End Result Will Blow Your Mind!!

[@raihanfitriansyah2372]

I just made the two square same size, so the corner of both square is in the middle of the semi circle, since diagonal of the square is the radius which is 8, using Pythagorean theorem, we get the both side of one square as 4√2, so the area of one square will be (4√2)² which is 32, since both square is equal so area of both square will be 32 × 2 which is 64

[@ankurclimbs]

You deserve that ❤

[@dougnettleton5326]

You don't even need to rationalize the denominator. The length of the sides of each square is 8/√2. The area of each square is 64/2, and the area of both is 64.

[@Hello_myselfhamza]

Just use the formula :- half * diagonal^2

[@alexandergutfeldt1144]

The area of a square is half the product of its diagonals. Since you have two identical squares and the diagonal is the radius of the circle (16/2=8) , the result is 2* (8 * 8 / 2) = 64.

[@taito404]

my trick was, since the two squares could be anywhere, I just assumed it doesn't matter. So I made it so that the two squares are perfectly equal then made some simple trigonometry process. Of course it's not a reliable method but it's what I would have done if this question came up in a competition

[@taito404]

This approach took me like 20 seconds to arrive at the answer. To be fair, this only works if my initial assumption was correct. But I was confident that my assumption was correct since if it wasn't, then the problem would lack specifications

[@ramharshdandekar5048]

This solution is only applicable if sides of both the square is not equal. If you figured out that 😉 Then react with 👍🏻

[@BrainStationAdvanced]

I cancelled (x-y) from both sides of the equation

[@nishisingh227]

Very nice !

[@jl2you]

3:42 : to cancel out (x² - y²) , you must first assume that this quantity 0, so that xy. Then, in the case of x=y, the proposed solutions in the comments can be applied.

[@sebastianki-t2m]

This way is easy to solve: r² = x² + (x-k)² = y² + (y+k)² x²+x² - 2*x*k + k² = y²+y² + 2*y*k + k² 2*x² - 2*y² = 2*x*k + 2*y*k x² - y² = x*k + y*k x² - y² = (x+y)*k (x+y) * (x-y) = (x+y)*k | 1/(x+y) x - y = k insert into r² = x² + (x-k)² r² = x² + (x-[x-y])² r² = x² + y² or insert into r² = y² + (y+k)² r² = y² + (y+[x-y])² r² = y² + x² A = x² + y² = r² = (16/2)² = 64 unit²

[@bryanalexander1839]

This eliminates the concern someone else pointed out regarding dividing out (x-y), which could be 0, since (x+y)>0 as lengths are always positive.

[@KuaterCraft]

Lets assume that both the square's are present in mid point of circle therefore the diagonalswill be equal length of 8 cm as the radius of circle because the are will be the same no matter where it is. We know ddiagonal of a circle can be given by D = √2 a where a is the side of the square 8 = √2a There fore a = 4√2 Area of 1 square= (4√2)² Area= 32 Area of both square= 32 x2 = 64 units

[@tmlen845]

If the position of the squares doesn't matter, they can simply be placed so that they are equal sizes and meet at the center. Then a square's diagonal equals the radius 8, and the side length and area can be calculated using Pythagoras.

[@JUGNUMEHROTRANEETASPIRANT]

Now , I just saw the thumbnail , &/Simply assumed R = radius = 8(units) , (&) (a,b) are the sides of the smaller & Bigger square Respectively , (&) Assumed the space between the vertice of the bigger square , & the center of the semicircle ,to be Q. Thus , I used the exact same method , as you and found 2.(a+b).Q = 2.(b^2-a^2) , OR => Q = (b-a) , which meant (3.b^2 + a^2 -2.a.b. + 2.a.b - 2.b^2) = b^2 + a^2 = R^2

[@mrmangoheadthemango]

I assumed the squares to be equal. This means the diagonal of each of the squares is a radius. Since in the diagram it is shown the circle already has a radius of 16 / 2 = 8, we know the side length of the square is 8 / sqrt(2) = 4sqrt(2). After that, I did 2 * (4sqrt(2) ^ 2) which equals 64

[@aligaming5583]

Omg my mind holy moly😮😂❤😅

[@drdegenerate814]

I just draw 2 equal squares with diagonal 8 So you have a big rectangle with mid point of one side meets at the 2 opposite ends of other side(this make a equilateral right angeled triangle) By this i got the lengths 8√2 and width 4√2 Just multiply 8√2 x 4√2 = 64 units

[@thehorizonx1540]

This is a really interesting problem. And thank you for explaining it so well.

[@Aayush-242]

I thought of a simple idea i assumed that first square's area is 0 unit sq then the other square sides will be in it's highest length and highest length in the semicircle is radius so the side of the square becomes 8 so it's area becomes 64 unit sq. Now, 0 unit sq + 64 unit sq = 64 unit sq.

[@thatsmetube]

I had that idea too. But then I realized that even with the biggest possible single square inscribed in the semicircle, you still have a good size second square possible to inscribe. And a square of sidelength 8 can not be inscribed in this case…

[@Aayush-242]

@@thatsmetube The question does not include that square is inscribed in semicircle. It only said that the base lies on that diameter and two vertices touch the boundary of the semicircle.

[@suzk1234]

I took the 2 squares to be equal, and let them share vertices at the centre of the semi circle. Thereafter I drew a diagonal having radius "r", and we know tis half the diameter, I.e, 8. (I forgot to mention.....diagonals would be equal and so will be the areas). Instead of Pythagoras theorem I tried out the ½d² formula, and when multiplied by two (two equal squares), the answer came 64.....m not exactly sure if I should've assumed them to be equal at the first place😅

[@g-z-L-2039]

Diameter=16 => Radius=8 Assume that both squares are equal So the area of a square is (r^2)/2 = (8^2)/2 =32 So the total area is 32*2 =64 square unit

[@Valkies-zj2bi]

64 just make them equal

[@RishanAjayFJ]

I solved this by trigonometry similar like you thought

[@marioalb9726]

A = R² = 8² = 64 cm² ( Solved √ ) WHAT ?????? A = 2 (½R²) = R² = 64cm² Same as above ????? A = 5/4 . (2/√5 R)²

[@marioalb9726]

We have to know that the required area doesn't change with the dimensions of those two squares, as long as the original conditions are meeting We can choose, at our convenience, the position and dimension of those two squares !! I chose the blue square centered in the semicircle. The green square, consequently, will have an area equal to 1/4 of blue square area : A₁ = 4 A₂ A = A₁+A₂ = A₁+ ¼A₁ = 5/4 A₁ and A₁=s² , and : s = R.cosα s = R cos(atan 1/2) s = 2/√5 . R Therefore: A = 5/4 * (2/√5 R)² !!! A = R² !!!

[@marioalb9726]

Other option is to match areas of those two squares In this case, the diagonal "d" of each square, is the radius of the semicircle. And area of square is: A = ½d² = ½R² and there are two squares A = 2 (½R²) = R²

[@proboiz_50]

There's no way a primary schooler can do this in seconds!! And my answer was 128 units , i know its wrong I just assumed that no matter where the squares are , the area will remain same , and i moved them a little so that they will be of equal area And then i assumed that the side is 8 units (don't ask me why) and the 8² × 2 = 128 units I should not multiply it!!

[@thatsmetube]

If you’re stupid like me you may think that “if it works with all two squares, I can set one square to area of zero and the other to be “the single square of max sidelength that fits into the semicircle”. Of course you can not set one square to be zero since even with the max size inscribed single square, you have a good size non-zero neighboring square (and a max size single inscribed square can not have sidelength 8 = area 64…

[@bryanalexander1839]

That is not actually stupid. The maximum square contained within the semi-circle would be the one where two vertices are on the semi-circle perimeter and that happens when the angles from the diameter to the diagonals of the square are at 60° (or π/3) and 120° (or 2π/3). Any angle larger than π/3 for the rightmost square or smaller than 2π/3 for the leftmost square causes one of the vertices to appear outside of the semi-circle. Continuing to move the vertices that lie on the perimeter of the semi-circle around will make a single square with the one vertex outside of the semi-circle, one directly below it at an intersection of the diameter and semi-circle, one at the center of the diameter, and one directly above it in the center of the semi-circle perimeter. Obviously, it is a square with r^2 area. The other square, having been rendered dimensionless, will no longer exist, just as you had thought to do. Your only problem was wanting to inscribe versus allowing it to circumscribe half of the semi-circle.

[@ZDTF]

This problem ain't nothing It won't blow my mind 🤯

[@BrainStationAdvanced]

is it blown or not?

[@jordierinsyaputra1932]

Believe in algebra baby!

[@RishanAjayFJ]

Solve this 0°c =273°F 0°c+0°c = 273°F + 273°F => 0°c = 546°F But how is it possible?

[@Lll-b8j]

1. I don't believe 0 Celsius is 273 Fahrenheit. 2. That's not how adding temperatures work. You need to convert them to an absolute zero based unit such as Kelvin and then convert it back.

[@xabixps]

One detail, 0ºc = 32ºF = 273.15ºK. In the second line when you add 273ºK you are not adding a number but a temperature which is absolute zero. Absolute zero + absolute zero is zero.

[@ZDTF]

Let's see

[@usmanchughtai1751]

Took me a few minutes because I am tired and just got off work, but this isn’t really anything remarkable.

[@jskwwkdkek3830]

If squares can be anywhere suppose they are symmetrical and have same side lenght and solve it ez solution

[@malaykumarsingh2808]

64 sq units