First: The preliminary concept is quite evident. With your notations, and if L is the length of the square, then we have: a + b = (1/2).( L). (hight of traingle a) + (1/2).(L). (hight of triangle b) = (1/2).(L).(L) = (1/2). (area of the square), and so a + b = c + d = (1/2). (area of the square). Now I finish just as you did. X + a +36 = (1/2). (area of the square), and also a + 81 = (1/2). (area of the square) , so we have X + a + 36 = a + 81, and finally X= 45.
Thank you! My aproach is to calculate the area of the triangle ( red+big white) according to a ( the side of the square. But I think that your way is much better.
Let's do it: . .. ... .... ..... The sum of the areas of the right triangle (area 81cm²) and the lower white triangle (unknown area) is half of the area of the square. The sum of the areas of the upper triangle (area 36cm²), the red triangle and the already mentioned lower white triangle is half of the area of the square as well. So we have the following identity: A(right triangle) + A(lower white triangle) = A(upper triangle) + A(red triangle) + A(lower white triangle) ⇒ A(red triangle) = A(right triangle) − A(upper triangle) = 81cm² − 36cm² = 45cm² Best regards from Germany
Nice problem! I labelled the height as 𝒉 and the base width as 𝒘. Then, the intersection point is some height 𝒉𝒌 where 𝒌 is the proportion of 𝒉 that represents the problem. Recognizing that the upper [2] △ and lower [8] △ are proportionatly similar (same as the video), their bases (well, top, for the [2]△) are in a (√(2 ÷ 8) → √(¼) → ½) relationship. 𝒌𝒉 is height of [8]△ (1 - 𝒌)𝒉 is height of [2]△ Represent as areas 8 = ½𝒘𝒌𝒉 16 = 𝒘𝒌𝒉 2 = ½(½ 𝒘)(1 - 𝒌)𝒉 expands to 8 = 𝒘𝒉 - 𝒘𝒌𝒉 Note that the last term is same as previous group: 8 = 𝒘𝒉 - 16 … move 16 24 = 𝒘𝒉 Well, now we have the full square. Cool! Just subtract out all the given areas to find the remainder. 𝒙 = 24 - (8 ⊕ 6 ⊕ 2) 𝒙 = 8 And we're done. ⋅-⋅-⋅ Just saying, ⋅-⋅-⋅ ⋅-=≡ GoatGuy ✓ ≡=-⋅
Thanks for the Areas lesson!
You are very welcome!
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Excellent!
Thank you, especially for the providing the basic concepts prior to solving the problem!
You are very welcome!
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Awesome
Very nice!
❤ good sharing❤
Thank you! Cheers!
Wow, that's amazing!
First: The preliminary concept is quite evident. With your notations, and if L is the length of the square, then we have:
a + b = (1/2).( L). (hight of traingle a) + (1/2).(L). (hight of triangle b) = (1/2).(L).(L) = (1/2). (area of the square),
and so a + b = c + d = (1/2). (area of the square).
Now I finish just as you did. X + a +36 = (1/2). (area of the square), and also a + 81 = (1/2). (area of the square) , so we have X + a + 36 = a + 81,
and finally X= 45.
Thanks ❤️
It's been a long while since you presented a puzzle that stumped me, but this one nearly did. GOOD JOB!!
Keep 'em coming.
Thanks, will do!
36+r+w=81+w, so r=81-36=45.😊.
Amazing puzzle, unexpected simple solution ❤
Excellent!
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Similar to problem posted 4 days ago with a very cool twist equating 36 + x + a to 81 + a. I absolutely love it! 🙂
Perfect!
Glad to hear that!
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Thank you! My aproach is to calculate the area of the triangle ( red+big white) according to a ( the side of the square. But I think that your way is much better.
@@PreMath, 🎉🎉😂🎉😂🎉😂😂🎉🎉
Mind blowing sir ❤❤❤❤❤
Thank you so much 😀
Love it!!!!!!!!!!!!!
Excellent!
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I believe the solution would be the same more generally for a rectangle and even more generally for a parallelogram?
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Let's do it:
.
..
...
....
.....
The sum of the areas of the right triangle (area 81cm²) and the lower white triangle (unknown area) is half of the area of the square. The sum of the areas of the upper triangle (area 36cm²), the red triangle and the already mentioned lower white triangle is half of the area of the square as well. So we have the following identity:
A(right triangle) + A(lower white triangle) = A(upper triangle) + A(red triangle) + A(lower white triangle)
⇒ A(red triangle) = A(right triangle) − A(upper triangle) = 81cm² − 36cm² = 45cm²
Best regards from Germany
Excellent!
Thanks ❤️
If I'm not mistaken, the sides of the square have a minimum length of 3 (√3 + √15) ≈ 16.8151 cm
Fantástico
Excellent!
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Гениально!
challenge puzzle 🎉
Yes!
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Merci infiniment pour ce problème de génie
You are very welcome!
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Nice problem!
I labelled the height as 𝒉 and the base width as 𝒘. Then, the intersection point is some height 𝒉𝒌 where 𝒌 is the proportion of 𝒉 that represents the problem.
Recognizing that the upper [2] △ and lower [8] △ are proportionatly similar (same as the video), their bases (well, top, for the [2]△) are in a (√(2 ÷ 8) → √(¼) → ½) relationship.
𝒌𝒉 is height of [8]△
(1 - 𝒌)𝒉 is height of [2]△
Represent as areas
8 = ½𝒘𝒌𝒉
16 = 𝒘𝒌𝒉
2 = ½(½ 𝒘)(1 - 𝒌)𝒉 expands to
8 = 𝒘𝒉 - 𝒘𝒌𝒉
Note that the last term is same as previous group:
8 = 𝒘𝒉 - 16 … move 16
24 = 𝒘𝒉
Well, now we have the full square. Cool! Just subtract out all the given areas to find the remainder.
𝒙 = 24 - (8 ⊕ 6 ⊕ 2)
𝒙 = 8
And we're done.
⋅-⋅-⋅ Just saying, ⋅-⋅-⋅
⋅-=≡ GoatGuy ✓ ≡=-⋅
Thanks ❤️