Can you find area of the Yellow shaded region? | (Rectangle) |

  Рет қаралды 27,467

PreMath

PreMath

19 күн бұрын

Learn how to find the area of the Yellow shaded region. Important Geometry and Algebra skills are also explained: area of the circle formula; area of the rectangle 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 Yellow shaded region? | (Rectangle) | #math #maths | #geometry
Olympiad Mathematical Question! | Learn Tips how to solve Olympiad Question without hassle and anxiety!
#FindYellowArea #Rectangle #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
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: 85
@bhikhubhavaghela1507
@bhikhubhavaghela1507 17 күн бұрын
Good explanation
@PreMath
@PreMath 17 күн бұрын
Thanks for liking❤️
@ChuzzleFriends
@ChuzzleFriends 17 күн бұрын
Label the radius of the semicircle with center Q (small) as r. The gaps between the top of the semicircles and the ceiling are 7 (small) and 4 (medium). So, the semicircle with center O (large) has a radius of r + 7. Using the radius of the large semicircle, we can find the radius of the semicircle with center P (medium). Medium semicircle radius = Large semicircle radius - 4 = (r + 7) - 4 = r + 3 The ceiling is the length of rectangle ABCD. So, by the Parallelogram Opposite Sides Theorem, AB = 38. The bottom side of ABCD also is formed by the diameters of the semicircles. So: 2r + 2(r + 3) + 2(r + 7) = 38 2r + (2r + 6) + (2r + 14) = 38 6r + 20 = 38 6r = 18 r = 3 So, the radii of the semicircles are as follows: Small: 3 Medium: 6 Large: 10 But the radius of the large semicircle is also the width of the rectangle. Find the area of the yellow region. Yellow region area = Rectangle ABCD Area - Total Area of Three Semicircles A = lw = 38 * 10 = 380 A = (πr²)/2 = (π * 3²)/2 = (9π)/2 A = (π * 6²)/2 = (36π)/2 = 18π A = (π * 10²)/2 = (100π)/2 = 50π Total Area = (9π)/2 + 18π + 50π = 4.5π + 18π + 50π = 72.5π = (145π)/2 Yellow region area = 380 - (145π)/2 So, the area of the yellow region is 380 - (145π)/2 square units, a. w. a. (760 - 145π)/2 square units (exact), or about 152.23 square units (approximation).
@PreMath
@PreMath 17 күн бұрын
Excellent! Thanks for sharing ❤️
@tamirrahman7929
@tamirrahman7929 17 күн бұрын
It was quite easy 😊😊
@PreMath
@PreMath 17 күн бұрын
Thanks for the feedback ❤️
@misterenter-iz7rz
@misterenter-iz7rz 17 күн бұрын
Thanks for always original and exciting puzzles,🎉. a+b+c=19, a=c-7, b=c-4, so 3c-11=19, c=10, a=3, b=6, therefore the area is 380-1/2(100+36+9)pi=380-145/2 pi.😊
@PreMath
@PreMath 17 күн бұрын
Great job! You are very welcome! Thanks for the feedback ❤️
@Ibrahimfamilyvlog2097l
@Ibrahimfamilyvlog2097l 17 күн бұрын
wow very nice sharing sir✨
@PreMath
@PreMath 17 күн бұрын
Thanks for visiting🌹
@anatoliy3323
@anatoliy3323 16 күн бұрын
Nice to meet you again,Mr PreMath. Thanks for your lesson!
@manuelantoniobahamondesa.3252
@manuelantoniobahamondesa.3252 11 күн бұрын
Excelente!!!
@johnjones8580
@johnjones8580 17 күн бұрын
The math is easier if you define r from the *smallest* circle. That way, you do not have to be concerned with negative values. 2(r+(r+3)+(r+7) = 38 2(3r+10)=38 6r+20=38 6r=18 r=3 r1=3, r2=6, r3=10
@PreMath
@PreMath 17 күн бұрын
Excellent! Thanks for the feedback ❤️
@yalchingedikgedik8007
@yalchingedikgedik8007 16 күн бұрын
Very nice and enjoyable Thanks PreMath Thanks prof. ❤❤❤❤❤ With my glades
@MrPaulc222
@MrPaulc222 17 күн бұрын
Radii of R, R-4, and R-7. 2R + 2(R-4) + 2(R-7) = 38 6R - 22 = 38, so R=10 R=10=height of rectangle. Total rectangle area = 38*10=380 Semicircle areas are ((100pi)+(36pi)+9pi))/2 = 72.5pi. Yellow shaded area = 380-72.5pi un^2. 380-227.77=152.23un^2.
@PreMath
@PreMath 17 күн бұрын
Excellent!👍 Thanks for sharing ❤️
@quigonkenny
@quigonkenny 17 күн бұрын
Let the radius of semicircle O be r. Let T be the point of tangency between the circumference of semicircle O and CD. As CD is tangent to semicircle O at T, ∠CTO = 90°, so as BC and OT are thus parallel, then BC = OT = r. Carrying this over the other two semicircles, we can see that the radius of semicircle P is r-4 and the radius of semicircle Q is r-7. Looking at AB and CD, which as opposite sides of the rectangle will be equal in length, the length of AB is 2(r-7)+2(r-4)+2r and the length of CD is 38. 2(r-7) + 2(r-4) + 2r = 38 6r - 14 - 8 = 38 6r = 60 r = 60/6 = 10 The yellow shaded area eill be equal to the area of the rectangle ABCD minus the areas of the three semicircles. Yellow shaded area: Aₛ = hw - πr₁²/2 - πr₂²/2 - πr₃²/2 Aₛ = BC(CD) - (π/2)(r₁²+r₂²+r₃²) Aₛ = 10(38) - (π/2)((r-7)²+(r-4)²+r²) Aₛ = 380 - (π/2)(3²+6²+10²) Aₛ = 380 - (π/2)(9+36+100) Aₛ = 380 - 145π/2 ≈ 152.23 sq units
@PreMath
@PreMath 17 күн бұрын
Excellent! Thanks for sharing ❤️
@himo3485
@himo3485 17 күн бұрын
FO=OB=r EP=PF=r-4 AQ=QE=r-7 2(r-7)+2(r-4)+2r=38 2r-14+2r-8+2r=38 6r=60 r=10 Yellow shaded area = 38*10 - (3r*3*π*1/2 + 6*6*π*1/2 + 10*10*π*1/2) = 380 - 72.5π
@PreMath
@PreMath 17 күн бұрын
Excellent! Thanks for sharing ❤️
@murdock5537
@murdock5537 17 күн бұрын
Very nice, many thanks, Sir!
@PreMath
@PreMath 17 күн бұрын
Glad to hear that! You are very welcome! Thanks for the feedback ❤️🙏
@KipIngram
@KipIngram 16 күн бұрын
Ok, let the circle radii, from left to right, be A, B, and C. So, we have A + B + C = 19 -A + C = 7 - B + C = 4 I hope it's obvious how we got those. The solution is A = 3, B = 6, C = 10. So the full yellow rectangle would have area 38*10 = 380. We need to subtract pi/2 times 3^2 = 9 plus 6^2 = 36 plus 10^2 = 100, so 145. So the answer is 380 - 72.5*pi = 152.23.
@jamestalbott4499
@jamestalbott4499 17 күн бұрын
Thank you!
@PreMath
@PreMath 17 күн бұрын
You are very welcome! Thanks, my dear friend James ❤️🙏
@Birol731
@Birol731 17 күн бұрын
My way of solution ▶ radius small semicircle: r₁ radius middle semicircle: r₂ radius large semicircle: r₃ ⇒ 2r₁+2r₂+2r₃= 38 r₁+r₂+r₃= 19 r₃= r₂+4 r₂+4= r₁+7 r₂= r₁+3 ⇒ r₁+r₂+r₃= 19 r₁+ r₁+3 + r₂+4= 19 2r₁+ (r₁+3)= 19-7 3r₁= 9 r₁= 3 r₂= 6 r₃= 10 for the rectangular: a= 38 length units b= r₃ b= 10 Arectangular= a*b = 38*10 = 380 square units Ayellow= 380- πr₁²/2 - πr₂²/2 - πr₃²/2 = 380 - π/2(3²+6²+10²) = 380- π/2(9+36+100) Ayellow = 380- 145π/2 Ayellow = 380 - 72,5 π Ayellow ≈ 152,23 square units
@PreMath
@PreMath 17 күн бұрын
Excellent! Thanks for sharing ❤️
@devondevon4366
@devondevon4366 16 күн бұрын
380 - 72.5 pi or 152.2345 Let the width of the rectangle = n then the diameter of the largest semi-circle = 2n the diameter of the middle semi-circle = n-4 + n-4 = 2n-8 the diameter of the smallest semi-circle = n-7 + n-7 = 2n -14 Hence, 2n + 2n-8 + 2n-14 =38 6n - 22 =38 6n = 60 n=10 Hence, the radius of the largest semi-circle = 10. Hence , its area = 10^2pi/2 = 50pi Hence, the radius of the middle semi-circle = 6 (10-4), Hence, its area = 36pi/2 = 18pi Hence, the radius of the smallest semi-circle =3 (10-7). Hence, its area = 9 pi/2 = 4.5 pi Hence, the TOTAL area of all the semi-circles is 72.5 pi (50 + 18 + 4.5) Since n= 10, then the width of the rectangle = 10 Hence, its area = 380 The shaded region is the difference between the two Hence, area of shaded region = 380 - 72.5 pi ot 152.2345
@PreMath
@PreMath 16 күн бұрын
Super!👍 Thanks for sharing ❤️
@cyruschang1904
@cyruschang1904 17 күн бұрын
2(a - 7) + 2(a - 4) + 2a = 38 6a = 60 a = 10 area sought = 38 x 10 - 9π/2 - 36π/2 - 100π/2 = 380 - 145π/2
@PreMath
@PreMath 17 күн бұрын
Excellent Thanks for sharing ❤️
@cyruschang1904
@cyruschang1904 17 күн бұрын
@@PreMath Thank you 🙂
@htvz
@htvz 15 күн бұрын
interesante
@santokhsidhuatla7045
@santokhsidhuatla7045 17 күн бұрын
Let large circle radius =R3 Middle R2,small. R1 R3=R2+4 R2=R1+3 2(R1+R2+R3)= 38 2(R1+R1+3+R1+3+4)=38 3R1+10=19 3R1=19-10=9 R1=9/3=3 Now area of Rectangle =38x10=380 square unit Deduction = Pi /2(3^2+6^2+10^2) Pi/2(9+36+100)=pi/2x 145= 227.765 square unit Net yellow area=380-227.765 =152.235 square unit
@PreMath
@PreMath 17 күн бұрын
Excellent! Thanks for sharing ❤️
@shakirhamoodi5009
@shakirhamoodi5009 12 күн бұрын
380- (pi / 2) x 145 Which is approximately 152.13
@sergeyvinns931
@sergeyvinns931 17 күн бұрын
R3=R1+7, R3=R2+4, 2R3=R1+R2+11, 2R3=38-2(R1+R2), 3(R1+R2)=38-11, 3(R1+R2)=27, R1+R2=9, 2R3=R1+R2+11, 2R3=20, R3=10, R1=R3-7=3, R2=R3-4=6, Area of the Yellow shaded region = 380-(100+36+9)pi/2=380-145pi/2=152,234.
@PreMath
@PreMath 17 күн бұрын
Excellent! Thanks for sharing ❤️
@giuseppemalaguti435
@giuseppemalaguti435 17 күн бұрын
r1+7=r2+4=R...2r1+2r2+2R=38..r1+r2+R=19..(2R-11)+R=19..R=10..Ay=38R-(π/2)(R^2+(R-7)^2+(R-4)^2)=38R-(π/2)(3R^2-14R+49-8R+16)..Ay=380-(π/2)145
@PreMath
@PreMath 17 күн бұрын
Excellent! Thanks for sharing ❤️
@Jorick_73
@Jorick_73 17 күн бұрын
I'm proud that i did all the same in my mind just whilst looking at the picture in 5 minutes. 51y.o.
@PreMath
@PreMath 17 күн бұрын
@@Jorick_73 Bravo!
@guerrino50
@guerrino50 15 күн бұрын
R1=3×2=6 R2=6×2=12 R3=10×2=20 H=10×38=380 Area
@AmirgabYT2185
@AmirgabYT2185 17 күн бұрын
S=5(152-29π)/2≈152,14
@PreMath
@PreMath 17 күн бұрын
Excellent! Thanks for sharing ❤️
@hakopajemyan334
@hakopajemyan334 17 күн бұрын
R1=10. R2=6. R3=3 S=152.35
@PreMath
@PreMath 17 күн бұрын
Excellent! Thanks for sharing ❤️
@wackojacko3962
@wackojacko3962 17 күн бұрын
I would've been the Appointed Water Commissioner of the Roman Empire building Aqueducts. Now I accept the post Era Titular position. Knowing how to do this problem I think I've earned it. 🙂
@PreMath
@PreMath 17 күн бұрын
👍😀 Thanks for the feedback ❤️
@sergiosereno1489
@sergiosereno1489 17 күн бұрын
da = r1 + 7 da = r2 + 4 da = r3 3*da = r1 + r2 + r3 + 11 3*r3 = r1 + r2 + r3 + 11 2*r3 = r1 + r2 + 11 -> r1 + r2 = 2*r3 - 11 ab = 2*(r1 + r2 + r3) ab = 2*(r1 + r2 + r3) 38 = 2*(r1 + r2 + r3) r1 + r2 + r3 = 19 r1 + r2 + r3 = 19 2*r3 - 11 + r3 = 19 3*r3 = 30 r3 = 10 r3 = r2 + 4 r2 = 10 - 4 r2 = 6 r3 = r1 + 7 r1 = 10 - 7 r1 = 3 ret_area = da*38 = r3*38 = 10*38 = 380 yellow_area = 380 - (9*p1 + 36*pi + 100*pi)/2 yellow_area = 380 - (pi + pi*49 + pi*225)/2 yellow_area = 380 - (145*pi)/2 yellow_area = 152,3
@PreMath
@PreMath 17 күн бұрын
Excellent! Thanks for sharing ❤️
@unknownidentity2846
@unknownidentity2846 17 күн бұрын
Let's find the area: . .. ... .... ..... From the known length of AB we can conclude: AB = 38 AQ + QE + EP + PF + FO + OB = 38 R(left) + R(left) + R(middle) + R(middle) + R(right) + R(right) = 38 2*R(left) + 2*R(middle) + 2*R(right) = 38 R(left) + R(middle) + R(right) = 19 [R(right) − 7] + [R(right) − 4] + R(right) = 19 3*R(right) − 11 = 19 3*R(right) = 30 ⇒ R(right) = 10 R(middle) = R(right) − 4 = 10 − 4 = 6 R(left) = R(right) − 7 = 10 − 7 = 3 Now we are able to calculate the size of the yellow area: A(yellow) = A(rectangle) − A(semicircle,left) − A(semicircle,middle) − A(semicircle,right) = AB*BC − π*R²(left)/2 − π*R²(middle)/2 − π*R²(right)/2 = AB*R(right) − (π/2)*[R²(left) + R²(middle) + R²(right)] = 38*10 − (π/2)*(3² + 6² + 10²) = 380 − (π/2)*(9 + 36 + 100) = 380 − (145/2)π ≈ 152.23 Best regards from Germany
@PreMath
@PreMath 17 күн бұрын
Excellent! You are the best! 😀 Thanks for sharing ❤️
@Dr_piFrog
@Dr_piFrog 13 күн бұрын
152.24
@Martin-hi7wx
@Martin-hi7wx 14 күн бұрын
But we were not told the main figure was a rectangle.
@sergioaiex3966
@sergioaiex3966 17 күн бұрын
2a + 2b + 2c = 38 (÷2) a + b + c = 19 ... ¹ a + 7 = b + 4 a - b = 4 - 7 a - b = -3 ... ² a + 7 = c ... ³ a + b + c = 19 a - b = -3 =========== 2a + c = 16 ... ⁴ 2a + a + 7 = 16 3a = 9 *a = 3* *b = 6* *c = 10* *A Rectangle = 380 Square Units* A SCa = π 3²/2 = 9/2 π A SCb = π 6²/2 = 18π A SCc = π 10²/2 = 50π Yellow Shaded Region = 380 - (9/2 π + 18π + 50π) *YSR = 380 - 72,5π Square Units* YSR = 152,2345326147 Square Units *YSR = 152,2345 Square Units*
@PreMath
@PreMath 17 күн бұрын
Excellent! Thanks for sharing ❤️
@NurayQismt
@NurayQismt 17 күн бұрын
İm Azerbaijani but im understand this so well thank you so much❤
@PreMath
@PreMath 17 күн бұрын
Glad to hear that! You are very welcome! Thanks dear❤️
@sorourhashemi3249
@sorourhashemi3249 17 күн бұрын
152.35
@PreMath
@PreMath 17 күн бұрын
Excellent! Thanks for sharing ❤️
@mindless-pedant
@mindless-pedant 13 күн бұрын
Sorry. Switched off. "SemI". Chalk scraping across a blackboard.
@ianweckhorst3200
@ianweckhorst3200 15 күн бұрын
After finding it myself before the video, I found out the decimal actually went 2345 which was funny
@sandytanner9333
@sandytanner9333 17 күн бұрын
380-72.5π
@PreMath
@PreMath 17 күн бұрын
Excellent! Thanks for sharing ❤️
@Xyz-sd7ub
@Xyz-sd7ub 17 күн бұрын
152.20 Sq. Units
@PreMath
@PreMath 17 күн бұрын
Excellent! Thanks for sharing ❤️
@LuisdeBritoCamacho
@LuisdeBritoCamacho 17 күн бұрын
STEP-BY-STEP RESOLUTION PROPOSAL : 01) DC = AB = 38 02) Let Radius of Big Semicircle = R(1) = CB = R 03) Radius of Medium Semicircle = R(2) = (R - 4) 04) Radius of Small Semicircle = R(3) = (R - 7) 05) AE = 2 * (R - 7) = 2R - 14 06) EF = 2 * (R - 4) = 2R - 8 07) FB = 2 * R = 2R 08) AB = 2R - 14 + 2R - 8 + 2R 09) 2R - 14 + 2R - 8 + 2R = 38 ; 6R - 22 = 38 ; 6R = 60 ; R = 60 / 6 ; R = 10 10) Rectangle [ABCD] Area = 38 * 10 = 380 sq un 11) Big Semicircle Area with Radius equal 10 = 50Pi sq un 12) Medium Semicircle Area with Radius equal 6 = 18Pi sq un 13) Small Semicircle Area with Radius equal 3 = 9Pi / 2 sq un 14) Sum of all 3 Semicircles Area = (50Pi + 18Pi + 9Pi/2) = (50 + 18 + 9/2)*Pi = 145Pi / 2 ~ 228 sq un (~ 227,77 sq un) 15) Yellow Are = 380 - 228 ~ 152 sq un ANSWER : The Yellow Shaded Area approx. equal to 152 Square Units. (Exact Form = (380 - 145Pi/2) Square Units or more precisely 152,2345 Square Units). Greetings from Cordoba Caliphate!!
@PreMath
@PreMath 17 күн бұрын
Super job! Thanks for sharing ❤️
@LuisdeBritoCamacho
@LuisdeBritoCamacho 17 күн бұрын
@@PreMath , I am here to learn Maths, wich is the Love of My Life. Thanks for sharing your knowledge and the Importance of Mathematical Thinking in our Lives.
@PreMath
@PreMath 17 күн бұрын
@@LuisdeBritoCamacho Thank you so much, my dear friend. Take care🌹
@Nothingx303
@Nothingx303 15 күн бұрын
It was quite juicy 😋 😊😊
@jackreacher.
@jackreacher. 17 күн бұрын
I did this in grade six.
@PreMath
@PreMath 17 күн бұрын
Excellent!👍 Glad to hear that! Thanks for sharing ❤️
@jackreacher.
@jackreacher. 16 күн бұрын
@@PreMath Math Teachers rule.
@georgecooke9010
@georgecooke9010 16 күн бұрын
NO!
@hakopajemyan334
@hakopajemyan334 17 күн бұрын
152.35
@PreMath
@PreMath 17 күн бұрын
Thanks for sharing ❤️
@kalavenkataraman4445
@kalavenkataraman4445 17 күн бұрын
152.35 Sq. Units
@PreMath
@PreMath 17 күн бұрын
Excellent! Thanks for sharing ❤️
Spain | A Nice Algebra Problem | Math Olympiad
11:27
Math Hunter
Рет қаралды 34 М.
Cool Items! New Gadgets, Smart Appliances 🌟 By 123 GO! House
00:18
123 GO! HOUSE
Рет қаралды 16 МЛН
Best father #shorts by Secret Vlog
00:18
Secret Vlog
Рет қаралды 22 МЛН
Looks realistic #tiktok
00:22
Анастасия Тарасова
Рет қаралды 103 МЛН
An incredible semicircle problem!
7:27
MindYourDecisions
Рет қаралды 227 М.
FABIANO SACRIFICES 2 ROOKS AND WINS IN 9 MOVES!
6:07
Chess.com
Рет қаралды 380 М.
Russian Math Olympiad | A Very Nice Geometry Problem
14:34
Math Booster
Рет қаралды 90 М.
Cool Items! New Gadgets, Smart Appliances 🌟 By 123 GO! House
00:18
123 GO! HOUSE
Рет қаралды 16 МЛН