Can you find area of the Blue shaded region?

No video

Can you find area of the Blue shaded region? | (Equilateral triangle) |

Рет қаралды 8,429

3 ай бұрын

Learn how to find the area of the Purple shaded region. Important Geometry and Algebra skills are also explained: Pythagorean theorem; area of the circle formula; area of the triangle formula. 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
• Can you find area of t...
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
Can you find area of the Blue shaded region? | (Equilateral triangle) | #math #maths #geometry
Olympiad Mathematical Question! | Learn Tips how to solve Olympiad Question without hassle and anxiety!
#FindBlueArea #EquilateralTriangle #Semicircles #GeometryMath #PythagoreanTheorem
#MathOlympiad #IntersectingChordsTheorem #RightTriangle #RightTriangles
#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 #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
Andy Math
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.

Пікірлер: 58

@PreMath 3 ай бұрын

Correction: The answer is supposed to be 12√3−6π. My sincere apologies for that!

@marcelowanderleycorreia8876 3 ай бұрын

I was mad trying to find my mistake....kkkkkk Very goos professor!!

@CharlesB147 3 ай бұрын

Everyone makes mistakes. Keep on submitting good work, though, I really enjoy doing these problems.

@PreMath 3 ай бұрын

@@CharlesB147 Thanks dear ❤🙏

@lillaschwarte8760 3 ай бұрын

Thank you! I was wondering, because the blue area cannot be bigger than the area of the three semicircles.

@PreMath 3 ай бұрын

@@lillaschwarte8760 Thanks for your understanding. ❤🙏

@cristi3383 3 ай бұрын

Area of triangle ABC is actually 12sqrt3, not 24sqrt3, the mistake is only the calculation

@RK-tf8pq 3 ай бұрын

Yes, I noticed that too.

@unknownidentity2846 3 ай бұрын

Let's find the area: . .. ... .... ..... We start with the calculation of the radius r of the yellow semicircles: A = πr²/2 (2π)cm² = πr²/2 4cm² = r² ⇒ r = 2cm Since every pair of semicircles has exactly one point of intersection, the distances of their centers are all the same: PQ = PS = QS = r + r = 4cm It is obvious that PQ (PS, QS) is perpendicular to BC (AB, AC). Since all interior angles of the equilateral triangle ABC are 60°, the triangles APS, BPQ and CQS are congruent 30°-60°-90° triangles. So we can conclude: PQ = PS = QS = 4cm ⇒ AP = BQ = CS = 4cm/√3 ∧ AS = BP = CQ = 8cm/√3 Therefore the side length and the area of the equilateral triangle ABC turn out to be: s(ABC) = AP + BP = AS + CS = BQ + CQ = 4cm/√3 + 8cm/√3 = 12cm/√3 A(ABC) = (√3/4)*[s(ABC)]² = (√3/4)*(12cm/√3)² = (√3/4)*(144/3)cm² = (12√3)cm² Now we can calculate the size of the blue area: A(blue) = A(ABC) − 3*A(semicircle) = (12√3 − 6π)cm² ≈ 1.935cm² Best regards from Germany

@hongningsuen1348 3 ай бұрын

It is obvious that PQ (PS, QS) is perpendicular to BC (AB, AC) but a proof is needed.The proof of angle PQB = 90 is made by constructing triangle PQE which is a right-angled triangle as QE is centre Q to point of tangency E. In triangle PQE, sin(angle EPQ) = 2/4 hence angle EPQ = 30 and then angle PQB = 180 - 60 - 30 = 90. Similarly angles APS and CSQ are proven to be 90.

@unknownidentity2846 2 ай бұрын

@@hongningsuen1348 Thanks for your feedback and you are absolutely right. I only used symmetry considerations to derive that PQ (PS, QS) is perpendicular to BC (AB, AC). Indeed this is not enough. After obtaining the side lengths PQ=4 and EQ=2 and the angle ∠PEQ=90° it can be concluded that EPQ and therefore also BPQ are 30°-60°-90° triangles. That should have been shown first before using symmetry arguments. Best regards from Germany

@Mathphysics-iy4nh 3 ай бұрын

You are absolutely unforgotten I always see you both making myself free and challenged to your question.

@flesby 3 ай бұрын

Area of Equilateral Triangle (A) = (√3/4)a² So all the is needed is the side length of the equilateral triangle to calculate its area an subtract 6pi. Draw the line PQ, since the radius is 2 that line must have a length of 4. This line now is one side of the right triangle PQB, where the other 2 sides are parts of the sides of the given triangle. By the information provided we know, that this right triangle we just constructed must have angles of 90, 60 and 30 degrees. - We know (maybe from previous Videos on this channel) that the side lengths of an 30-60-90 Triangle have a ratio of 1:√3:2 Since PQ (the second longest side) has a length of 4, QB equals 4/√3 and PQ = 8/√3 Thus the side length = 12/√3 Therefor Area of Equilateral Triangle (A) = (√3/4)*(12/√3)² = (√3/4)*(144/3) =(√3/4)*48 = 20,785 (rounded to the 3rd decimal) => 20,785 - 6pi is the blue area. (roughly 1.935) So we do not need to fuss around with sin

@Fan_di_Euclide-1234 3 ай бұрын

After finding the lenght of the radius, we know that the triangles CSQ and BPQ are congruent, because they have a right angle, a congruent acute angle and a congruent side( SQ=PQ)----> CB= 4square root of 3. And Area blu= 12square root of 3 - 6pi.

@timothyprice5240 3 ай бұрын

I agree. PreMath made an error somewhere

@unknownidentity2846 3 ай бұрын

I agree as well. The expression (1/2)*(4√3)*(4√3)*(√3/2) of the area of the triangle ABC (in cm²) is correct, but that expression turns out to be 12√3. I guess this is good news: It means that PreMath is a human being.🙂 Best regards from Germany

@CharlesB147 3 ай бұрын

Right on the tail end, your area calculation should result in 12sqrt(3), not 24sqrt(3). The rest up to there is good, you just forgot to divide by 2 twice. 😅 So the final answer should be 12sqrt(3)-6pi. Thank you for the video as always, though.

@ludosmets2018 3 ай бұрын

I first calculated the area of the blue sector in the center (area of triangle PQS minus 3 circle sectors of 60°). The total area of the other blue parts is 2 times the area of the center blue part: 3 (1/2) + 3 (1/6) = 2. So multiply the area of the center blue sector by 3 and you have the answer.

@quigonkenny 3 ай бұрын

Blue area = area of ∆ABC minus three yellow semicircle areas. - semicircle areas: given (= 6π) - ∆ABC area : unknown - Need: ∆ABC side length x AB = BC = CA = x Semicircle S: A = πr²/2 2π = πr²/2 r² = (2/π)2π = 4 r = √4 = 2 Let M be the poont of tangency between semicircle P and CA, N be the point of tangency between semicircle Q and AB, and T be rhe point of tangency between semicircles P and Q. As ∆ABC is an equilateral triangle, ∠ABC = ∠BCA = ∠CAB = 60°. Draw MP, QN, and QP. As the centers of two tangent circles and their point of tangency are colinear, and as all three semicircles are fongruent, point T is the midpoint of QP. PM = QN = PT = QT = r = 2. As ∠MAP = ∠NBQ = 60°, ∠PMA = ∠QNB = 90°, and PM = QN, ∆PMA and ∆QNB are congruent 30-60-90 special tight triangles. In a 30-60-90 special right triangle, if the short leg (opposite 30°) is length k, the long leg (opposite 60°) is length √3k and the hypotenuse (opposite 90°) is 2k. Thus PM = QN = √3k = 2. √3k = 2 k = 2/√3 NB = k = 2/√3 PA = 2k = 4/√3 As QN = r, QP = 2r, and ∠PNQ = 90°, ∆PNQ is also a 30-60-90 special right triangle, and PN = √3r = 2√3. AB = PA + PN + NB s = 4/√3 + 2√3 + 2/√3 s = √3(4/3+2+2/3) s = 4√3 Blue area = ∆ABC - 6π A = bh/2 - 6π A = (4√3)(2√3(√3))/2 - 6π A = (4√3)(6)/2 - 6π A = 12√3 - 6π cm² ≈ 1.935 cm²

@Ddntitmattrwhtuthnk 3 ай бұрын

I'm glad I finally read your comment about the correction. I checked and rechecked my math. Thank you.

@MMmaths8800 3 ай бұрын

King of math "Premath" 👑👑

@LuisdeBritoCamacho 3 ай бұрын

1) Radius (R) of Yellow Semicircles: 2) Pi*R^2 = 4*Pi ; R^2 = 4 ; R = 2 cm 3) Area of Equilateral Triangle [PQS] (A) with Side Length = 2*R = 4 cm; using Heron's Formula: 4) A ~ 6,93 Square Cm 5) Area of 3 Equal Right Triangles : [CSQ] ; [APS] ; [BPQ]. The Triangles are (30º ; 60º ; 90º) 6) BQ = (4 * tan(30º)) cm ~ 2,31 cm 7) Area of each Triangle (AT) 8) 2*AT = (4 * [4 * tan(30º)]) ; 2*AT = 16*tan(30º) ; AT = 8*tan(30º) ~ 4,62 Square Cm 9) Area of all 3 Right Triangles = AT [CSQ] + AT [APS] + AT [BPQ] = 24*tan(30º) ~ 13,86 Square Cm 10) Area of the Great Triangle (GT) = 24*tan(30º) + 6,93 ; GT ~ 13,86 + 6,63 ~ 20,79 Square Cm 11) Blue Region Area (BRA) = 20,79 - 6*Pi ~ 20,79 - 18,85 ~ 1,94 Square Cm. 12) Answer : The Blue Region Area is equal to 1,94 Square Centimeters.

@darknesssqueeze 3 ай бұрын

SQP is equilateral triangle so reverse calculation to find PQ,PE,(QB=AP),EB and then use sqrt3/4 *( AB*AB) to find ABC triangle and minus 6pi will be answer

@jamestalbott4499 3 ай бұрын

I took a different, exploiting the two tangent theorem and special triangles. But, I got there in the end. Thank you!

@StuartSimon 3 ай бұрын

It should have been stated in the given information that each semicircle was tangent to one of the sides. Else we need to prove it. It is not obvious from the diagram.

@AmirgabYT2185 3 ай бұрын

6(2√3-π)≈1,92

@sergeyvinns931 3 ай бұрын

Drav a line PS, SQ, PQ. Area trangle PSQ = \/p(p-2r)^3. p=3r, r=\/2, area PSQ=4\/3, area ASP=area CQS=area BPQ =8\/3 Area blue trangle = 4\/3+8\/3=12\/3, area of the blue shaded region = 12\/3-6(3,14)=1,9356!

@juanalfaro7522 2 ай бұрын

The area of the ABC triangle is actually 12sqrt(3) not 24sqrt(3). This makes the Blue Area = 12sqrt(3) - 6Pi =1.935

@pedllz 3 ай бұрын

I did 3 equilaterals triangles of side 2/sin60 per semicircle...so the final answer is 18/sin60 - 6pi

@dinusha7127 3 ай бұрын

Wow

@misterenter-iz7rz 3 ай бұрын

Interesting figure 😊, radius of semicircle are all 2 clearly, thus CS=4/sqrt(3) and CQ=4/(sqrt(3)/2)=8/sqrt(3), therefore the size of the triangle is 12/sqrt(3) and the area is 1/2 (12/sqrt(3))^2 sqrt(3)/2-6pi=1/2 144/3 sqrt(3)/2-6pi=12sqrt(3)-6pi.😊

@desrajminhas4509 3 ай бұрын

Area of triangle ABC will be 12underroot3not 24 underroot3

@marcgriselhubert3915 3 ай бұрын

All right.

@murdock5537 3 ай бұрын

6(2√3 - π)

@billferrol4202 3 ай бұрын

Once you've learned that it's an equilateral triangle its area is (length of the sides squared times √3) divided by 4, which is the same as 1/2absin60º therefore the area is 12√3 NOT 24√3!

@maxforsberg8852 3 ай бұрын

The introduction of point E is redundant. We can figure out AP by dividing the radius by cos30 and likewise we get PB by dividing 2 times the radius by cos30. After that you can get the side length of the triangle by adding AP and PB and the rest follows. Keep up the good work.

@hongningsuen1348 3 ай бұрын

E is not redundant. In your calculation, you assume angle PQB = 90 such that triangle PQB is a right-angled triangle. The proof of angle PQB = 90 is made by constructing triangle PQE which is a right-angled triangle as QE is centre to point of tangency. In triangle PQE, sin(angle EPQ) = 2/4 hence angle EPQ = 30 and then angle PQB = 180 - 60 - 30 = 90.

@addisuanbessie3615 3 ай бұрын

Please check triangle EBQ, its an isosceles triangle since eq and qb is a radius of semi circle ,so it can't be right angle triangle

@zehradiyab3439 3 ай бұрын

I found AB in another way *The triangle SPQ is equilateral triangle which its side=r=2 *angel SPB=90°(because AB is tangent of the circle which SP is diameter. Angle(QPB)=Angel(SPB)-Angel (SPQ) 30°=60°-90°= Similar angle(BQP)=90° In the right angled triangle (BQP) PB=y, QB=x, y=2x x²+4²=y² x²+16=(2x)² 4x²-x²=16 3x²=16 x=4sqrt(3)/3 AB=x+y=x+2x=3x=4sqrt(3)

@pedllz 3 ай бұрын

ABC area is 12*Sqr(3) ...not 24*