By the same way, we can see the sum of the blue region and the green region equals the some of the yellow region and the pink region. So the answer is (6x+5)+(6x+2)-(12x-1)=8 square units
You are correct but haven't proved that the two sums are equal. Using PreMath's diagonal lines, you have 8 triangles, 2 of each color. You can find the blue and green triangles with vertical sides have the same combined area as the yellow and pink pair with vertical sides. Similarly, the blue and green triangles with horizontal sides have the same combined area as the yellow and pink pair with horizontal sides. So, the combined area of the blue and the green region equals the combined area of the yellow and the pink region. Alternatively, you can connect the midpoints of the sides with line segments that meet the sides at 45° angles. You get 4 right isosceles triangles of equal area, one in each corner. Now consider the other 4 triangles and treat the 45° line segments as their bases, which will be equal in length. The sum of heights of the blue and green triangles is equal to the sum of heights of the red and yellow triangles so the combined areas of the pairs is the same. Add the same colored corner triangles and the combined blue and green areas is equal to the combined red and yellow areas.
Connect the 4 side midpoints with diagonal line segments, angled 45° with the sides of the square, producing 4 congruent isosceles right triangles. The hypotenuses form a smaller square inside the larger square. Let the lengths of the hypotenuses, which are also the lengths of sides of the small square, be designated b. The point where the four colored areas meet may lie inside that smaller square, in one of the isosceles right triangles or on a side. Either 3 or 4 of the isosceles right triangles will be solidly filled in with the same color. Choose a pair of colors opposite each other that solidly fill their isosceles right triangles. Let's say those colors are red and yellow. The combined area of the red and yellow isosceles right triangles is equal to half the area between the sides of the large square and the sides of the small square. There is another red triangle and another yellow triangle. (There are special cases where one of these triangles may be missing. If so, consider its height and area to be 0.) Let the sides that are also hypotenuses of the isosceles right triangles be considered their bases, thus having length b. Let's say the red triangle has height h₁, thus area (1/2)bh₁ and yellow triangle h₂, thus area (1/2)bh₂ and the combined area is (1/2)bh₁ + (1/2)bh₂ = (1/2)b(h₁ + h₂). But (h₁ + h₂) = b, so the combined area is (1/2)b², or half the area of the smaller square. The combined area of the red and yellow isosceles right triangles and the other red and yellow triangles is half the area of the large square. The combined green and blue areas must be the other half of the large square's area, So, the combined red and yellow areas must equal the combined green and blue areas. Let y = the blue area, then y + (12x - 1) = (6x + 5) + (6x + 2) y + 12x - 1 = 12x + 7 y - 1 = 7 y = 8 So the blue area is 8 sq. units, as PreMath also found.
In this situation we can easily prove that both, the sum of the yellow plus the pink area and the sum of the green an the blue on are 50% of the total size of the square (=a²), if the side length is a. Let us shortly show this with yellow and pink area, if we have divided each of them by connecting the point in the middle with the left down edge or the upper right one. Now we look for the two combinations of triangles in our yellow and pink area from left to right and from bottom to up. We have two height values in each case, for example h and a-h. The size is in this situation... (1/2)*(a/2)*h + (1/2)*(a/2)*(a-h) = (1/4)*a² Therefore the sum of both is a²/2 as mentioned before. With this we can easily determine the size of the blue area (=BA). 12x-1 + BA = 6x+2 + 6x+5 12x-1 + BA = 12x+7 BA = 8
Trazar lineas desde el vértice central a las cuatro esquina del cuadrado → Los triángulos que tienen por base el mismo lado tienen igual superficie → Serie de superficies triángulares resultantes: a, a (6x+5-a), (6x+5-a), (6x-6+a), (6x-6+a), (8-a), (8-a) → La suma del primer y último valor de la serie es la superficie azul → Azul =8-a+a=8 Gracias y saludos.
Connect all four corner points with the intermediary point P. Herein NW = a + b, NE = b + c SE = c + d, SW = d + a Hereby NW = NE + SW - SE = 6 x + 5 + 6 x + 2 - (12 x -1) = 8 sqr unit
Very nice video, sir
Respect
So nice of you, dear
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@@PreMath
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By the same way, we can see the sum of the blue region and the green region equals the some of the yellow region and the pink region. So the answer is (6x+5)+(6x+2)-(12x-1)=8 square units
Thanks for your feedback! Cheers! 😀
You are awesome. Keep it up 👍
You are correct but haven't proved that the two sums are equal. Using PreMath's diagonal lines, you have 8 triangles, 2 of each color. You can find the blue and green triangles with vertical sides have the same combined area as the yellow and pink pair with vertical sides. Similarly, the blue and green triangles with horizontal sides have the same combined area as the yellow and pink pair with horizontal sides. So, the combined area of the blue and the green region equals the combined area of the yellow and the pink region.
Alternatively, you can connect the midpoints of the sides with line segments that meet the sides at 45° angles. You get 4 right isosceles triangles of equal area, one in each corner. Now consider the other 4 triangles and treat the 45° line segments as their bases, which will be equal in length. The sum of heights of the blue and green triangles is equal to the sum of heights of the red and yellow triangles so the combined areas of the pairs is the same. Add the same colored corner triangles and the combined blue and green areas is equal to the combined red and yellow areas.
Very well explained 👍
Thanks for sharing 😊
تمرين جميل جيد . رسم واضح مرتب . شرح واضح مرتب . شكرا جزيلا لكم والله يحفظكم ويرعاكم ويحميكم جميعا . تحياتنا لكم من غزة فلسطين .
Let the area of the regions be a+b, a+c, c+d, d+b, where a+c=6x+5, c+d=12x-1, d+b=6x+2, a+b=(a+c)-(c+d)+(d+b)=6x+5-12x+1+6x+2=8.😊
Excellent!
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Superb. What a technique!
Thanks sir for the solution
Thanks for video.Good luck sir!!!!!!!!!
Thank you too, dear
You are awesome. Keep it up 👍
Great one!
Thanks for your feedback! Cheers! 😀
You are awesome. Keep it up 👍
تمرين جيد. رسم واضح مرتب. شرح واصح مرتب . شكرا جزيلا لكم والله يحفظكم ويحميكم ويرعاكم جميعا. تحياتنا لكم من غزة فلسطين
Fantastic
Connect the 4 side midpoints with diagonal line segments, angled 45° with the sides of the square, producing 4 congruent isosceles right triangles. The hypotenuses form a smaller square inside the larger square. Let the lengths of the hypotenuses, which are also the lengths of sides of the small square, be designated b. The point where the four colored areas meet may lie inside that smaller square, in one of the isosceles right triangles or on a side. Either 3 or 4 of the isosceles right triangles will be solidly filled in with the same color. Choose a pair of colors opposite each other that solidly fill their isosceles right triangles. Let's say those colors are red and yellow. The combined area of the red and yellow isosceles right triangles is equal to half the area between the sides of the large square and the sides of the small square. There is another red triangle and another yellow triangle. (There are special cases where one of these triangles may be missing. If so, consider its height and area to be 0.) Let the sides that are also hypotenuses of the isosceles right triangles be considered their bases, thus having length b. Let's say the red triangle has height h₁, thus area (1/2)bh₁ and yellow triangle h₂, thus area (1/2)bh₂ and the combined area is (1/2)bh₁ + (1/2)bh₂ = (1/2)b(h₁ + h₂). But (h₁ + h₂) = b, so the combined area is (1/2)b², or half the area of the smaller square. The combined area of the red and yellow isosceles right triangles and the other red and yellow triangles is half the area of the large square. The combined green and blue areas must be the other half of the large square's area, So, the combined red and yellow areas must equal the combined green and blue areas.
Let y = the blue area, then
y + (12x - 1) = (6x + 5) + (6x + 2)
y + 12x - 1 = 12x + 7
y - 1 = 7
y = 8
So the blue area is 8 sq. units, as PreMath also found.
Doesn't area of Green + Blue = Yellow + Red
12x-1 + B = 6x+2+6x+5
B = 12x+7 - (12x-1)
B = 8
Excellent!
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You are awesome. Keep it up 👍
Interesting sum
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Almost comedic! Thanks Professor!❤🥂❤
Glad you enjoyed it!
You are very welcome!
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Thank you!
simply great, you are master of triangle, Sir, many thanks! 🙂
Wow, thanks
You are the best. Keep it up 👍
Very Very nice sir.
So nice of you, dear
Thanks for your feedback! Cheers! 😀
You are awesome. Keep it up 👍
In this situation we can easily prove that both, the sum of the yellow plus the pink area and the sum of the green an the blue on are 50% of the total size of the square (=a²), if the side length is a. Let us shortly show this with yellow and pink area, if we have divided each of them by connecting the point in the middle with the left down edge or the upper right one. Now we look for the two combinations of triangles in our yellow and pink area from left to right and from bottom to up. We have two height values in each case, for example h and a-h. The size is in this situation...
(1/2)*(a/2)*h + (1/2)*(a/2)*(a-h) = (1/4)*a²
Therefore the sum of both is a²/2 as mentioned before. With this we can easily determine the size of the blue area (=BA).
12x-1 + BA = 6x+2 + 6x+5 12x-1 + BA = 12x+7 BA = 8
Brillint solution.
Trazar lineas desde el vértice central a las cuatro esquina del cuadrado → Los triángulos que tienen por base el mismo lado tienen igual superficie → Serie de superficies triángulares resultantes: a, a (6x+5-a), (6x+5-a), (6x-6+a), (6x-6+a), (8-a), (8-a) → La suma del primer y último valor de la serie es la superficie azul → Azul =8-a+a=8
Gracias y saludos.
Excellent!
Thanks for sharing! Cheers!
You are awesome. Keep it up 👍
Connect all four corner points with the intermediary point P.
Herein NW = a + b, NE = b + c
SE = c + d, SW = d + a
Hereby
NW = NE + SW - SE
= 6 x + 5 + 6 x + 2 - (12 x -1)
= 8 sqr unit
Tricky at first but easy at the end
Very weird problem! Interesting to watch the circuitous (literally) solution.
What is most interesting is how the equations for the Areas of pink green and yellow came about...6x+5 12x-1 and 6x+2 respectively. 🤔
Yellow+red-green=light blue
area of Blue region : S
S+12x-1 = 6x+5+6x+2
S = 8
好神奇
Bu soruyu kafadan çözdüm