE = alpha.
4) The angle DCP = POE, therefore the triangle pair CDO, CPO are similar to OEA, OPA.

[Stereomoo]

I had angle DCA = CAB (Z angles), triangle OPC = ODC (side, side, side). So let θ = angle OCD, angle DCA = 2θ = CAB, angle EAP = 180 - CAB = 180 - 2θ. Triangle EOA = POA so angle EAO = (180-2θ)/2 = 90-θ. Hence angle EOA is θ and triangle ODC is exactly twice AEO (same angles, OE = half DC). By similar triangles, EA is half DO, and thus a quarter of the square's side. So AB = 3/4 L, BC = L, triangle formula L*3/4*L*1/2 = 3/8 L^2 = 96. ∴ L = 16. Making the circle area 8^2 π / 2 = 32π

Your way's definitely tidier on the back half, same similar triangles to get into it though.

[marcgriselhubert]

Let's use an adapted orthonormal. C(0;0) D(a;0) E(a;a) B(0;a) O(a;a/2), R beeing the diameter of the circle.
The equation of the circle is: (x-a)^2 + (y-a/2)^2 = (a/2)^2, or: x^2 + y^2 -2ax -ay +a^2 = 0 when developped.
The straight line (CPA) has an equation of the type y = mx, with m un unknown parameter at present.
At the intersection of this line with the circle we have: x^2 + (mx)^2 -2ax -amx + a^2 = 0, or (1 + m^2).x^2 - (2a+am).x + a^2 = 0.
At the point P (tangency), this second degree equation has a double solution, so its delta is equal to 0.
delta = (2a +am)^2 -4.a^2.(1 +m^2) = 4.a^2.m -3.a^2.m^2, which is equal to 0 when m = 0 or m = 4/3.
m=0 is corresponding to point D and m = 4/3 to point P. So the equation of the straight line (CPA) is y = (4/3).x
Its intersection with (BE) (which equation is y = a) is point A ((3/4).a; a), and so the length AB is (3/4).a
We finish. The area of triangle ABC is then ((3/4).a).a. (1/2) = (3/8).a^2 = 96, then a^2 = (96). (8/3) = 256, and finally a = 16.
The area of the green semi disk is then (1/2).Pi. (a/2)^2 = (1/2).(8^2).Pi = 32.Pi

[alster]

Wow, that's amazing!
After seeing the area of the square in the Pythagorean Theorem alongside the diameter being the square's side measure, I knew where this was going.

[soniamariadasilveira]

Obrigada querido professor por seus ensinamentos! Desejo ao senhor um Ano Novo maravilhoso, cheio de bênçãos e realizações! Viva 2024!!!

[wackojacko]

This problem is cool! And too me Fundamentally Binary. 🙂

[phungpham]

1/ 1/2 a.b=96 so a.b= 192
2/ Let the angle DCA = 2 alpha so the angle DCO= alpha and tan alpha = 1/2 tan 2 alpha= 1 /(1-1/4)= 4/3
3/ the angle CAB= 2 alpha so a/b= 4/3 b =3a/4 a.b= 3a/4 . a= 3/4 sqa= a=16
Area of the semicircle= 1/2 pi . 64= 32 pi =100.53 sq cm

[MrPaulc]

Great stuff. I made harder work of that than necessary through using values and r rather than your a and b variables. I had AB as 96/r and AE and AP as 2r - 96/r. It made very hard work of it all, so thank you for enlightening me :)

[ybodoN]

DC is twice DO and EO is twice EA as △OEA ~ △CDO ⇒ AB = ¾ BC ⇒ ABC is a 3:4:5 Pythagorean triangle.
96 cm² is 4² times the area of a 3, 4, 5 cm triangle, so BC = 16 and the area of the semicircle is 32 π cm².

[FlatEarthMath]

Fantastic, using the "two tangents" principle. I went a completely different way, using r for radius, and constructing OP = r, then constructing a segment (with length 2r) rightward from O. This creates multiple similar triangles. :-)

[flashg]

I first found myself constructing the circle with OC as diameter, in order to make use of Thales' theorem, and algebraically finding the intersection of this and the given semicircle. Once I got to the point where you introduce the two-tangent theorem, I saw that (since a circle tangent is perpendicular to the radius to the point of tangency) APO and OPC are similar triangles, and used the ratios AP : OP and OP : CP to determine b in terms of a.

[LuisdeBritoCamacho]

Since BC > AB we have (Integer Solutions):
Let BC = x and AB = y ; x > y
96 = x * y
Factors of 96 = 2 * 2 * 2 * 2 * 2 * 3 = 2^5 * 3
Only Integer Solutions, when x > y:
Dividers of 96 = {1 ; 2 ; 3; 4 ; 6 ; 8 ; 12 ; 16; 24; 32; 48; 96}
(96 ; 1) ; (48 ; 2) ; (32 ; 3) ; (24 ; 4) ; (16 ; 6) ; (12 ; 8)
Excluding (98 ; 1) and (48 ; 2) ; (24 ; 4) ; (32 ; 3) and (12 ; 8) because the Area of the Square is too big: 98^2 = 9.604 cm^2 ; 48^2 = 2.304 cm^2 ; 24^2 = 576 cm^2 ; 32^2 = 1.024 cm^2 and the Area of the Square is too small: 12^2 = 144 cm^2
Remaining Area of the Square : 9.604 - 96 = 9.508 cm^2 ; 2.304 - 96 = 2.208 cm^2 ; 1.024 - 96 = 928 cm^2 ; 576 - 96 = 480 cm^2 and 144 - 96 = 48 cm^2
In my opinion the only Possible and Elegant Integer Solution is (16 ; 6).
Size of the Square = 256 cm^2
Side of the Square = 16 cm
Radius of the Circle = 8 cm
Green Area = (8^2*Pi) / 2 cm^2 = (64*Pi) / 2 = 32*Pi cm^2
There are too many non Integer Solutions within a certain Interval, where the Remaining Area of the Square is bigger than the Rectangle Area.
P.S. - Please let me know if I'm wrong.

[Ben-wdin]

Quickest soln is to recognise that with an area of 96 sqcm, the sides are most likely comprised of two integers. One can then skip most of the geometry and proceed directly to the answer 😂