Find area of the Blue shaded region | Three circles given

Find area of the Blue shaded region | Three circles given | Important Geometry skills explained

Рет қаралды 21,929

Жыл бұрын

Learn how to find the area of the Blue shaded region. Important Geometry and Algebra skills are also explained: Similar triangles, area of the circle, Thales's Theorem. Step-by-step tutorial by PreMath.com
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!
Step-by-step tutorial by PreMath.com
• Find area of the Blue ...
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
/ premath
Find area of the Blue shaded region | Three circles given | Important Geometry skills explained
Olympiad Mathematical Question! | Learn Tips how to solve Olympiad Question without hassle and anxiety!
#CalculateTheBlueRegion #GeometryMath #AreaOfCircle #PythagoreanTheorem
#MathOlympiad #ThalesTheorem #RightTriangle #RightTriangles #CongruentTriangles
#PreMath #PreMath.com #MathOlympics #HowToThinkOutsideTheBox #ThinkOutsideTheBox #HowToThinkOutsideTheBox? #FillInTheBoxes #GeometryMath #Geometry #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
#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
Area of circle
Semicircle
Pythagorean Theorem
Thales's Theorem
Right triangles
Similar triangles
Subscribe Now as the ultimate shots of Math doses are on their way to fill your minds with the knowledge and wisdom once again.

Пікірлер: 73

@pralhadraochavan5179 Жыл бұрын

Good evening sir

@PreMath Жыл бұрын

Same to you, dear So nice of you. Thank you! Cheers! 😀

@timeonly1401 Жыл бұрын

@8:33 once you got A(blue shaded) = pi/2 (CE)(ED) , you can get (CE)(ED) more directly by using the Intersecting Chord Theorem: (CE)(ED) = (AE)(EB) = (10)(10) = 100. So that A(blue shaded) = (pi/2) (100) = 50 pi.

@nineko Жыл бұрын

When you got to ①, couldn't you just use the chord theorem to say that CE•ED = AE•EB? Since AE = EB, it yields the same result of course.

@Micboss1000 Жыл бұрын

I came here to say this, you beat me to it.

@PreMath Жыл бұрын

Many approaches are possible to find the solution to this problem! Thank you! Cheers! 😀

@massimookissed1023 Жыл бұрын

You absolutely could, but PreMath also quickly showed *_why_* that chord theorem is true.

@madhusudangupta3661 Жыл бұрын

PreMath is solving with the Geometry basics method without using theorems. Well done.

@flash24g Жыл бұрын

@@madhusudangupta3661 Several theorems are used in this solution.

@himo3485 Жыл бұрын

(Chords Theorem) CE＊ED=AE＊EB=10＊10=100 Blue region : π/2＊100=50π=157

@PreMath Жыл бұрын

Excellent! Thank you! Cheers! 😀

@madaunn42221 11 ай бұрын

absolutely i also solved your way easily

@georgebliss964 Жыл бұрын

The position of line AB is not defined, so can be moved to the right to be the diameter of the large circle equal to 20 or 10 radius. The yellow and green circles will then be congruent, with diameters equal to10 or 5 radii. Then area of large circle = Pi x 10^2 = 100 Pi. Area of the 2 small circles = 2 x Pi x 5^2 = 50 Pi. Then area of shaded area = 100Pi - 50Pi = 50 Pi.

@PreMath Жыл бұрын

Excellent! Thank you! Cheers! 😀

@TimBoulette Жыл бұрын

Brilliant

@ybodoN Жыл бұрын

Now suppose this is a trick question and the area of the blue region is not constant🤔

@fadetoblah2883 Жыл бұрын

@@ybodoN If the area of the blue region were not constant, then the problem, as formulated, could not be solved at all. That's the point. There are many such problems (in geometry especially) where the absence of certain data (in this case, the exact positioning of AB relative to the center of the larger circle) just "tells" you that the answer is constant within a set of specified parameters. Here, so long as AB=20 and intersects the diameter of the larger circle at a 90 degree angle, the area of the blue region will remain constant. You can safely and reliably infer that from the way the problem is set up. It's a good trick to know when you just need to quickly get to the answer, and it comes up fairly often.

@alantucker3014 Жыл бұрын

Yes , that's the way I did it. Took me about 20 seconds with no pen and paper 😊

@Abby-hi4sf 11 ай бұрын

Important Geometry skill here! Love it.

@j.r.1210 Жыл бұрын

I solved this with no triangles used in any way. I defined the radius of the big circle in terms of the radii of the small circles, so I could create a formula for the area of the blue region in terms of pi and the two smaller radii. Conveniently, this formula reduced to a single term in which the two smaller radii were multiplied: 2pi*Ry*Rg. I calculated that product using the chord theorem: since AB = 20, then 10 x 10 = 100, so 2Ry × 2Rg also = 100. 4RyRg = 100, so RyRg = 25. Finally, 2piRyRg = 50pi.

@flash24g Жыл бұрын

I personally started with the intersecting chord theorem, giving me CE.ED = 100 hence xy = 25, where x and y are the radii of the inner circles. I then applied the area formula to all 3 circles and thereby arrived at the answer.

@mohabatkhanmalak1161 Жыл бұрын

Briliant! I just love how you solve these geometry problems. Geometry has always been a favorite subject and I can see the wisdom on why you sometimes take the longer method - so that the student/observer gets to know the applications of the various theorems and methods. Cheers!

@raya.pawley3563 Жыл бұрын

Thank you!

@KAvi_YA666 Жыл бұрын

Thanks for video.Good luck sir!!!!!!!!!

@MultiYesindeed Жыл бұрын

moreover I think your videos are 100% truely brilliant

@shrikrishnagokhale3557 Жыл бұрын

Very good

@richardsullivan1655 10 ай бұрын

Is there a way to solve for the radius of the two smaller circles? I can see great responses here but i don't know how to solcve the smaller circle diameters.

@SAHIRVLOGCLP 11 ай бұрын

Thank you so much Proffesor for your sharing

@PreMath 11 ай бұрын

You are very welcome

@wackojacko3962 Жыл бұрын

Now I'm closer too figuring out the meaning of crop circles.🙂

@PreMath Жыл бұрын

Excellent! Thank you! Cheers! 😀

@mahinnazu5455 Жыл бұрын

You are my boss... Mahin from Bangladesh..

@PreMath Жыл бұрын

So nice of you, Mahin dear Thank you! Cheers! 😀

@amimaalam3357 Жыл бұрын

fantastic sum purely based on basic theorems

@parthtomar6987 8 ай бұрын

Sir inspite of construction of lines , we can also use intersecting chords theorem. Anyways the solution is awesome

@procash1968 Жыл бұрын

Very complicated problem. Beautifully explained bringing clarity Thanks PreMath Guru Ji

@alantucker3014 Жыл бұрын

See above comments , it's not complicated at all

@lk-wr2yn Жыл бұрын

Question: Is that independant how big is CE is, we get always 50Pi? (Only that ED and CD change)

@soli9mana-soli4953 Жыл бұрын

Prof, your proportions on ACD triangle is the second Euclid theorem

@burkenersesian8263 11 ай бұрын

Is there some intuitive way to understand that the difference of the area of the outer circle and the two inner circles remains constant as the diameter of the larger inner circle goes from equal to the smaller inner circle to equal to the outer circle?

@williamwingo4740 Жыл бұрын

Since the only measurement given is the 20-unit chord, the specific dinensions of the circles may be irrelevant. Move the vertical line AB to the right, so that it crosses CD at O, changing the various circle sizes as necessary. This makes the diameter of the big circle 20 units, and the two small circles are now identical with diameters of 10 units each. The area of the big circle is 100 pi and the smaller ones are 25 pi each for a total of 50 pi. The blue shaded area is the difference, or 100 pi -- 2(25 pi) = 50 pi, the same as your answer. Carpe Diem! 🤠

@Copernicusfreud Жыл бұрын

I came up with the solution using a data plot. y=radius of yellow circle. g = radius of green circle. b = radius of blue circle. In the diagram 2y + 2g = 2b. dividing both sides of the equation by 2 gives y + g = b. With the intersecting chord theorem, 2y * 2g = 10 * 10. 4yg = 100 and gy = 25. I picked 3 points where g < 25. g = 10 means y = 2.5 and b = 12.5. Calculating all the areas and subtracting out the yellow and green gives 50*pi for the answer. The same with g = 8 which means y = 25/8 and b = 89/8. Calculating all the areas and subtracting out the yellow and green gives 50*pi for the answer. It is exactly the same answer with g = 25/4 which means y = 4 and b = 1681/16. Calculating all the areas and subtracting out the yellow and green gives 50*pi for the answer. I wonder if this ties back into Thales Theorem for a circle and right angles. The answer is consistently 50*pi. 50*pi is the answer.

@marioalb9726 11 ай бұрын

Blue Shaded Area: A = ⅛ π c² A = ⅛ π 20² A = 50π cm² ( Solved √ ) Easier !!!

@kaliprasadguru1921 Жыл бұрын

Why similar triangles theory . Why not intersecting chords theory . CE.ED = AE.EB (=AE²)

@JSSTyger Жыл бұрын

I got snagged trying to figure out OE. Anyone know what that is in terms of the radius of the big circle?

@devindermehta5577 11 ай бұрын

Once you got the equation for area of blue shaded area, could have used intersecting cords theorem.

@marioalb9726 11 ай бұрын

Blue Shaded Area: A = ⅛ π c² A = ⅛ π 20² A = 50π cm² ( Solved √ )

@terelelli01 Жыл бұрын

you could solve it much more it easily using the power theorem on a circle

@noahvale2627 Жыл бұрын

I also prefer intersecting chords theorem.

@spiderjump Жыл бұрын

Let the radii of the yellow and green circles be a and b respectively. Radius of the largest circle = a+b Area of blue shaded region = pi(a+b)^2 - pi•a^2 - pi•b^2 =pi•2ab Using intersecting chord theorem, 2a•2b=20/2•20/2 4ab= 100 2ab=50 Therefore, area of blue region .=50pi

@MarieAnne. Жыл бұрын

I used two methods, one specific and one more general. In both cases, we make use of the information you used in video: - Centres and points of tangency are collinear, i.e. the centers of all 3 circles lie on CD. Therefore, CD is diameter of larger circle, and CE and ED are diameters of the smaller circles. - If a radius of a circle is perpendicular to a chord, then the radius bisects the chord. Therefore, AE = EB = AB/2 = 10 Specific solution: Since the position of chord AB seems irrelevant to solution, we assume AB is a diameter of larger circle. So larger circle has radius = 10, and the two smaller circles have radius = 5 Area of blue region = π(10)² − 2 × π(5)² = 100π − 50π = 50π General solution: Let r and R be radius of yellow and green circles respectively: CE = 2r, ED = 2R → CD = 2r+2R → radius of larger circle = r+R By intersecting chord theorem, we get: CE × ED = AE × EB 2r × 2R = 10 × 10 4rR = 100 rR = 25 Area of blue region = Area of larger circle − Areas of the 2 smaller circles = π(r+R)² − πr² − πR² = π(r² + 2rR + R²) − πr² − πR² = (r² + 2rR + R² − r² − R²)π = (2rR)π [ but rR = 25 from above ] = 50π

@marioalb9726 11 ай бұрын

Blue Shaded Area: A = ⅛ π c² A = ⅛ π 20² A = 50π cm² ( Solved √ ) Easier !!!

@lavrentizapadni747 Жыл бұрын

Here's an interesting observation! AS LONG AS THE CHORD AB = 20 UNITS, then it doesn't matter what the sizes of the circles are - the blue shaded area will always be 50pi square units!

@misterenter-iz7rz Жыл бұрын

It seems to be a very challenge puzzle.😊First y,g,b be the radii of yellow, green and blue circles, then OA=b, b=y+g, OE=b-2y=g-y, thus 100=AE^2=b^2-(g-y)^2=(g+y)^2-(g-y)^2=4gy, so gy=25. Our answer is (b^2-y^2-g^2)pi=((y+g)^2-y^2-g^2)pi=2gyxpi=2x25pi=50pi=157.1.😃

@PreMath Жыл бұрын

Excellent! Thank you! Cheers! 😀

@anestismoutafidis4575 Жыл бұрын

At first, we have to make the right construction of the three circles. Then we can recognize that the little circle has a diameter of 46,32, the middle circle has 132,72, and the big one 334,58. For calculating the blue shaded area, we have to make the subtraction of the area of the little circle and the middle circle, from the area of the big circle. That means: 334,58 - (132,72+46,32)= 155,5. The calculation can be fulfilled with the known calculation-type: A=(π•d^2)÷4. The blue shaded area has 155,5 area units.

@marioalb9726 Жыл бұрын

A=Ac-(A₁+A₂) A= ¼πD²-¼π(D₁²+D₂²) Intersecting chords theorem: (AB/2)²=D₁.D₂=10²=100 (D₁+D₂)²=D₁²+2D₁D₂+D₂²=D² D₁²+D₂²=D²-2D₁D₂ = D²-2.100 A= ¼πD² - ¼π(D²-200) A= ¼πD² - ¼πD² +50.π A= 50π =157,08cm² ( Solved √ )

@samuelraj4992 Жыл бұрын

I have a short method Sir.

@amiramy48 Жыл бұрын

I believe the following solution is simpler: note that R=r1 + r2 (where R , r1, r2 are the radii of the big, small, medium circles). Then note that OE = R-2r1 = r2-r1 Apply the Pythagorean theorem to OEA : (r1+r2)^2 = (r2-r1)^2 = 10^ 2 The terms of r1^2 and r2^2 on both sided cancel each other, yielding 4 r1 r2 = 100, so r1 r2 = 25 Now, the blue area is Pie (R^2 - r1^2 - r2^2) , and applying R= r1 + r2, again the square Terms disappear, yielding Area = pie * 2 r1 r2 = 50 pie

@shyamalarajagopalan2074 Жыл бұрын

Exactly , my solution was the same, too.

@marioalb9726 11 ай бұрын

Blue Shaded Area: A = ⅛ π c² A = ⅛ π 20² A = 50π cm² ( Solved √ ) Easier !!!