Can you find area of the Green shaded Trapezoid? | (Trapezium) |

Подписаться 412 тыс.

Просмотров 14 тыс.

50% 1

Learn how to find the area of the Green shaded Trapezoid in the semicircle. Important Geometry and Algebra skills are also explained: Pythagorean Triples; Pythagorean theorem; area of the trapezoid; similar triangles. 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 Green shaded Trapezoid? | (Trapezium) | #math #maths #geometry
Olympiad Mathematical Question! | Learn Tips how to solve Olympiad Question without hassle and anxiety!
#GreenTrapezoidArea #Trapezoid #Trapezium #Semicircle #GeometryMath #PythagoreanTheorem #SimilarTriangles #PythagoreanTriples
#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.

Опубликовано:

30 сен 2024

Ссылка:

Скачать:

Готовим ссылку...

Добавить в:

Мой плейлист

Посмотреть позже

Комментарии : 57

@mathbynisharsir5586 7 месяцев назад

Very Very useful video sir 👍

@PreMath 7 месяцев назад

Thanks and welcome🌹❤️

@Ensign_Cthulhu 6 месяцев назад

2:44 No. You cannot assume that the two triangles are congruent because you cannot assume that AF = EB unless OF = OE. And that was not stated as a starting condition of the problem. You have to prove it and show why it is the case. The appearance of visual symmetry isn't good enough, especially since you've already warned us that the diagram is not to scale.

@johnbrennan3372 6 месяцев назад

Triangles dfo and coe are congruent so |fo|= |oe|=7/2.Then in triangle dfo fo=7/2, do=(2x/7)/2 and df = sqroot (225- x^2)? Using Pythagorous you end up with x=9 etc.

@71littlerabbit 6 месяцев назад

@@johnbrennan3372you can't say that, unless O is located exactly in the middle of EF...

@71littlerabbit 6 месяцев назад

CB can't be equal to AB!!!! The only case when this happens would be if O is at the same distance from F and from E. Otherwise, the assumption of CB=AD is wrong!

@md.abdurrahmantarafder2256 6 месяцев назад

Right.

@johnbrennan3372 6 месяцев назад

Triangles dfo and Coe are congruent,so fo=oe.

@robertlynch7520 7 месяцев назад

I like your solution, yet I approached it with a different 'knowns' point of view. Basically, I noted that the DC segment, if divided by a not-shown [O→top] vertical line, is always symmetrically divided in half. Always. Each half, I called (𝒂) [0.1] 2𝒂 = 7 [0.2] 𝒂 = 3.5 Therefore, can write [1.1] 15² - (𝒓 - 𝒂)² = 𝒉² = 𝒓² - 𝒂² … with numbers [1.2] 225 - (𝒓 - 3.5)² = 𝒓² - 3.5² … expanding [1.3] 225 - 𝒓² + 7𝒓 - 3.5² = 𝒓² - 3.5² … cancelling and rearranging [1.4] 225 - 2𝒓² + 7𝒓 = 0 … is quadratic so [1.5] 𝒓 = [12.5 or -9] With the positive 𝒓, 𝒉 is [2.1] 𝒉 = √(12.5² - 3.5²) [2.2] 𝒉 = √144 [2.3] 𝒉 = 12 So, the area is the sum of the square, and the triangle [3.1] area = (□ = 7 × 12) + (△ = ½ (𝒓 - 𝒂 = 9) × 12) [3.2] area = 84 + 54 [3.3] area = 138 And that's the solution! ⋅-⋅-⋅ Just saying, ⋅-⋅-⋅ ⋅-=≡ GoatGuy ✓ ≡=-⋅

@PreMath 7 месяцев назад

Super GoatGuy! Thanks ❤️

@scottdort7197 6 месяцев назад

i used the proportions of the sides of the semi circle. That give y^2 = (r+3.5)*(r-3.5). That yields y=(r^2-12.25)^1/2. Plug that into Pythagoras you get (r-3.5)^2 + r^2 -12.25 =225. Simplify that and solve for r you get r equals 12.5 of -9. Of course the -9 is invalid. If r = 12.5, y = 12. Plug that into the trapezoid area equation and you get 138 (of course). Good problem Premath. You keep my 61 year old brain in shape.

@PreMath 6 месяцев назад

Excellent! Glad to hear that! Thanks for the feedback❤️

@kaziummeruman524 7 месяцев назад

Me be like😂- If i use pythagoras theory,, so the length and hight of the triangle 🔺DBA will be 9 and 12 So,the area of the triangle is 54 And the area of the rectangle is 12×7 = 84 So, GREEN TRAPEZOID AREA IS 54+84= 138

@PreMath 7 месяцев назад

Excellent! Thanks ❤️

@prossvay8744 7 месяцев назад

Area of the green trapezoid=1/2(7+9+7)(12)=138 square units. ❤ Thanks dir

@PreMath 7 месяцев назад

Excellent! You are very welcome! Thanks ❤️🌹

@marcgriselhubert3915 7 месяцев назад

I propose something different (I don't say that it is better). Let'a name t = angleAOD and t' = angleDOC, we have t + t' + t = Pi radians, so t' = Pi - 2.t, and R is the radius of the circle. In triangle AOD: AD^2 = AO^2 + DO^2 - 2.AO.DO.cos(t), or 225 = 2.R^2 - 2.R^2.cos(t) or 225 = 2.R^2. (1 -cos(t)) In triangle DOC, in the same way we have 49 = 2.R^2.(1 -cos(t')). From these two equations we get by division that 225/49 = (1 -cos(t))/(1 -cos(t')) or 1 - cos(t) = 2.sin(t/2)^2 and 1 -cos(t') = 2.sin(t'/2)^2, so we simplify and obtain that 15/7 = sin(t)/sin(t'). Now let's note sin(t) = x As t' = Pi -2.t, we have t'/2 = Pi/2 -t and sin(t') = cos(t) = 1 - sin(t/2)^2 = 1 -2.x^2. So we have the equation 15/7 = x/(1 -2.x^2) That gives: 30.x^2 +7.x -15 = 0. Delta = 49 +4.30.15 = 1849 = 43^2, so x = (-7 -43)/60 (rejected as beeing negative) or x = (-7 +43)/60 = 36/60 = 3/5 Now we have that sin(t/2) = 3/5. Then 1 -cos(t) = 2.sin(t/2)^2 = 2.(9/5) = 18/25. We replace in the first equation we had: 225 = 2.R^2.(18/25) That gives R^2 = (225.25)/36 and R = (15.5)/6 = 25/2. Now in triangle ODF: DF = OD.sin(t) = R.2.sin(t/2).cos(t/2) = (25/2).2.(3/5).(4/5) = 12 The area of the trapezoid in now: ((AE + DC)/2). DF ,with AE = R + 7/2 = 11, so this area is ((16 +7)/2). 12 = 138

@PreMath 7 месяцев назад

Excellent! Thanks ❤️

@phungpham1725 7 месяцев назад

I solved it almost the same as you. 1/ ADCB is an isosceles trapezoid. We have: sq h = (7+x) .x ( by using the right triangle altitude theorem) (1) and sq h= sq AD- sq x = sq15 - sq x (Pythagorean theorem) (2) From (1) and (2) we have: 2sq x + 7x -225= 0----> x= 9 and h= 12 and AE=16 Area= 1/2 (7+16) .12 =138 sq units

@PreMath 7 месяцев назад

Excellent! Thanks ❤️

@WernHerr 7 месяцев назад

My solution with 2x Pythagoras (r=Radius, h=CE): OCE: h² = r² - 3.5² AFD: h² = 15² - (r - 3.5)² Solve r² - 3.5² = 15² - (r - 3.5)² and get r=12.5 and h=12 A=0.5*(r+3.5 + 7)*h = 138

@PreMath 7 месяцев назад

138 Thanks ❤️

@misterenter-iz7rz 7 месяцев назад

Let AE be x, then AF=x-7, so the diameter is x+(x-7)=2x-7, the radius is x-7/2, thus the square of the height h^2 will be (x-7/2)^2-(x-(x-7/2))^2=(x-7/2)^2-(7/2)^2, it is also equal to 15^2-(x-7)^2, thus x^2-7x=15^2-x^2+14x-49, 2x^2-21x-176=(x-16)(2x+11), so x=16, reject negative root, and h=sqrt(225-81)=sqrt(144)=12, therefore the area is 1/2×(7+16)×12=23×6=138.😊

@PreMath 7 месяцев назад

Excellent! Thanks ❤️

@AmirgabYT2185 7 месяцев назад

S=138

@PreMath 7 месяцев назад

Thanks ❤️

@LuisdeBritoCamacho 6 месяцев назад

I could make a lucky guess knowing that 15^2 = 225 and x^2 + y^2 = 225 can be equal to 9^2 + 12^2 = 225 ; 81 + 144 = 225 Then I could say that the Area of Trapezoid is equal to 138 Square Units, because A = ((9 + 7) + 7) * 12/2 = 23 * 6 = 138 Sq un. But I didn't!! What have I done? Defining Point F: Perpendicular Line passing point D and intersecting Line AB. Let's call DF = h (height) and Radius = R A) h^2 = 15^2 - (R - 3,5)^2 B) h^2 = R^2 - 3,5^2 225 - (R^2 - 7R + 12,25) = R^2 - 12,25 225 - R^2 + 7R - 12,25 = R^2 - 12,25 225 +7R -2R^2 = 0 R = - 9 or R = 12,5 R = 12,5 lin un AE = (12,5 + 3,5) li un = 16 lin un AF = (12,5 - 3,5) lin un = 9 lin un Now I know that h^2 = 225 - 81 = 144 ; h = sqrt(144) = 12 A = [(16 + 7) * (12 /2)] sq un = 23 * 6 sq un = 138 sq un Answer: The Trapezoid Area is Equal to 138 Square Units. QED.

@MarieAnne. 6 месяцев назад

Alternate solution: Draw vertical line through point O that intersects chord CD at point M. Any line that passes through center of circle and is perpendicular to a chord, bisects that chord. Therefore, CM = DM = 7/2 and since CDFE is a rectangle and OM is parallel to CE and DF, then OM also bisects EF Therefore, OE = OF = 7/2 Let AF = x, DF = h Since OA and OD are both radii, then both are equal and OD = OA = x + 7/2 Using Pythagorean Theorem in △ODF, we get h² + (7/2)2 = (x+7/2)² h² + 49/4 = x² + 7x + 49/4 h² = x² + 7x Using Pythagorean Theorem in △ADF, we get x² + h² = 15² x² + (x² + 7x) = 225 2x² + 7x − 225 = 0 2x² −18x + 25x − 225 = 0 2x (x − 9) + 25 (x − 9) = 0 (x − 9) (2x + 25) = 0 Since x is a length, then x > 0 x = 9 h² = x² + 7x = 81 + 63 = 144 h = 12 Now we can calculate area top base = 7 bottom base = 7+x = 7+9 = 16 height = h = 12 A = 1/2 (7 + 16) * 12 = 138

@raymondfrey9465 6 месяцев назад

I agree with the solution, but i dont agree with skipping to the assumption that triangle CEB is necessarily congruent with ADF based on "symmetry"... For younger learners, i would've "proved" that by reminding that the radius is the same from the center to any point along the perimeter of the circle THEN moving to prove symmetry from there... Just an observation rather than assume symmetry.....ESPECIALLY, when you go out of the way to say the drawing isn't to scale. Love all these videos though.

@murdock5537 7 месяцев назад

AD = BC = 15 → AC = BD = 20 → AB = 25 → r = 25/2 → cos⁡(δ) = 4/5 = AE/20 → AE = 16 → CE = h = 12 → green area = (1/2)(16+7)h = 138; ∆ ABD = ∆ ABC = pyth. triple = 5(3 - 4 - 5) → ∆ AEC = pyth. triple = 4(3 - 4 -5)

@phungpham1725 7 месяцев назад

Could you explain how AC=20?

@PreMath 7 месяцев назад

Excellent! Thanks ❤️

@murdock5537 6 месяцев назад

@@phungpham1725 AO = BO = CO = DO = r sin⁡(BDA) = 1 ↔ AD = 15 → ABD = pyth. triple = 5(3 - 4 - 5) → 2r = 25 → BD = 20

@ybodoN 6 месяцев назад

@@murdock5537 with sin⁡(BDA) = 1 ↔ AD = 15, DB could have any length and we would still have AO = BO = CO = DO = r 🧐

@murdock5537 6 месяцев назад

@@ybodoN or try this: φ = 30°; DF = h = NO → DN = CN = CD/2 = 7/2 → ∆ AOD → AO = AF + FO = (r - 7/2) + 7/2 = r = DO AOD = α; ODF = β → α + β = 3φ → sin⁡(α) = cos⁡(β) → cos⁡(α) = sin⁡(β); AD = 15 DAO = δ → AO = DO = r → DAO = ODA = δ → AD = AT + DT ↔ AT = DT = AD/2 → AOT = TOD = α/2 → sin⁡(α/2) = cos⁡(δ) = (r - 7/2)/15 = (2r - 7)/30 → cos⁡(α/2) = sin⁡(δ) = h/15 → sin⁡(α) = h/r = 2sin⁡(α/2)cos⁡(α/2) → 1/r = 2(2r - 7)/450 → r1 = 25/2; r2 = -9 < 0 ≠ solution → h = 12 → r - 7/2 = 9 → green area = (1/2)(7 + 16)12 = 6(23) = 138

@giuseppemalaguti435 7 месяцев назад

Intanto calcolo r...15^2=r^2+r^2-2r^2cos(90-arcsin3,5/r)...(Teorema dei coseno)...r=12,5..h=√(12,5^2-3,5^2)=12...A=(7+√(15^2-12^2)+7)*12/2=(7+9+7)*6=23*6=138

@PreMath 7 месяцев назад

Excellent! Thanks ❤️

@StephenRayWesley 6 месяцев назад

(15)^2=225 (7)=49 2x(45°)=90°x. 2x(45°)=90°x. 3x(15°)=45°x 3x(15°)=45°x (90°x+90°x+45°x+45°x)!=,270°x^4 (49+225)=474 (270°x^4-474)=√196x^4 √10^√108^√12 x^4√ 5^√2√5√^2 √8^3^√4 x^√4 √1^√1√ 1^√1√ 2^√3 3^√2^√2 x^√2^2. √1^√1 3^√1^√1x√1^2 32 (x+2x-3)

@PreMath 6 месяцев назад

Thanks ❤️

@elmurazbsirov7617 7 месяцев назад

Bakıdan salamlar.Azərbaycan.Əla həll etdiniz.Təşəkkürlər.

@PreMath 7 месяцев назад

Bunu eşitməyə şadam! Çox xoş gəlmisiniz! Təşəkkür canım ❤️🌹 ABŞ-dan sevgi və dualar! 😀

@JSSTyger 6 месяцев назад

I'll say 138

@hongningsuen1348 6 месяцев назад

There is no need to construct triangle BCE (Its congruence to triangle ADF by symmetry is just by intuition and not a proper congruence test.) to get BE = AF. Perpendicular line from centre to chord is a line bisector. Hence chord DC can be bisected with perpendicular line from centre O into two equal segments. In the 2 equally divided rectangles formed from the rectangle CDFE, OF = OE (opposite sides equal). AF = radius - OF and BE = radius - OE. Hence AF = BE.

@PreMath 6 месяцев назад

Thanks ❤️

@jamestalbott4499 7 месяцев назад

Thank you Thale!

@PreMath 7 месяцев назад

😀 Excellent! Thanks ❤️

@misterenter-iz7rz 7 месяцев назад

this is much more difficult compared with yesterday's one.😅😅😅

@PreMath 7 месяцев назад

Yes! Thanks ❤️

@marcelowanderleycorreia8876 6 месяцев назад

Great!

@PreMath 6 месяцев назад

Excellent! Thanks ❤️

@wackojacko3962 7 месяцев назад

😎

@PreMath 7 месяцев назад

Excellent! Thanks ❤️

@quigonkenny 7 месяцев назад

Let F be the point on AB where DF is perpendicular to AB. As CD is a chord, and is parallel to diameter AB, ∆DFO and ∆OEC are congruent and OF = OE = 7/2. Let DF = CE = h. Triangle ∆DFO: OF² + DF² = OD² (7/2)² + h² = r² h² = r² - 49/4 h = √[r²-(49/4)] Triangle ∆AFD: AF² + FD² = DA² (r-(7/2))² + [h²] = 15² r² - 7r + 49/4 + [r² - 49/4] = 225 2r² - 7r - 225 = 0 2r² + 18r - 25r - 225 = 0 2r(r+9) - 25(r+9) = 0 (2r-25)(r+9) = 0 2r - 25 = 0 | r + 9 = 0 r = 25/2 ✓ | r = -9 ❌ r > 0 h² = (25/2)² - 49/4 = 625/4 - 49/4 h² = 576/4 = 144 h = √144 = 12 Trapezoid AECD: A = h(a+b)/2 A = 12(7+(7+9))/2 A = 6(7+16) = 6(23) A = 138 Edit: Interesting how with the quadratic equation in this problem, if you solve for r, the discarded solution is -x, and if you solve for x, the discarded solution is -r. They're effectively the same quadratic equation, just with opposite signs on the x coefficient and constant.

@PreMath 7 месяцев назад

Thanks ❤️