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

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
Olympiad Mathematical Question! | Learn Tips how to solve Olympiad Question without hassle and anxiety!
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[@MMmaths8800]

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

[@PreMath]

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

[@ritwikgupta3655]

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]

Exactly.

[@jimlocke9320]

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]

Super! Thanks for the nice feedback ❤️

[@shahdmohammed4597]

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

[@prossvay8744]

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]

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

[@soniamariadasilveira7003]

Thank you Sir. I always like your explanations!

[@ramanivenkata3161]

Very well explained

[@redfinance3403]

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]

Nice work! Thanks for the feedback ❤️

[@derwolf7810]

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]

[@CloudBushyMath]

Nice one👍

[@PreMath]

Thank you! Cheers!🌹❤️

[@jamestalbott4499]

Thank you!

[@PreMath]

You are very welcome!🌹 Thanks ❤️

[@hanswust6972]

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

[@klexosia]

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

[@ercantulunoglu]

very good

[@user-cm7zz8zn4h]

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

[@waheisel]

He does say it is a rectangle.

[@Deribus575]

Which is an assumption not given in the problem

[@soli9mana-soli4953]

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]

Excellent! Thanks for sharing ❤️

[@richardbloemenkamp8532]

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]

Excellent! Thanks for the feedback ❤️

[@santiagoarosam430]

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]

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

[@think_logically_]

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]

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

[@think_logically_]

@@santiagoarosam430 Por nada

[@think_logically_]

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.

[@marcgriselhubert3915]

Fine.

[@PreMath]

Glad to hear that! Thanks for the feedback ❤️

[@alster724]

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

[@PreMath]

Bravo🌹

[@SirKaftar_Requiem]

❤❤❤

[@PreMath]

Thanks dear 🌹❤️

[@wackojacko3962]

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

[@PreMath]

😀 Thanks ❤️

[@shahdmohammed4597]

نشكرك

[@misterenter-iz7rz]

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

[@PreMath]

Excellent! Thanks for sharing ❤️

[@manojmiya9141]

Proof of (5+6+7)²-4*5*6

[@tamarshahverdyan2723]

# 44 #

[@veby_ff]

خیلی زیبا و جالب

[@PreMath]

ممنون فاطمه عزیزم 🌹❤️

[@stephenbrand5779]

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

[@goranbrankovic9283]

14,283 cm2

[@shreedhanmehta3553]

Why is 18-2√51 scenario rejected?

[@DeathZebra]

Too small. Area>18 (the combined area of the coloured triangles)

[@bfelten1]

@@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.

[@LuisdeBritoCamacho]

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]

Excellent! Thanks for sharing ❤️

[@amudangopal]

Why the negative root is not possible?

[@user-vf4lm1lp2n]

面積必大於0,而18-2×51＾1／2＜0，故不合

[@ritwikgupta3655]

​@@user-vf4lm1lp2n 18-2(√51) approx 18-2*7.14=18-14.28=3.72 which is +ve, but this will give a negative white area, only later.

[@rohandasgupta5365]

my answer is 6 cm square

[@lijiancz2066]

bad number selected. furhter ,when calculating delta, yoiu'd better keep facters(in this case, 4^2 rather than calculate it out.

[@JobBouwman]

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)

[@RajendranPK]

√(5+6+7)^2---4×5×6

[@PreMath]

Thanks for sharing ❤️