What Is The Square's Area?
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- Опубликовано: 7 фев 2025
- Can you solve this delightful problem? Thanks to Elmer for the suggestion!
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0:51 "Thou who shall not be named"
🤣
shalt*
Yes, and his name is "Baudhayan" 🙏
.
and
3:00
Thou who has not been credited to create the perception
"all math was done by Greeks"
by Voldemort's theorem ...
@@ShekaranJagadeesan U enlightened me bro👍 thanks for information
“My favourite right triangle theorem” - the comments have gotten to him
I guess, everyone is fine with him finding a neutral way out. XD
finally!
Yes, but at what cost? He’s now labeled the quadratic equation.
Yes! That is so sad and funny
@@ehess1492 it’s been called that for a looooong time
It looks like 3D to me...
I'm glad you said something. That is exactly how I interpreted the question at first (before clicking on the video) and was therefore confused how the square had side lengths of sqrt(21) and sqrt(12) haha
Same here bro 👍
@@Andyg2g that's what I thought so!
@@Andyg2g same!! except then i thought it might not be square after all xD
It's alright, but be reminded that a 3D figure like that should contain some dotted lines for establishing the 3D sense of the figure 😉
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)))
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!
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
Hii! You may have to visit this channel for more harder questions #onlymathlovers ..
@@madhukushwaha4578 You should also visit mine!
X is less than 4 and greater than 2. Area about 9? Close enough for a physicist.
I am tired, but I got x between 3 and 6 at first glance... didn't start working on the problem yet.... brb
NVM, if the kids both take a breath at the same time, which is rare, the effing dog barks. This is why I don't grade papers at home....
Hii! You may have to visit this channel for more harder questions #onlymathlovers ....
Unless it's space tryrants we're dealing with
No. You are a fail physicist for testing.
Actually you can say it is greater than 3 and less than 4 if you consider the triangle inequality theorem in both triangles. So the area is somewhere between 9 and 16.
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.
Whoa you are from 6 days in the future!
@@michaeldaunicorn5314 omg !! Time Traveler
Cool
@@Oylesinebiri58 He's a supporter on his patreon…
@@Bry10022 No! HE TIME TRAVELED!
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.
This one was easy. E is the point where three circles meet: that with center at (0,0) radius 2; that with center at (L,0) radius 4, that with center at (0,L) radius 5.
Solving for x and y you find x=(L^2-12)/2L and y=(L^2-21)/2L. Substitute into x^2+y^2=4 and solve for L.
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.
Hii! You may have to visit this channel for more harder questions #onlymathlovers ..
2:59
Proud to be a Indian
👍
Lots of love from India❤
Hii! You may have to visit this channel for more harder questions #onlymathlovers ..
After the success of Gougu's theorem, we have Brahmagupta's quadratic formula!
It's sri Dharacharya's
@@vinayakpandey1032 lekin vo to Brahmagupta bol raha hai.
😂😂😂😂😂😂
It's sridhar acharya's formula
I've Learned that growing up doesn't make me learn things I previously couldn't. Thanks.
This question looks very easy but very difficult to solve. From JAPAN 🇯🇵
Excellent solution! Another way to disambiguate the solution at the end is by looking at the bottom triangle. Since the hypotenuse of the triangle is 4, and x is one of the sides, we know that x has to be < 4, and x^2 has to be < 16.
Hii! You may have to visit this channel for more harder questions #onlymathlovers ..
“My favourite right triangle theorem” that's a step in the right direction!
One small step for normal people - one giant leap for Presh.
The "right" direction 😂
Why are you so triggered? It's a step in the right direction for your narcissism.
I got to the same equation in 2:41 using law of cosine and using "x" and "270-x" for the angles.
Likewise!
I have the same idea
Yeah solved in this way only
Yes, exactly. So much faster.
Can you elaborate plz. I have difficulties to compose proper the cosine formula -equations that lead to the equation in 2:41.
My first thought was "put it into the Cartesian coordinate system ez" (basically what he did) but then I forgot and began to think about Carnot and other strange angle-related systems of equations...
Thank you Cartesio, much easier.
Great video and funny problem, this time you got me XD
I remember solving this problem but with E inside the square. You can rotate E 90° counterclockwise around B to obtain the point E'. Then triangle BEC is congruent to triangle BE'A, thus E'A=4. Then apply law of cosines in triangle EAE' to obtain the cosine of angle BE'A.
I solved this question using cosine formula..But I think your approach is even better
I used the Pythagorean theorem and got 8.041 but the sides wouldn't make it a square.
What's your real name?
I never try to solve these, only try to see if I can think of how to solve them. Then I watch to see if you use what I came up with. Today I had the right triangles in the right places
Let the intersection point be E
Square side x
El kashi theorem:
x²=20-16cosCÊB (1)
x²=29-20cosAÊB (2)
From (1) we can deduct that:
x²
beautiful. here i am pondering for 2 days trying to solve with trig and playing around with ratios.
Thanks for posting these. It's great to work the mind with things we've either forgotten, or not yet learned.
I'm a little disenchanted by a lot of the comments. Apparently they don't teach some pretty basic geometry/trig/calculus in HS any more.
Hii! You may have to visit this channel for more harder questions #onlymathlovers ..
Alternative methode using x, y coordinates: consider the left lower point of the square the origin point (0,0), use the law of distance between two points, you get three equations of three variables, solve it the same he completed the solution.
Let m(angle(ABE)) = theta implies m(EBC) = 270 - theta.
Using the law of cosines on triangle(AEB) and triangle(BEC) we obtain:
4x * cos(theta) = x^2 - 21
4x * sin(theta) = 12 - x^2
implies
16x^2 * cos^2(theta) =x^4 - 42x^2 + 441
16x^2 * sin^2(theta) = x^4 - 24x^2 + 144
Adding the two equations above we obtain:
2x^4 - 82x^2 + 585 = 0 implies x^2 = (41 + - sqrt(511))/2.
For (+) sin(theta) = (12 - x^2)/(4x) < -1, therefore drop (+) and choose
x^2 = (41 - sqrt(511))/2.
Looking at it in the morning... how the hell he is going to do that... he starts... my brain... well why havent i thought of that... xD
I solved this, but used Law of Cosines.
I tried solving it like that, I got a formula with one unknown in it, but it was quite complicated and I didn't know how to solve it.
Yes I too solved with cosine formula
Yeah did the same let the angle between side of square and and side length 4 be theta
Similarly we get that that the angle between side length 5 and side is 270-theta
Get two equations in side length and theta and thus solve
@@yashvardhan2093 Exactly
me too!
Another solution, rotate the 2|4|x triangle so that the x of each triangle align. Create a 2|2|2root2 triangle in the new gap. SSS calculate the angle between 2root2|4. Subtract 45, SAS x and then square.
Here's a trigonometry solution:
Let ∠EBA = θ. cos θ = (x^2 - 21)/(4x) and cos (270° - θ) = -sin θ = (x^2 - 12)/(4x).
Using sin² θ + cos² θ = 1 solves the value of x².
the simpler way to find the answer:
a^2 + 2^2 = 5^2
a^2 + 4 = 25
(subtract 4 on both sides)
a^2 = 21
(take the square root of both numbers)
a = 4.58
(now to find the area)
4.58 * 2 = 9.16 (or 9.2)
You always come up with 👍👍👍 unpredictable solutions
Hii! You may have to visit this channel for more harder questions #onlymathlovers ...
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.
The quadratic formula is known as 'completing the square 'formula.
The quadratic formula is KNOWN AS the quadratic formula!
I used a slightly different approach. I took three triangles, one with sizes 2,4 and x, one with sizes 5,2 and x and one with sides 4,5 and x*sqrt(2). Then I applied the cosine rule on the three equations. This lead to three equations and three unknowns (x, corner alpha and corner beta) from which x could be solved (by computer).
Presh- My favourite right triangle theorem.
Me- my favourite channel
Hii! You may have to visit this channel for more harder questions #onlymathlovers ..
Excellent method and answer,there is no easier way out of this.If you can get a relationship between the angles in a simple way there might be another way.
2:47 You'll let it be equal to.... Me? 💓
Aw gosh, I'm gonna blush ☺️🤭🥰
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.
If ACD were to inscribe a circle with a center in B, that would require BD to be a radius of the circle, which would mean x = 2.
@@Umbra451 yup I agree but what is wrong with my solution?
@@ttyagraj9554 that is what’s wrong with your solution. You incorrectly assume that the circle ADC has its center in B, which would be required for the rest of your solution to work.
@@Umbra451 thanks a lot , I realised my mistake.👍👍
2:42 You can get to this point quicker by using law of Cosine of the 2 triangles. The two angles joint with the square (call them i,j) add up to 270 so cos(i) = -sin(j). Eliminate them by square them and add give you the above equation.
Yep that's what I did!
Al-kashi’s law? 😝
i think cosine law can be utilized to solve this problem as well. Please hear me out. Draw line from top left to bottom right corners of square, lets say this line has length l. let side of square equal to x. using cosine law the angle between 5 & 4 is the sum of the angles between 5 & 2 and 2 & 4 which can be written as equations in terms of x. finally equate that to the angle between 5 & 4 by using the cosine law again but this time, in terms of length l, which is merely sqrt(2 * x^2 ). I think you can solve it that way too. so we have:
let a = angle between 5&2
let b = angle between 2&4
let c = angle between 5&4
cos(a) + cos(b) = [(5^2 + 2^2 - x^2)/(2 * 5 * 2)] + [(2^2 + 4^2 - x^2) / (2 * 2 * 4) ]
cos(c) = (5^2 + 4^2 - 2*x^2) / (2 * 5 * 4)
but to get a + b = c you gotta take inverse cosine of the terms, and you get quite a disgusting mess to work with after. But it is a try and can be solve with a bit of knowledge in complex numbers.... but putting the equation in wolfram alpha gives x ≈ ± 3.03271255554209831828048143... and so x^2 is about 9.19
Your solution is quite creative. I solved used cosine law and the "complementary" relationship of the two angles in the triangles (cosine of one is the negative sine of the other).
Very beautiful solutionnnn.....
Hii! You may have to visit this channel for more harder questions #onlymathlovers ....
What I find it interesting, that indeed this has two solutions for point E, when you don't give E has to be outside of the square.
Hii! You may have to visit this channel for more harder questions #onlymathlovers ....
Your solution was very elegant.
I figured the diagonal of the square was 3, making a 3-4-5 right triangle.
Alas, it was not stipulated that that angle was right
close is not correct
The diagonal can't be 3, because the square side would be 3 over sqrt 2, and this added to 2 would be less than 5, making a impossible triangle that way
@@hlcsilva yea, i see i was wrong, it just looked like the sort of trick a lot of these problems use.
Im not sure if i calculated the area correctly.
I was thinking of A^2 + B^2 = C^2 (the Pythagorean theorem) and i just saw the number ''5'' as 5^2 is C = 25 and number ''2'' as 2^2 as A = 4. C^2 minus A^2 = B^2 which is 21, then take the square root of √21 = 4.58 multiplied by 2 (the base length of A) gives me a total area of 9.16 ≈ 9.2.
Not sure if this makes any sense whatsoever...(My knowledge on math is very basic) but i paused the video, looked at it and came to this conclusion.
Love your channel by the way
I solved the problem a different away but arrived at the same answer. I got to say, your solution is really good, and I really love your videos. Anyways, the way that I solved it was using coordinate geometry. I used the point (0,0) for B, (0,x) for A, and (x,0) for C. I used the point (a,y) to represent E. (The point D is not needed, because it does not connect to E). Anyways from this, you can create a triple systems of equation with 3 variables using the distance formula. The equations are. √(a^2 + y^2) = 2, √(a^2 + (y-x)^2) = 5, √((a-x)^2 + y^2) = 4. Solving this system of equations for x, there is only one reasonable value of x, and squaring this answer would get the area.
i watch your videos to relax my self.
Hii! You may have to visit this channel for more harder questions #onlymathlovers ..
I solved that using analytical geometry... It´s fascinating how beautyful math can be
But there is a probability that the value of "a" may be negative.
In 3d, the point E may be right side to the bottom left corner point of square.
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You could have directly applied cosine rule taking external angles as X and 270-x
Can you please explain how can we say that? I'm not given in maths so i didn't understand.
That was also my reasoning. It is shorter. On the other hand, the given reasoning does not need cosine rule, making it longer, but, more creative.
@TheAW 100% agreed: as we *all* confirmed already: the given solution is more creative and elegant (because it does not need a formula, which, like you also correctly argue, may be unknown to lower grade students). But, as many others claim, cosine rule *plus* fact that ABE+CBE=270, gives a different solution... and ... it is not because someone writes down a longer solution, that nobody can write down a shorter solution :-)
Your videos teach me to think literally out of the box
This is how I thought about it, but used another theorem that helped me get AC.
The triangle of ACE is a 3-4-5 Triangle, which leads to Angle ACE having 90 degrees
Using the 3-4-5 Triangle of ACE, this leads to AC = 3. Any triangle with sides of 4 and 5 must have the other side as 3, there just isn't another way to make a triangle with sides of 4 and 5.
Using the concept that a square has all 4 sides the same and now creates a triangle of ABC, with AC = 3, we can use Pythagorean Theorem to get the sides of AB or AC, since they are equal
a^2 + b^2 = c^2
AB^2 + BC^2 = AC^2
AB = BC = x
x^2 + x^2 = 3^2
2(x^2) = 9
x^2 = 4.5
x = sqrt(4.5)
x * x = 4.5
Since we are looking for x * x anyways (area of a square is x^2), the answer for the area is 4.5
Can anyone help me out where I went wrong, and why this isn't the right answer?
Edit: The more I try to sketch it out to scale, i just don't think that this is geometrically possible with these dimensions as shown in the problem.
Solved it. Initially, I put the square on the coordinate plane, and then used analytic geometry to derive 3 equations similar to those shown in the video. From then, it is just algebra.
Hmm... this is what "Think outside the box" means!
If you set origin at B and using vectors solve for distance for AE, BE and CE you get similar equations and solve to get the solution
Hii! You may have to visit this channel for more harder questions #onlymathlovers .....
I like that this problem can be solvable in 2 and 3 dimensions and have such radically different answers
@MindYourDecisions I tried also an approximat solution to get a feel (don't know if it is correct): since a side of the triangle must be less than the sum of the other two, callingthe known sizes of triangle EBC a and 2a and the size of the square (which is also a ssize of EBC), x, we know that x2. Putting all together we know: 3
Your videos are addictive for me😂
Simple approach:
Rotate the lower triangle by 90 degree counter clockwise. Then the result is a triangle with sides 5, 4, and 8^0.5. From where the result can be attained.
Cool. Did with Cos rule comparing the angles EBC and EBA using the fact that they add up to 270, same quadratic comes.
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Using Cayley-Menger determinant, we can set the determinant of the matrix ([[0,1,1,1,1],[1,0,a**2,a**2,4],[1,a**2,0,2*a**2,25],[1,a**2,2*a**2,0,16],[1,4,25,16,0]]) to zero and solve for a.
I was an English major. Stuff like this makes me feel like rocks are rolling around in my head.
Please keep making these videos , THANKU :)
I used the two angles outside of the bottom left corner of the square, and called them "A" and "270° - A", and then I used the Law Of Cosines for both angles, turned "cos(270° - A)" into "-sin(A)", squared both equations, added the left and right sides, used the fact that cos²(A) + sin²(A) = 1, and then solved the resulting equation.
That's EXACTLY what I did
We could have used cosine rule in two triangles. Since one angle is 90, sum of other angles would be 270. Hence could have easily got x by eliminating theta.
Exactly... 👌👌👌💡💡💡
@@yashuppot3214 Let angle opposite to side 4 = Theta. Then angle opposite to side 5 will be 360-(90+theta)= 270-theta. Now apply cosine rule twice in both triangles and eliminate theta to find x.
But, as Presh showed, the problem can be solved with no need of trig.
What an Interesting Answer
I solved it by applying Heron's formula and the "unspeakable theorem" to the triangle with sides 5, 4 and 2*sqrt(2) that shares its shortest side with an isosceles triangle with two other sides equal to 2 and the third vertex inside the first triangle. The distance between the two distinct vertices of the triangles is x.
Consider EA = AC, which mean AC is also equal to 5 and EC = BC = 4, we have triangle ABC. AC = 5, BC = 4 , use Pythagorean’s theorem to find AB :
AB^2 = AC^2 - BC^2
AB^2 = 5^2 - 4^2
Finally, AB = 3 and the surface of the square is equal to 9. This is the second method to solve the square without use equations
Great problem. Loved it
the simple way to solve ( firstly we find the value of point B,Cand after we find the value of AB by the PYTHAGORAS'S THEOREM using in the triangle BCE & ABE ) thanks SIR
This is one of the most surprising problem resolutions i have seen in your channel 😱
Hii! You may have to visit this channel for more harder questions #onlymathlovers ..
This question is just remarkable
I got this one EXACTLY how Presh solved it but it was not easy. It took me a while to figure out how to solve the 3 equations in 3 unknowns involving higher order terms, but I arrived at what Presh did, which I believe is the only way to solve this. I also concluded the smaller answer had to be the one to represent the area of the square.
Thanks for featuring my problem presh!
Hii! You may have to visit this channel for more harder questions #onlymathlovers ..
Here's the process I used
The bottom left of the square is (0,0), or the origin
n is the length of the square
E is somewhere along x² + y²= 4
It's also somewhere along (x-n)² + y² = 16
Lastly, it's also along x² + (y-n)² = 25
So
1) x² + y² = 4
2) x² + y² - 2xn + n² = 16
3) x² + y² - 2yn + n² = 25
Set them to zero
1) x² + y² - 4 = 0
2) x² + y² - 2xn + n² - 16 = 0
3) x² + y² - 2yn + n² - 25 = 0
Now take 2) and use 1) to simplify
n² - 2xn - 16 = 0
And solve for x
2xn = n² - 16
x = (n²-16)/2n
Apply the same process for 3) to get
y = (n²-25)/2n
Then plug into 1)
[(n²-16)/2n]² + [(n²-25)/2n]² = 4
[(n²-16)² + (n²-25)²]/4n² = 4
m = n²
Now notice that the area of the square is n², so solving for m gets us the answer
[m² - 32m + 256 + m² - 50m + 625]/4m = 4
[2m² - 82m + 881]/4m = 4
2m² - 82m + 881 = 16m
2m² - 98m + 881 = 0
m = [-(-98)±√[(-98)²-4(2)(881)]]/2(2)
= [98±6√71]/4 = (49±3√71)/2
Now, just from the diagram it's easy to see the n
I got the same answer using the law of cosine on tringle ADE and CDE
In tringle CDE we can assume angle D = theta and angle D in the other tringle is 270- theta.
After using law of cosine and angle difference formula for cosine
We get one equation has cos and the other has sin we solve for cos and sin
And use sin^2 + cos^2 =1
4^2=x^2+2^2 -2(2) x cos t
5^2 = x^2+2^2-2(2) cos (270°-t)
Simplifying
12= x^2 -4x cos t
21= x^2+ 4x sin t
cos t = (x^2-12)/(4x)
Sin t = (21-x^2)/(4x)
Using sin^2 t + cos^2 t = 1
And sub the value for sin and cos
We get the same two answer
We know that t is angle in tringle which mean 0 0 since 4x has to be positive because x is side of square
x^2
We can also use 2 laws of cosines. There is 25=x^2 +4 -4x cos beta & 16=x^2+ 4-4x cos(360-90-beta). Then we have eq.
21=x^2 -4x cos beta & 12=x^2 +4x sin beta. We have cos beta=(x^2 -21)/4x & sin beta=(12-x^2)/4x . We see, that beta>90, there is second quadrant. If we go to first quadrant and we use Pythagor theory, we get eq. 16x^2=(21-x^2)^2 + (12-x^2)^2. After the substitution z=12-x^2 we have 2 z^2 +34z-111=0. z1=2,802; z2=-19,8 ;(x1)^2=9,2;(x2)^2=31,8. The root x2 is not posible, we have there sin beta
Turning the left obtuse traingly by 90 degrees to the right I obtained a triangle with 3 known sides (if you complete to 90-45-45 triangle). After that someone may use cos rule twice to get a side of the square.
2:59 I learned this in school as "THE quadratic formula".
Yeah I know, he also has a strange name for the cosine rule, and of course the famous name for the Pythagorean theorem.
In ncert of India it is named sridhhacharya quadratic equation.
Fellow Indian. Not everyone will understand what NCERT is.
@@zackwhite6199 I think he wants to promote that mathematician who created that formula to show his respect to them.
Not the English word instead of Cosine rule, created by Al - Kashi or Quadratic Formula, created by Brahmagupta proved by Shri Dharacharya (both are 🇮🇳) and Right Triangle Theorem is little bit controversial as who is the first person who created it between chinese, indians & pythagoras .
And I think we should support his this act.
It is the Pythagorean theorem - PERIOD!
Gougo theorem×××
My fav theorem✓✓✓
🤣
gougou? it's Pitagora
@@PierAntonioBrunello He calls it gougo theorem every video I can't remember the reason but he does that
@@hayimshamir8426 I believe Gougu was a Chinese(?) mathematician who discovered the theorem independently of Pythagoras and was technically first.
What about using the carpenter's right triangle first thing to figure length AC? AE is 5, EC is 4, angle ECA is a right angle and AC is 3. From there back out calculate the length of the blue square using your "sum of the squares" formula of choice. Ch.
I really like ur videos , unique way to get ans , it is really mind ur decisions
I hope that one day, I can see those videos, without being stressed because of the heartbeat
What's the significance of the 31.8 solution? Another configuration of the problem where point E is on the interior of the square?
yes, the distances to three of the corners would still be 2, 4 and 5
Brahmagupta formula😲..even indians aren't proud of their man🙁
3:02
its not brahmagupta's formula
its sridaracharya's formula
@@venkatswamy3763 Brother Brahmagupta created while Sri Dharacharya proved that formula.
You can say it what you want.
Just brilliant!
5squared - 2squared=21AB 4squared - 2squared= 12BC.
Square Root of 21 X square root of 12
4.583 x 3.464= 15.875512 is area of the rectangle ABCD. Accuracy of components rounded up to thousandth.
If a2 + b2 = c2, then c2 - b2 = a2. 25 - 4 = 21 and 16 - 4 = 12. Square root of 21 = 4.582. Square root of 12 = 3.464. 3.464 x 4.582 = 15.874 which is the area of the rectangle. When there are differing hypotenuse like in this scenario, the length of both sides of the rectangle has to be different.
Its my solution.
0:51 Gougu's theorem 😁😁😁
I solved this by the law of cosines. Finished like you.
You can use the cosine rule on each of the two triangles with reference to the two angles formed at the bottom left of the square. For the 5,2,x triangle, use Cos(270-ø) = -Sin(ø) and then sin^2 = 1-cos^2. (This gives two equations in two unknowns rather than your three equations in three unknowns.) A bit of manipulation and the quadratic formula yields exactly the same radical answer as in the video. No quicker than yours though.
I solved broo
Exactly the answer
But my way was sooo easier
I did using cosine theorum
(Cos ¥ )^2 + (-Sin ¥)^2 = 1
Simple
I first tried to define the angles at B and use the cosine rule, so CE^2 = BC^2 + BE^2 - 2*BC*BE*cos(theta) and AE^2 = AB^2 + BE^2 + 2*AB*BE*sin(theta) (the angle on this side is equal to 3*pi/2 minus the angle on the other side triangle, and cos(3*pi/2-theta) = -sin(theta)). It gave me two equations with two unknowns: angle theta and the side length of the square. But it was very difficult to solve and I couldn't quite work it out that way.
thats how I did it too but when u do solve it it becomes a simple quadratic equation for the area
you will want to write sin(theta) = sqrt(1-cos^2(theta)) so it becomes 2 equations with the unknowns of cos(theta) and side length
At first I thought it was an infinitely tall skyscraper. This is how it'd be modeled actually!
Finally solved it after 15 minutes of work.
Hmmm… Another 'interesting' way is to use some trigonometry (and a PERL successive approximation algorithm) to solve this.
If ∠θ is from the lower left point to the lower left of the square, and ∠φ is to the lower right point, and ∠γ is to the upper left point then,
Let H₁ be height of the θ angle, given hypotenuse of 2;
Let H₂ be height of the φ angle, which MUST be same as H₁;
Let B₁ be baseline of ∠θ
Let B₂ be baseline of ∠φ
Using trig…
H₁ = 2 sin θ
φ = asin( H₁ ÷ 4 );
B₁ = 2 cos θ
B₂ = 4 cos φ
𝒙 = ( B₂ - B₁ ) … 'side' of square
H₃ = 𝒙 + H₁; (equals 5 sin γ)
γ = asin( H₃ ÷ 5 );
B₃ = 5 cos γ
But of course we have no idea what the θ φ and γ ought to be. So, we set dθ = 0.01 radians, and start iterating. We haven't converted on the answer when B₃ is greater than B₁. When it is less, then remove one dθ from θ, divide dθ by 2.718281828 (because I like it), and continue until dθ is less than 10⁻¹¹ (again, because I like it);
Here is the output of this algorithm:
θ=1.330000 hθ=1.946969 bθ=0.457506 φ=0.508357 bφ=3.494183 d=3.036677 hγ=4.983646 γ=1.489894 bγ=0.404070
θ=1.337358 hθ=1.947442 bθ=0.455488 φ=0.508492 bφ=3.493919 d=3.038431 hγ=4.985873 γ=1.495608 bγ=0.375587
θ=1.337358 hθ=1.946378 bθ=0.460015 φ=0.508187 bφ=3.494512 d=3.034497 hγ=4.980875 γ=1.483303 bγ=0.436907
θ=1.337358 hθ=1.945983 bθ=0.461680 φ=0.508074 bφ=3.494732 d=3.033052 hγ=4.979035 γ=1.479189 bγ=0.457395
θ=1.337541 hθ=1.945923 bθ=0.461936 φ=0.508057 bφ=3.494765 d=3.032829 hγ=4.978752 γ=1.478573 bγ=0.460464
θ=1.337608 hθ=1.945900 bθ=0.462030 φ=0.508051 bφ=3.494778 d=3.032748 hγ=4.978648 γ=1.478347 bγ=0.461588
θ=1.337633 hθ=1.945892 bθ=0.462065 φ=0.508048 bφ=3.494782 d=3.032718 hγ=4.978610 γ=1.478264 bγ=0.462001
θ=1.337651 hθ=1.945893 bθ=0.462060 φ=0.508049 bφ=3.494782 d=3.032722 hγ=4.978615 γ=1.478276 bγ=0.461941
θ=1.337655 hθ=1.945892 bθ=0.462064 φ=0.508048 bφ=3.494782 d=3.032718 hγ=4.978610 γ=1.478265 bγ=0.461997
θ=1.337655 hθ=1.945891 bθ=0.462069 φ=0.508048 bφ=3.494783 d=3.032714 hγ=4.978606 γ=1.478255 bγ=0.462046
θ=1.337655 hθ=1.945891 bθ=0.462070 φ=0.508048 bφ=3.494783 d=3.032713 hγ=4.978604 γ=1.478251 bγ=0.462064
θ=1.337655 hθ=1.945891 bθ=0.462070 φ=0.508048 bφ=3.494783 d=3.032713 hγ=4.978604 γ=1.478251 bγ=0.462067
θ=1.337655 hθ=1.945891 bθ=0.462071 φ=0.508048 bφ=3.494783 d=3.032713 hγ=4.978603 γ=1.478250 bγ=0.462070
θ=1.337655 hθ=1.945891 bθ=0.462071 φ=0.508048 bφ=3.494783 d=3.032713 hγ=4.978603 γ=1.478250 bγ=0.462070
θ=1.337655 hθ=1.945891 bθ=0.462071 φ=0.508048 bφ=3.494783 d=3.032713 hγ=4.978603 γ=1.478250 bγ=0.462070
θ=1.337655 hθ=1.945891 bθ=0.462071 φ=0.508048 bφ=3.494783 d=3.032713 hγ=4.978603 γ=1.478250 bγ=0.462071
θ=1.337655 hθ=1.945891 bθ=0.462071 φ=0.508048 bφ=3.494783 d=3.032713 hγ=4.978603 γ=1.478250 bγ=0.462071
θ=1.337655 hθ=1.945891 bθ=0.462071 φ=0.508048 bφ=3.494783 d=3.032713 hγ=4.978603 γ=1.478250 bγ=0.462071
θ=1.337655 hθ=1.945891 bθ=0.462071 φ=0.508048 bφ=3.494783 d=3.032713 hγ=4.978603 γ=1.478250 bγ=0.462071
θ=1.337655 hθ=1.945891 bθ=0.462071 φ=0.508048 bφ=3.494783 d=3.032713 hγ=4.978603 γ=1.478250 bγ=0.462071
area of square is ( side = 3.032713 ) … 9.197345 units
The algorithm does NOT print out all the values far from the convergence point. Here is the code. (below my signature)
⋅-⋅-⋅ Just saying, ⋅-⋅-⋅
⋅-=≡ GoatGuy ✓ ≡=-⋅
#!/usr/bin/perl;
# include your favorite trig library if needed, such as 'use Math::Trig;'
my $h_theta;
my $h_phi;
my $d;
my $h_gamma;
my $b_theta;
my $b_phi;
my $b_gamma;
my $dtheta = 0.01;
my $theta;
my $phi;
my $gamma;
for( $theta = $dtheta; $theta < pi/2; $theta += $dtheta )
{
$h_theta = 2 * sin $theta;
$phi = asin( $h_theta / 4 );
$b_theta = 2 * cos $theta;
$b_phi = 4 * cos $phi;
$d = $b_phi - $b_theta;
$h_gamma = ( $d + $h_theta );
$gamma = asin( $h_gamma / 5 );
$b_gamma = 5 * cos $gamma;
next if $b_gamma > $b_theta;
$theta -= $dtheta;
$dtheta /= 2.71828182;
last if $dtheta < 1e-11;
printf "th=%f hth=%f bth=%f phi=%f bph=%f d=%f hg=%f g=%f bg=%f
",
$theta, $h_theta, $b_theta, $phi, $b_phi, $d, $h_gamma, $gamma, $b_gamma;
}
printf "area of square is ( side = %f ) ... %f units
", $d, $d**2;
solved, thanks for creating such channel.
3:06 - THAT answer would be if those lines drawn were inside the square! Fascinating.
Now consider this a three-dimensional model where the square is on the x and y coordinates and the point is on the z.
Who would like to solve that one?
Ans: Area is 20 1/2, so how do you get there? :D Maybe I'll do the video on my channel.
Thanks, Presh!
I'm not sure where you got your answer. If the square is on the x-y plane, suppose I construct a right triangle ABE with base 2 and hypotenuse 5. That would make the side of the square =√21. Now rotate the triangle about the leg AB tracing the point E. This would create a circle in the x-z plane. Without completing the math, I can easily create a triangle with sides √21, 2 , 4. The coordinates for point E would be (x>0, 0, z0) and would not fall on the square. The area of this square is obviously 21.
It turns out that the two answers presented in his video are the lower and upper bounds of the area of the square. The upper bound is strictly less than, while the lower bound is greater than or equal to.
9.2
@@freewilly5794 Sorry. You are correct. I had yet another puzzle that yields 20.5 ; I'll have to get back to my notes. Thank you for the feedback.
@@thomashughes4859 No problem. Your answer of 20.5 isn't incorrect, it's just one of an infinite number of possibilities.
Regarding the shape or path of all the possible points for e I haven't figured out if it's a circle or a hyperbolic ellipse (like the edge of a Pringles potato chip).
@@freewilly5794 Understood. I was using the idea of a leg of 4 and a leg of 5 coming out of a diameter, and you are right, the diameter has infinite lengths to a min of 1 and it becomes a line. Thanks again!