95% Failed to solve the Puzzle | Can you find area of the White Triangle? |
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- Опубликовано: 12 май 2024
- 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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There is no one like you, you are the best teacher in the world🥰
Let AD=BC=a ; AB=CD=b
Very nicely done by isolating and reducing variables to only a single "ab"
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.
Thank you Sir. I always like your explanations!
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!
Very well explained
Nice one👍
Let be the area of triangles 7 = A, 6 = B and 5 = C.
Thank you!
Thank you! I tried to do this on my own before watching the video, and I got the correct answer 🙂
❤❤❤
Alternatively:
2sqrt51
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.
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....
Fine.
very good
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)
by formula, A^2=(5+6+7)^2-4×5×6=18^2-120=204, A=2sqrt(51).😊