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

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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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[HarHarMahadev-]

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

[ritwikgupta]

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.

[jimlocke]

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).

[prossvay]

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
(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.

[redfinance]

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!

[williamorrick]

A conceptual explanation of why it is not possible to solve for the individual lengths: if the figure is rescaled vertically by a factor of S and horizontally by a factor of 1/S, none of the areas in the figure (and areas are the only pieces of information provided) will change. Hence we are free to choose the scaling so that a selected line segment, say AB, has length 1.

With this choice, AE has length 10. Let BC have length a, so that a*1 = a is the desired rectangle area. Then ED has length 10-a, FC has length 12/a, and DF has length 1 - 12/a. So (10-a)(1-12/a) = 14, from which we can find a.

[santiagoarosam]

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 Área BEF =(18+2√51)-5-6-7=2V51 =14, 2828....
Gracias y saludos.

[alscents]

Since there are no variables, the answer needs to be a constant.
Area of big rectangle = constant = A
Area of white triangle = constant = A - 5 - 6 - 7 = A - 18

Assuming D as the origin (0, 0), let B = (a, b)
The locus of B is
a b = A
Without loss of generality, one can assume the special case when a = b, giving us
a^2 = A

Also
(1/2) (a - 10/a) (a - 12/a) = 7
a^2 - 12 - 10 + 120 / a^2 = 14
A - 22 - 14 + 120 / A = 0
A^2 - 36 A + 120 = 0
A^2 - 36 A + 18^2 + 120 = 18^2
(A - 18)^2 + 120 = 324
(A - 18)^2 = 204
A - 18 = sqrt(204) = 2 * sqrt(51) = 14.282856857 square units

[derwolf]

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]

[hanswust]

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

[solimana-soli]

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

[bankston]

I would have added a little reasoning why x=18-2sqrt(51) should be rejected. It's still positive, but is less than 4. We're given the area of the rectangle is at least 18.

[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.

[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)

[misterenter-izrz]

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

[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.

[richardbloemenkamp]

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)

[markuskonway]

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

[soniamariadasilveira]

Thank you Sir. I always like your explanations!

[klexosia]

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