Calculate area of the Blue shaded region | Fun Geometry | Important Geometry skills explained

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Calculate area of the Blue shaded region | Fun Geometry | Important Geometry skills explained

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[YHoitink]

I solved it differently. Intuitively, I thought that the opposite shapes must combine to make the same number: yellow + blue = pink + green. Whenever you move the middle point where the shapes meet, it will add surface for one area but subtract surface for the opposite area, meaning the sum will be the same. The middle point is shifted from the central point, but that won't affect the surface of two opposite shapes. You can use this to calculate blue = pink + green - yellow = 16+32-28=20.

[alexniklas]

Let's denote the vertices of the square, starting from the upper left corner, clockwise - A, B, C, D. The midpoints of the sides of the square - E, F, G and H, the point of intersection of the lines inside the square - O. Let's connect the midpoints of the sides of the square ABCD with lines EF, FG, GH and HE. The lines will divide the original quads into a series of triangles. Areas of triangles
EFO+HOG=HEO+FOG, since their bases are equal and equal to the sum of their heights.
Required area
32+16-28=20.
Thanks sir!❤

[santiagoarosam]

Uniendo el punto central con los cuatro vértices del cuadrado, éste queda dividido en ocho triángulos; los dos que corresponden a cada lado del cuadrado tienen la misma superficie puesto que su altura y base son iguales → Si calculamos la superficie de cada uno de ellos comenzando por los correspondientes al lado superior del cuadrado y continuamos en sentido horario, obtenemos la serie siguiente: a → a → 32-a → 32-a → 28-(32-a)=a-4 → a-4 → 16-(a-4)=20-a → 20-a → Área azul = Suma del primer y último valor de la serie anterior =a+(20-a)=20
Un saludo cordial.

[Escviitash]

My approach was that both horizontally and vertical the difference in the area is the same.´. i.e green - yellow = blue - pink and also green - blue = yellow - pink.
The video indirectly uses the third way to split 4 parts into to groups i.e green + pink = blue + yellow.
Plug the known values into either equation and you will get that blue = 20.

[peterdevos]

I just went clockwise: A= 32-D -> A-32 - (38-C) etc. Ends up with A= 32 - (28-(16 - (X - A))) where X is the blue area. A disappears from the equation and one is left with X = 20

[tombufford]

At a quick glance, assuming integers, then 16 + 28 +32 =76. The next square integers are 81 and 100. 81-76 =5, this does not give an integer side length. using 100, sqrt(100) = 10, . Area of the blue region is 100 -76 = 24. This gives a slightly larger square than yours.

[n.]

Think you! It's need draw for to understand.
I don't now English, I'm sorry)

[aeroscout]

I find it fascinating to see what you don't have to determine to solve some of these problems. The individual values for A, B, C, and D were not needed to solve the problem at hand. I wonder how many students get fixated on trying to find them.

[tassiedevil]

Perhaps worth noting that this proof of the equality of the combined areas of the pairs of opposing regions, using the subdivision into triangles, i.e. (A+B)+(C+D)=(A+D)+(B+C) as presented also works for similiar partitioning of any quadrilateral, since it only requires common bases (i.e. the bisected edges) for each same lettered pair of triangles.

[normanc]

Thank you sir. You are smart to think of 2 triangles with equal base and height are equal. That is the key point to solve the unknown.

[WaiWai-qvwv]

You are very best man
I always respect

[mikemanh]

I dont know any rule about it, but if you connect the corners of a square with a random pont inside a square, it divides the area to 4 part and the sum of the opposite parts are always equal. Like Red+green=yellow+blue.

[misterenter-izrz]

A very unimaginably puzzle, for possession of beautiful solution 32=a+b, 28=b+c, 26=c+d, the area of blue region is a+d, thus it is 32+16-28=20, done.🙂

[HappyFamilyOnline]

Great explanation👍
Thanks for sharing😊😊

[tombufford]

Your method clearly ensures the common point is fixed and the side lengths of each area are as described. Triangles 'A', have the same base length and height and hence they have equal area giving the desired result.

[dave]

It’s not the same when you don’t reference the formula for the area of a triangle…

[drlift]

I assumed that 32+16=28+x so blue area is 20.

[mikefon]

Can we say that wherever we put the common point inside the square we always have that the sum of areas of opposite quadrangles are equal? i.e.: is it always true that green area + pink area = yellow area + blue area? Is this a valid demonstration or another one more complete is needed?

[alster]

Wow! Another fantastic geometry problem.


Cheers from The Philippines 🇵🇭

[amn]

When I had been trying to solve this issue for 2 minutes, I unexpectedly realized that we'd better write four equations and four variables!On the contrary, since we had no clue about the blue region's area, it dawned on me the most suitable method to figure out this problem is to draft three equations, subtract one equation from another and at last, sketch one variable in terms of the other one!On the whole, thanks a lot for your own mind-challenging exercises!peace out!🙂