95% Failed to solve the Puzzle | Can you find area of the White Triangle? |

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Learn how to find the area of the white shaded triangle in the rectangle. Important Geometry and Algebra skills are also explained. Step-by-step tutorial by PreMath.com
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95% Failed to solve the Puzzle | Can you find area of the White Triangle? | #math #maths | #geometry
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12 май 2024

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Комментарии : 75

@MMmaths8800 Месяц назад

There is no one like you, you are the best teacher in the world🥰

@PreMath Месяц назад

Thanks dear for your continued love and support!❤️ You are the best!

@Uma-hr8cc 7 дней назад

Best teaching

@ritwikgupta3655 Месяц назад

Very nicely done by isolating and reducing variables to only a single "ab" However you cannot reject 18-2√51 outright as its value is 3.72, ie +ve. So, we should state: White area = 18+/-2√51 - (5+6+7) = 18+/-2√51 - 18 = +/-2√51. As area is a positive number the answer is 2√51.

@123rockstar2010 29 дней назад

Exactly.

@OverclockingCowboy 6 дней назад

The author rejected it outright being “not possible” because it would be less than the sum of the 3 given triangles (5+6+7=18).

@prossvay8744 Месяц назад

Let AD=BC=a ; AB=CD=b Area of rectangle ABCD=ab CF=12/a ; AF=10/b DE=AD-AE=a-10/b DF=CD-BF=b-12/a Area of triangle DEF=1/2(DE)(DF)=1/2(a-10/b)(b-12/a)=1/2ab(ab-10)(ab-12)=7 (ab-10)(ab-12)=14ab ab=18+2√51=32.28cm^2 White triangle area=32.28-(5+6+7)=14.28 cm^2.❤❤❤ Thanks sir Best regards.

@PreMath Месяц назад

Super! You are the best🌹 Glad to hear that! Thanks for sharing ❤️

@jimlocke9320 Месяц назад

The barrier students must overcome is that there are not enough equations to solve for a and b. PreMath was able to narrow it down to one equation and 2 unknowns (a and b). However, we only need to know the product of a and b to find the area of the white triangle, not the individual values of a and b. One solution approach would be to find the area of the white triangle for the special case of ABCD being a square. Then, letting s be the side of the square, a = b = s and you have one equation with one unknown. The problem statement does not rule out a square. So, you solve, get a value for s² and subtract the combined area of the red, green and blue triangles to get the area of the white triangle. The problem statement implies that the same solution applies to the general case of the rectangle, so you present that as your solution. If the problem statement is modified to require students to solve ABCD being the general case of a rectangle, students are effectively given a clue that the solution is valid for a range of values of a and b, in this case, whenever the product ab equals 18 + 2√(51).

@PreMath Месяц назад

Super! Thanks for the nice feedback ❤️

@shahdmohammed4597 Месяц назад

نشكر حضرتك للمجهود والمعلومة انا اوجدت الناتج بنظرية ايجاد المساحة الداخلية المظلله فى الشكل =مساحة الشكل الخارجى _مجموع مساحة الشكل الداخلى والناتج كان عدد صحيح وليس عدد عشرى. اسفة للاطالة وارجو التصحيح لو هناك خطا فى الحل

@redfinance3403 Месяц назад

Nice! I finsihed this one very quickly ⏰! I think what maybe made a lot of people fail to do this puzzle is labelling all the sides of the triangles with different variables instead of writing the height out in terms of the base, since you are given the area. Its a common pattern i have seen in other problems on your channel!

@PreMath Месяц назад

Nice work! Thanks for the feedback ❤️

@soniamariadasilveira7003 Месяц назад

Thank you Sir. I always like your explanations!

@ramanivenkata3161 Месяц назад

Very well explained

@klexosia Месяц назад

Thank you! I tried to do this on my own before watching the video, and I got the correct answer 🙂

@derwolf7810 Месяц назад

Alternatively: Using your labeling a, b, h, A to F and define i := |FC|. Copy the triangles, rotate them by 180 degree and align them along the bases of their counterparts to get colored rectangles. Note that the blue and the red rectangle overlap on a rectangle with area of i*h (smaller than both rectangles) and that the following is true. i*(b-h) * (a-i)*h = (a-i)*(b-h) * i*h (i*b - i*h) * (a*h-h*i) = (a-i)*(b-h) * i*h (2*6 - h*i) * (2*5-h*i) = 2*7 * h*i (h*i)^2 - 22*(h*i) + 18^2 - 324 + 120 = 14*(h*i) (h*i - 18)^2 - sqrt(204)^2 = 0 | /(h*i - 18 - sqrt(204)); < 0, because h*i < 10 h*i - 18 + sqrt(204) = 0 h*i = 18 - sqrt(204) ==> Area = 2*(5+6+7)-S - (5+6+7) = 18-S = 18-(18 - sqrt(204)) = sqrt(204) = 4 sqrt(51) ~= 14.28 [in cm^2]

@think_logically_ Месяц назад

AE=bp => DE = b(p-1). Also DF=aq => CF=a(1-q). So we have abp=10; ab(1-q)=12; ab(1-p)q=14. But ab=s - the area of rectangle. So p=10/s => 1-p=1-10s=(s-10)/s and 1-q=12/s => q = (s-12)/s. Thus from third equation (s-10)(s-12)/s² = 14/s with leads to quadratic equation s²-36s+120=0. The rest is clear.

@jamestalbott4499 Месяц назад

Thank you!

@PreMath Месяц назад

You are very welcome!🌹 Thanks ❤️

@soli9mana-soli4953 Месяц назад

2sqrt51 calling B the base of rectangle, H its height, FC = x, AE = y we can write: x*H = 12 y*B = 10 (B - x)*(H - y) = 14 => BH - xH - yB + xy = 14 BH = 36 - xy Now we can divide the rectangle in 4 smaller rectangles in which the upper right rectangle has area = xy, then tracing perpendicular to base and height. Calling xy=a we can calculate the areas of each rectangles as: 10 - a | a _____________ 14 | 12 - a doing the crossed product we have: 14 a = (10-a)*(12-a) a² - 36a + 120 = 0 a = 18 - 2sqrt51 = xy BH = 36 - (18 - 2sqrt51) = 18 + 2sqrt51 White area = 18 + 2sqrt51 - 5 - 7 - 6 = 2sqrt51

@PreMath Месяц назад

Excellent! Thanks for sharing ❤️

@CloudBushyMath Месяц назад

Nice one👍

@PreMath Месяц назад

Thank you! Cheers!🌹❤️

@richardbloemenkamp8532 Месяц назад

Ok got it correct. At first it seemed to become too complex with sqrt(816) and that there may be a simpler solution. But just continuing I got 2*sqrt(51)

@PreMath Месяц назад

Excellent! Thanks for the feedback ❤️

@ercantulunoglu Месяц назад

very good

@santiagoarosam430 Месяц назад

Dividimos ABCD en cuatro celdas trazando una horizontal por E y una vertical por F→ Área de las celdas: (2*5-a); a; 14; (2*6-a)→ (10-a)/14=a/(12-a)→ a=18-2√51→ Área ABCD =10+12+14-a=36-18+2√51=18+2√51→ Área BEF =(18+2√51)-5-6-7=2V51 =14,2828.... Gracias y saludos.

@PreMath Месяц назад

Excellent!🌹 You are very welcome! Thanks for sharing ❤️

@think_logically_ Месяц назад

Un error pequeño: en lugar de (10-a)/14=a=(12-a) debe ser (10-a)/14=a/(12-a). But it's clear anyway (sorry my Spanish is far from perfect). Excellent solution. I was looking for something like that, but failed to complete.

@santiagoarosam430 Месяц назад

@@think_logically_ Gracias por advertirme de la errata en la ecuación. Corrijo y quedo muy agradecido. Hasta siempre.

@think_logically_ Месяц назад

@@santiagoarosam430 Por nada

@user-cm7zz8zn4h Месяц назад

Assumes that at angles at vertices of abcd are 90degrees, not given, anyone can make assumptions.

@waheisel Месяц назад

He does say it is a rectangle.

@Deribus575 Месяц назад

Which is an assumption not given in the problem

@hanswust6972 28 дней назад

Let be the area of triangles 7 = A, 6 = B and 5 = C. Then asked area = X *X = sqrt((A+B+C)^2 - 4BC)*

@mayihelpyou5557 23 дня назад

我也是這麼寫的結果拿零分老師要求我寫計算過程

@awolzz5720 11 дней назад

两位兄弟为什么得出这个看似有点匪夷所思的结论? 不过代入后的结果与答案不符啊，是不是哪里写错了？

@mayihelpyou5557 11 дней назад

@@awolzz5720 只有一個原因你代錯

@awolzz5720 7 дней назад

@@mayihelpyou5557 你是怎么得出这个结论的呢，能不能说一下

@mayihelpyou5557 7 дней назад

@@awolzz5720 我記得用畢氏定理就能證了你最好不要用太多未知數兩個未知數就夠了

@SirKaftar_Requiem Месяц назад

❤❤❤

@PreMath Месяц назад

Thanks dear 🌹❤️

@alster724 Месяц назад

The solution may be complex at first, but I was able to get it.

@PreMath Месяц назад

Bravo🌹

@marcgriselhubert3915 Месяц назад

Fine.

@PreMath Месяц назад

Glad to hear that! Thanks for the feedback ❤️

@misterenter-iz7rz Месяц назад

by formula, A^2=(5+6+7)^2-4×5×6=18^2-120=204, A=2sqrt(51).😊

@PreMath Месяц назад

Excellent! Thanks for sharing ❤️

@manojmiya9141 Месяц назад

Proof of (5+6+7)²-4*5*6

@wackojacko3962 Месяц назад

The top 5% that did solve this this International Mathematical Olympiad puzzle all went on too be Substitute Teachers! 🙂

@PreMath Месяц назад

😀 Thanks ❤️

@stephenbrand5779 Месяц назад

Great problem and solution. Afraid I am one of the 95%.

@goranbrankovic9283 Месяц назад

14,283 cm2

@sarojkumarbharti2906 13 дней назад

It's very easy

@shreedhanmehta3553 Месяц назад

Why is 18-2√51 scenario rejected?

@bfelten1 Месяц назад

@@DeathZebra@DeathZebra: Well, that motivation must be in the solution for it to pass the test. But you have DF = a - 12/b, which means that a > 12/b and therefore ab > 12 (likewise for ED, but only ab > 10). So 18 - 2 * sqrt(51) must be greater than 12 -- it is slightly less than 4, so dismisst. QED.

@kkkim4643 20 дней назад

ab > 5+6+7, ab > 18

@LuisdeBritoCamacho Месяц назад

1) Let's baptize things! 2) x = 5 sq cm 3) y = 6 sq cm 4) z = 7 sq cm 5) Fortunately we have a General Formula for these cases : White Area = "sqrt((x + y + z)^2 - 4xy)" 6) WA = sqrt((18)^2 - 4*(30)) ; WA = sqrt(324 - 120) ; WA = sqrt(204)sq cm ; WA = (2*sqrt(51)) sq cm ; WA ~ 14,3 sq cm 7) Final Answer : The Area of the White Square is equal to approx. 14,3 Square Cm. NOTE: I reach to the conclusion that the Domain of the Solution (White Area) should be somewhere between : 8 sq cm < WA < 16 sq cm. The Total of Possible Ordered Pairs (X ; Y) of Integer Solutions is : S {(18 ; 0) ; (20 ; 2) ; (22 ; 4) ; (24 ; 6) ; (26 ; 8) ; (28 ; 10) ; (30 ; 12) ; (32 ; 14) ; (34 ; 16) ; (36 ; 18)} X = Area of Rectangle [ABCD] and Y = Area of White Triangle [BEF]. And X - Y = X - (5 + 6 + 7) sq cm ; X - Y = 18 sq cm. The difference must always be 18 sq cm.

@PreMath Месяц назад

Excellent! Thanks for sharing ❤️

@Uma-hr8cc 7 дней назад

@shahdmohammed4597 Месяц назад

نشكرك

@tamarshahverdyan2723 Месяц назад

# 44 #

@veby_ff Месяц назад

خیلی زیبا و جالب

@PreMath Месяц назад

ممنون فاطمه عزیزم 🌹❤️

@JobBouwman Месяц назад

The area R of the whole rectangle is: R = AB*AE+ DE*DF + BC*CF - AE*FC = 10 + 14 + 12 - AE*FC (1) Since AE = 10/AB and FC = 12/BC, we get AE*FC = 120/(AB * BC) = 120/R Plugging this into (1) we get: R = 36 - 120/R R^2 - 36R + 120 = 0 This quadratic equation yields R = 18 + 2*sqrt(51) Now our answer = 18 + 2*sqrt(51) - 5 - 7 - 6 = 2*sqrt(51)