Can you find area of the Blue shaded region? | (Think outside the Box) | #math #maths | #geometry

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Can you find area of the Blue shaded region? | (Think outside the Box) | #math #maths | #geometry

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[yalchingedikgedik]

Very useful and nice
We are learning more from your cases
Thanks Sir
Good luck with respects
❤❤❤❤

[MegaSuperEnrique]

That is a specific case. But in general, if the triangle has sides a, b, 8, then the blue area =(a+b)² + (a-b)² = 2(a²+b²)=128

[carlofvo]

Area blu= (x+y)^2+(x-y)^2
=2(x^2+y^2)=2*64=128

[highlyeducatedtrucker]

"Assume the area of the small square is negligible."

Okay, I have an engineering degree, so I'm fine with that. But that makes the result an estimate, not a precise calculation. I am not seeing how the area of the small square + the big square equals *exactly* 128.

[georgebliss]

By similar thinking, if the red line is rotated anti clockwise to coincide with its base, then the result is 4 equal squares of side 8.

[TurquoizeGoldscraper]

That's very meta, using the lack of information to assume that the ratio of the square sizes can be ignored; I have done that with some of the puzzles.

For something that would get marked correct on a Math test:
Set the medium boxes to size a, and let b be the other side of the right triangle.
By Pythagorean Theorem: a^2 + b^2 = 8^2 = 64
The two squares have areas: (a+b)^2 and (a-b)^2
A = (a+b)^2 and (a-b)^2
A = a^2 + 2ab +b^2 + a^2 - 2ab +b^2
A = a^2 + b^2 + a^2 + b^2
A = 2 * (a^2 + b^2)
A = 2 * 64
A = 128

[marioalb]

Same words of video, with other position for segment "8"

"Dimensions of blue squares are not fixed"
"We are going to think outside the box "
"Let's assume that both blue square are equals"
"The result of this blue region is going to become a rectangle, same as white area, with minor side 8cm"
"And now we are going to calculate area of this new blue rectangle":
A = 2 s² = 2 . 8² = 128 cm² ( Solved √ )

[unknownidentity]

Let's find the area:
.
..
...
....


Let w, b and B be the side lengths of the white squares and the smaller and bigger blue square, respectively. From the diagram we can conclude:

b + B = 2w
w² + (B − w)² = 8²

(b + B)/2 = w
w² + (B − w)² = 64

[(b + B)/2]² + [B − (b + B)/2]² = 64
[(b + B)/2]² + [2B/2 − (b + B)/2]² = 64
[(b + B)/2]² + [(2B − b − B)/2]² = 64
[(b + B)/2]² + [(B − b)/2]² = 64
b²/4 + b*B/2 + B²/4 + B²/4 − b*B/2 + b²/4 = 64
b²/2 + B²/2 = 64
b² + B² = 128

Therefore the area of the blue region turns out to be A=b²+B²=128.

Best regards from Germany

[CliffordMorris-lslc]

I find this idea a problem. You make the small square become negligible which makes sense but you seem to maintain the square shape of the two smaller squares which I do not understand if the blue square shrunk would you not have the diagonal of a rectangle much smaller than the square. Suddenly this diagonal has become the diagonal of a small square . I understand the process after but cannot understand the deformation into your new out of the box diagram. Could you not explain it in more detail to see your reasoning.

[devondevon]

128
Let the length of the unshaded square = n
Let the length of the small leg of the triangle =p
then n^2 + p^2 = 64
Since the length of the unshaded = n
then the length of the small shaded = n-p
and the length of the large shaded = n+ p
their combined area = (n-p)^2 + (n+p)^2 = 2 (n^2 + p^2) = 2 ( 64) = 128 Answer

[bricepilard]

General formula is, (a+b)²+(a-b)² =128
with a and b sides of the right triangle.

But with your approach, it would be easier to do the opposite and maximise the small blue square.
You will get 2 identical squares, equal to the 2 white square with side=8.
Area will be 2x 8²= 128.

[timmcguire]

Since the top left square isn't fixed you can move the left end of the red line down creating four identical boxes and immediately see the area is 1/2 of a 16x16 square. No pencil needed.

[SkinnerRobot]

x² + y² = 8²

Blue area = (x + y)² + (x - y)²
= x² + 2xy + y² + x² - 2xy + y²
= 2(x² + y²)
= 128 square units

[marcgriselhubert]

No need to assume anything. That's very simple.
Be a the side length of the big blue square, and b the side length of the little blue square. We have to calculate a^2 + b^2 which is the blue shaded area.
Let's use an orthonormal center the left lower side of the big blue square, and horizontal first axis to the right.
The side length of both white squares is (a+b)/2, and if A is the left point of the segment whose length is 8 we have A(a; a) and if B is the right point of this segment we have B(a+(a+b)/2, (a+b)/2). Then VectorAB((a+b)/2; (b-a)/2) and AB^2 = (1/4).[(a+b)^2 + (b-a)^2] = (1/4).[2.(a^2 + (b^2)] = (1/2).[(a^2) + (b^2)]
We know that AB^2 = 8^2 = 64, so (1/2).[(a^2 +(b^2)] = 64 and a^2 + b^2 = 128, which is the solution of the problem.

[sergioaiex]

Solution:

Since we have 2 blue squares (one larger and one smaller) and 2 identical white squares, let's dimension the sides

Largest Blue Square
Side a

2 Identical White Squares
Side b

Smallest Blue Square
Side (2b - a)

Blue Area = a² + (2b - a)²
Blue Area = a² + (4b² - 4ab + a²)
Blue Area = 2a² + 4b² - 4ab
Blue Area = 2 (a² + 2b² - 2ab) ... ¹

Pythagorean Theorem

(a - b)² + b² = 8²
(a² - 2ab + b²) + b² = 64
a² - 2ab + 2b² = 64
a² + 2b² - 2ab = 64 ... ²

Replacing ² in ¹, we have:

Blue Area = 2 (a² + 2b² - 2ab)
Blue Area = 2 (64)
Blue Area = 128 u. a.

[AhmedMarzoug-fr]

small square (x-y)**2 + big=(x+y)**2 =2x**2 +2y**2=128

[michaelkouzmin]

Let me present an "exact" solution w/o assumptions ;)))
1. Let b = side of the large blue square, a = side of the small blue square;
2. side of the "white" square = run = (a+b)/2;
3. the "rise" of the triangle (with hypotenuse equal to 8) = b-(a+b)/2 = (b-a)/2;
4. Let us apply Pythagorean theorem: rise^2 +run^2 = 8^2;
((a+b)/2)^2 + ((b-a)/2)^2=8^2;
((a^2+2*a*b+b^2) + (b^2-2*a*b+a^2))/4=64;
(2*a^2+2*b^2)/4=64;
(a^2+b^2)/2=64;
Ablue = a^2+b^2 =128 sq unts.

[jimlocke]

The problem statement implies that blue area is the same regardless of the angle between the red line and the line common to the two, equal size, white squares, so long as a valid figure is produced. An angle of 0° is not valid because the triangle collapses into a line. The angle can not be 45° because the upper blue square disappears. An angle exceeding 45° does not produce a valid figure, so the angle must be less than 45°. PreMath's solution focuses on what I call the limiting case where the angle becomes closer and closer to 45° but does not reach 45°. There is another limiting case where the angle becomes closer and closer to 0° but does not reach 0°. The 2 blue squares approach having sides of length 8, area 64 square units each, total area 128.

The limiting case is great for a multiple choice test and also for checking your result after solving the general case. This is the first time I've seen someone like PreMath solve the special case and apply the result to the general case using the implication in the problem statement. The general case is not hard to solve and the solution can be found elsewhere in the comments.

[TheGreyfoo]

It's very interesting to look at the other extreme: where the triangle with 8 as the hypotenuse approaches an adjacent length of 8, ie, the extreme limit where the triangle has no height. In this case, the blue squares would be of equal size, each 8x8. These two combine to the same total area of 128 square units. I know there's some pythagorean stuff lurking here, as the hypotenuse is a fixed value (8), and equal to the sum of the squares of opposite/adjacent sides ... but now I wanna go animate it :)

[keithwood]

Would like to see this geometric relationship animated.