What Is The Square's Area?

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Press

[evolutionxbox]

“My favourite right triangle theorem” - the comments have gotten to him

[tchn]

0:51 "Thou who shall not be named"

[morodochable]

X is less than 4 and greater than 2. Area about 9? Close enough for a physicist.

[EastBurningRed]

I used heron's triangle area formula for the two triangles (with the side of the square as x) added the triangle with area 1/2*x^2 and set it equal to the combined area of the triangle with sides 4, 5, and sqrt(2)*x. I plugged it into wolfram alpha to solve the equation for x and got it to be sqrt(1/2*(41-sqrt(511)))

[bluerizlagirl]

From the diagram, it is clear that x must be less than 4. (The other solution corresponds to another, bigger square with the intersection point somewhere on the inside of it, but that is not what we drew here.) This is a quicker way to eliminate the unwanted answer.

[anishsrinivasan]

“My favourite right triangle theorem” that's a step in the right direction!

[MathPythonYT]

Thanks for sharing this problem! I always like channels who enjoy sharing knowledge, and I hope I can be a small part of that community

[machmudahabalhajad]

I’m pretty tired, so I just thought about how I would go about solving. My thought was to use the laws of sin and cos, as well as the fact that the sum of the angles in each triangle is 180, and the external angle of a square is 270. Idk if that’d even work if I’d tried it, but I like your solution so much more! Very nice!

[brossjackson]

What I did with this one:
Let the obtuse angles in those triangles be A and B. since A + B + 270, we have cosB = -sin(A) and thus (cosB)^2 + (cosA)^2 = 1. Applying the law of cosines to the two triangles (with x as the side of the square) gives the same quadratic you had.

[NestorAbad]

Nice problem, thanks for sharing it and for your solution!

There's a compass-and-ruler method to find the unknown square that we can also use to find its area by analytic means:
Consider 3 circles c₂, c₄ and c₅ with radius 2, 4 and 5 respectively, all with center at point E=(0, 0).
Without loss of generality we can take point B=(2, 0) on circle c₂. Now we know that C and A must be on c₄ and c₅ respectively. Furthermore, if we rotate C 90º with respect of B we land on A, so if we rotate c₄ 90º around B (call c₄' the new circle) then the intersection c₄'∩c₅ will give us A. Finally, AB² will be the area we are looking for.
Let's do all these steps:

c₄: x²+y²=16, c₅: x²+y²=25.
As B=(2, 0), c₄' is a circle of radius 4 centered at E'=(2, -2), so c₄': (x-2)²+(y+2)²=16.
Now to find c₄'∩c₅ we must solve the system

(x-2)²+(y+2)²=16
x²+y²=25

Expanding the first equation and regarding x²+y²=25, we get y=x-17/4. Replacing this y into the second equation gives us

x² + (x-17/4)² = 25

with solutions x=(17±√511)/8. Here, as you point in the video, we must ignore one of the solutions, in this case the one corresponding to the negative root.
Hence x=(17+√511)/8, y=x-17/4=(-17+√511)/8 and we've just found the point A=((17+√511)/8, (-17+√511)/8).

Finally, the area of the square is

AB² = ((17+√511)/8 - 2)² + (-17+√511)/8)² = (41-√511)/2.

[MKD]

After the success of Gougu's theorem, we have Brahmagupta's quadratic formula!

[Amor_fati.Memento_Mori]

I've Learned that growing up doesn't make me learn things I previously couldn't. Thanks.

[gvs]

I got to the same equation in 2:41 using law of cosine and using "x" and "270-x" for the angles.

[kaltkalt]

I figured the diagonal of the square was 3, making a 3-4-5 right triangle.

[Shalok_]

2:59
Proud to be a Indian
👍
Lots of love from India❤

[naturelover]

I solved this question using cosine formula..But I think your approach is even better

[マリタイム-wg]

This question looks very easy but very difficult to solve. From JAPAN 🇯🇵

[jonathancheng]

I solved this, but used Law of Cosines.

[ttyagraj]

I have a solution,

Let A, B, C be the three vertices of a square and D be the point joined to the 3 vertices.
AD=5 BD=2 CD=4 and AB=BC=x

Since AB=BC, we can say that 🔺ACD is inscribing a circle with centre B
Then by formula we know that angleADC=angleABC/2=90°/2=45°

Now that we know that AD=5, CD=4 and angle ADC=45°,
Thus, by cosine law

AC=root(41-20root2)
Since AC is the diagonal, we can conclude

2x^2=41-20root2
x^2= (41-20root2)/2 which is roughle equal to 6.35 sq. units

Thus we get the square's area . Pls can someone tell what is wrong with my method.

[aryadeep]

The quadratic formula is also known as Sreedharacharya's formula not Brahmagupta's formula.

Brahmagupta's formula is used to calculate the area of cyclic quadrilaterals.