Find the Area (with graphing)

[@1milliondogs]

I'm almost certainly never going to need to know how to do this. Yet...I can't stop watching these videos!

[@0004voltz]

Same here

[@TheAvgZombieKillers]

Bro, how you gona not hit us with the "How Exciting?!"

[@OotsMcGoots]

Ikr I was so sad at the end 😢

[@markmullery7059]

He said it in the middle!

[@joeld.k.7652]

he said it at 2:06

[@alharethsawan]

@@joeld.k.7652 2:03 not 2:06

[@Sidowse]

@@joeld.k.7652 * 2:02

[@umariqbal7557]

These problems with straightforward answers but long ones are pretty amazing and interesting. I also want to ask whether you know how to use matrices to solve multi-variable(3+) systems of equations. Edit: Thank you all for replying to my comment. I kind of forgot some of the restrictions matrices have when it comes to solving systems and your comments have led me to review some of them.

[@grim-reaper-2168]

Search up Gaussian elimination and row reduction, it’s the name of the process used to solve equations with matrices

[@AmazinCris]

how could you use matrices to solve the system of equations introduced this particular problem 2:43 ? i haven't seen cases where 2 variables like x and y are being multiplied together.

[@abhinavanand9032]

@@AmazinCris xy is not linear

[@AmazinCris]

that would mean you can't use matrices here right?@@abhinavanand9032

[@and_rotate69]

u cant do that with non-linear equations like the 3rd one i think

[@tommy-ij9nd]

You have an amazing teaching style…positive, simple, entertaining. I grew up hating math though at 65 years old I still marvel at how cool and useful it can be. I wish I would have had you as a teacher growing up. In school teachers always seemed to cater to and praise the students that just “got it” as if that was going to motivate the rest of us to do better. What I hated about math the most was that it seemed so boring as a kid. Take some numbers, feed them into some equation and you got some other number pooped out the other side. I didn’t understand trigonometry until I got into college and took a physics class and found out the hard way you needed that to solve vector equations. I was so thrilled to find an actual use for it that I almost took on a physics minor. I subscribed because although a lot of my basic math is buried under years of Biology training I find it entertaining to see how you find solutions and your easy going nature reduces my anxiety at not remembering all the basic geometry, etc as I should. Thanks for taking the time to put these videos out there!

[@jimmyh1139]

I found a pretty pretty simple solution: The big white triangle, and the small white triangle on the right among the two small white triangles, have bases of the same length. Therefore the ratio of their areas is the same as the ratio of their heights. Label their bases y. Label the height of the small white triangle to the right x. Then the height of the big white triangle is (48/18)x=(8/3)x. Now we can write the lengths of the sides of the small white triangle to the left in terms of x and y: the left side is y-(8/3)x and the top side is y-x. The area of said triangle: (y-(8/3)x)(y-x)/2=18 The area of the small white triangle on the right: yx/2=18. We can now isolate x in terms of y to solve the top equation: x=36/y. Solving said equation with rudimentary algebra we get y^2=144, which is the area of the square. Therefore, the area of the red triangle is 144-(48+18+18)=60.

[@whitetiger3258]

Just saw this comment, and am amazed at how much easier this method is. I actually used this method and was just about to comment it but you beat me to it. It's just amazing how there are so many different ways to get to the same solution!

[@bleizar7448]

I just tried to solve it and came up with the exact same solution! This is much easier as you just work with 2 variables. Well then we have to solve a quartic equation, but it's actually biquartic so it's much easier to solve. This equation is: y^4-168y^2+3456=0 We have really big numbers, but factoring 8 each of the two roots, it becomes really easy. And after this we get the solutions for y: ±2√(6) and ±12. We can the negative ones as a length can't be negative, and also 2√(6), as then x would be greater than y, which would also lead to a negative length. So finally, we have that y, or the side of the square, is 12 and the area of the red triangle is 60. :D

[@jimmyh1139]

​​​@@bleizar7448Your solution is logically valid, but it involves more work than it needs to. The extra step in your solution is solving for y. It's sufficient to just solve for y^2. Here's how I did it: The abc formula gives us two solutions in terms of y^2 for the equation y^4-168y^2+3456: y^2=144 and y^2=24. We can discard the second solution, since y^2 is the area of the square, and the area of the square must be larger than the sum of the areas of the white triangles: 48+18+18=84. The area of the red triangle is the sum of the areas of the three white triangles subtracted from the area of the square: y^2-84=144-84=60 meters squared.

[@jimmyh1139]

​​​@@bleizar7448 Your solution is logically valid, but involves more work than it needs to. You don't need to solve the equation for y, because when you've solved it for y^2 you're done. The abc formula gives the solutions y^2=144 and y^2=24. The second one can be discarded since y^2 represents the area of the square, and the area of the square is obviously greater than 24 square meters. Now we can just subtract the sum of the areas of the white triangles from the area of the square to get the area of the red triangle: 144-84=60 square meters.

[@lmarmisa]

I defined only 3 variables: S (the side of the square, really the only value we need); x (your x); y (your w). Then 3 equations. I got a biquadratic equation for S. S1=12 and S2=4 (rejected because x & y are negative). Finally A = 60 m^2. Very nice problem. Thanks a lot.

[@starcaptainyork]

96=z*(x+y) 36=y*(w+z) 36=w*x (w=z)=(x+y) so lets call that "S" for "Side length of the square" z*S=96 y*S=36 96/36=(z*S)/(y*S) and then we can eliminate "S" leaving: 96/36=z/y This gives us the ratio 8/3=z/y, coincidentally 8 and 3 are also their actual values. Just plug them into z and y and the rest comes easy. Love you're channel by the way, keep it up👍

[@brucearling9468]

I always love watching these video's. I was pretty good at math in highschool but all of this is too complicated for me but I love watching how you get to the answer!

[@GirishManjunathMusic]

taking the side length of the square as a, the height of the 48m² triangle as b, the height of the 18m² triangle on the left as c, the base of the same as d, and the base of the other 18m² triangle as e: b + c = d + e = a. ½ab = 48 ½cd = 18 ½ae = 18 This is a system in 5 variables of 5 equations: it is technially solvable. b + c = a, b = a - c ½a(a - c) = 48 a² - ac = 96 ac = a² - 96 c = (a² - 96)/a d + e = a e = a - d ½a(a - d) = 18 a² - ad = 36 ad = a² - 36 d = (a² - 36)/a ½cd = 18 cd = 36 (a² - 96)(a² - 36)/a² = 36 This is a quadratic equation in a², substituting a² = z (z - 96)(z - 36) = 36z z² - 132z + 3456 = 36z z² - 168z + 3456 = 0 z² - 144z - 24z + 3456 = 0 z(z - 144) - 24(z - 144) = 0 z = 24 or z = 144 However, as z is also the area of the square, it has to be greater than 48 + 18 + 18 = 84, therefore z = 24 is extraneous. Therefore taking z = 144: area of the red triangle is area of square - area of three white triangles: Area = z - 84 = 144 - 84 = 60 square meters.

[@Ludaris1]

thats a lot

[@supermax64]

Comparing the factors of 96 and 36 gets you the side of the square = 12 relatively easy. The "upper" factors of 36 are 9 and 12 (since we're looking for the longer side) and only 12 is a factor of 96.

[@chrisc6468]

this assumes that the side lengths are integers, which we were not told at the start.

[@TadRichard]

​@@chrisc6468 Yeah, but even though the question didn't state the values were integers, it's worth trying integers first -- especially if the question is on a timed test. Similarly, you should always try multiples of 3, 4, 5 and 5, 12, 13 when you see a right triangle.

[@Zanxos]

I solved it this way too and it's easy to check your answer by checking the triangle sides having the side of the square defined as 12.

[@mrbigg151]

This man's killing math equations, one triangle at a time

[@SympatheticMonster]

Even simpler: Area of triangle = 1/2 * b * h Top Right and Bottom Left have the same base, so the only variable that distinguishes the two triangles is the height, therefore the scale factor of height between the two triangles is 3:8. Therefore, 1/2 * 3x = 18 - or 3x = 36, and 8x = 96. In either case, x = 12. 144 - 84 = 60.

[@danielbrooke-gandhi]

You've missed out a step, you need to prove that there is no constant K affecting both heights. Conveniently k is 1, but it isn't necessarily.

[@HDB-e5r]

Hi Andy, I think this one can be solved quickly through simple symmetry observations: Construction: Take a big square and include one diagonal with divides it into two equal triangles. Now pick two random points A and B on this diagonal and draw 2x 2 lines parallel to the sides of our square through these points. This leads to 3 smaller squares and 2x3 rectangulars. The 2 outer squares and the 3 upper rectangulars form an L-shape. Analysis: For symmetry reasons, the 3 rectangulars left and right of the diagonal are congruent. Again for symmetry reasons, our L-shape has the same area as one of the triangles -- equal to half of the total area of our big square. Calculation: If the area of the L-shape is known (2*18 + 2*18 = 72), you only need to double it to arrive at the total area of the big square (2*72 = 144). Substracting all given areas from this total area results in the missing part (144-18-18-48=60). How exciting! :-) Best from Zurich, Harald

[@victorgorelik7383]

This is a quadratic equation. I published the solution in February in “281. 4 triangles in the rectangle” and today: “639. Rectangle and 4 triangles” (with your numbers). Notice that it is not necessarily a square. Like your posts!

[@rg7729]

Coming from a non-math guy, i have found your videos very stimulating! U have my subscription! I am proud to say i got this one on my own (i usually cant lol). Not as elegant as how u did it. Starting with the bottome left triangle: created a table of all possible multiples of the area 48x2=(96). Based on the visual proportions of the triangle, only 8x12 seemed correct. Knowing the base was 12 i subtracted it by 8 to find the height of the top left triangle (4). With the top left triangle height and area, its width came out to 9. Next i used 9+x=the known length of the square (12) and got 3. Now that i knew all the triangle lengths added up and were true to the square, i used area of all the triangles and subtracted them from the area of the square (144) and got 60!

[@giuseppe1912]

i'm a lover of math, and it's astonishing how you are able to use this game and make to feel very easy. i think that all the school should have such as you docent to teach the math and the powerful of math

[@jarnhus]

Lovely video! I did things slightly different by starting by introducing scaling factors. This way we can end up with a quadratic equation for one of the scaling factors and solve it like this: Let the side length of the square be denoted s. Let the factors a,b in ]0,1[ be defined such that * x = as * y = (1-a)s * w = bs * z = (1-b)s where x,y,w and z are the sides defined in the video. With this we are left with three equations for our triangles' areas (called A, B and C for future reference): * A = abs^2/2 = 18 m^2 * B = (1-a)s^2/2 = 18 m^2 * C = (1-b)s^2/2 = 48 m^2 As the first two triangles have the same area (A=B), we get the equality abs^2/2 = (1-a)s^2/2 which reduces to b = (1-a)/a Likewise, we have that 8B = 3C, which gives the equality 8(1-a) = 3(1-b) This simplifies to b = 8/3 a - 5/3 Combining the two expressions for b, we get: (1-a)/a = 8/3 a - 5/3 Simplifying, we get the quadratic equation 8a^2 - 2a - 3 = 0 This can now be solved to give (ignoring the negative solution) a = 3/4 With the area B, we now get B = (1-a)s^2/2 = (1-3/4)s^2/2 = s^2/8 = 18 m^2 Resulting in s^2 = 144 m^2 From this the area of the red triangle becomes s^2 - A - B - C = 144 m^2 - 18 m^2 - 18 m^2 - 48 m^2 = 60 m^2

[@44ThaNatos44]

Thank you so much for your content. It's perfect for a refresh after my lazy brain got rid of the logic for so long. Happy holidays !

[@MufaddalMansoor_913]

Man, never stop uploading, ur amazing ! keep it up!

[@hurricanefist]

I was just messing around with the numbers trying to figure out an easier way to get the area of the square from the given information to subtract from and came upon if you subtract 72 (18*2 + 18*2) from 132 (48*2 + 18*2) it comes to 60, the area of the triangle. I haven't studied math in nearly half my life, so I don't know if that's a fluke or an actual way and why that is. I'll ponder this more later.

[@lunstee]

I define edge lengths x,y, and a,b such that the triangle areas are: A = xa/2 B = yb/2 C = (x-b)(y-a)/2 = xy/2 + ab/2 -xa/2 - yb/2 = xy/2 + ab/2 -A - B Now note that AB = abxy/4, from which we get ab= 4AB/xy. Substituting this in the equation for C and rearranging a little gives us: 2(A+B+C)= xy +4AB/xy Multiplying this out by xy and putting all terms on one side gives us: (xy)^2 - 2(A+B+C)(xy) + 4AB = 0 This is a quadratic for xy, which solves as xy = A+B+C +- sqrt((A+B+C)^2 - 4AB) The area of the red triangle is just xy - A-B-C, which leaves us with: area = sqrt((A+B+C)^2 - 4AB) with the negative solution being superfluous, relating to triangle C extending outside the rectangle instead of inside, and one of triangles A or B extending out and overlapping with C to reach the far vertex.

[@HajimeruSatoru]

I love to watch these when eating for some reason lol. Maybe to make me looks smarter than I actually is

[@samueldeandrade8535]

*am. *than I actually am

[@DefeatLust]

This dude has been uploading for 7 years straight with dog shit views, and now all of a sudden the algorithm gods have blessed him. Wow. Can we make some sort of statistic math problem out of this?

[@ChordLord]

He didn’t say “how exciting!” The start of the apocalypse has begun

[@Larsbutb4d]

The fact that you got concerned that you did it in 20 minutes surprises me.

[@Larsbutb4d]

For reference, it took me half an hour to solve only one of Cristina Agg's puzzles.

[@muratyigitakbas]

for x+y and w+z I used "a" instead. x+y=w+z=a z.a/2=48 y.a/2=18 since "a" is the still in the both equation I focused more on z and y. Becouse diffirance between 48 and 18 isnt the "a" its the diffirance between z and y coused it. I just dont see the "a/2" in the equation and try to find ratio between z and y. z=48 y=18 if we divide both sides up to down: z 48 8 - = - = - y 18 3 as like that we found the ratio between z and y is 8/3 thats means if z is 8 then y have to be 3 so I write "8.k" instead of z and write "3.k" instead of y. our new equation becomes: 8k.a/2=48 3k.a/2=18 if we do the math in the both sides a is equal to "12/k" and since the a>z>y "k" have to be 1. and if "k" is 1 then we do not have to put it in the equation becouse we use it for the multply function. so if we write 1 instead of "k" we see that a=12 and the all sides of the square can be found since if a=12 thats means z=8 and y=3. in the end one side of the square is 12 so the m² of the square is 144 and all the white triangles equal to 84 if we subtract 144 and 84 result is 60.

[@alasanof]

Before watching the video: I factored what lengths would give the given areas of triangles. I got the triangle on the right would be a 3 by 12 triangle, the top one is 9 by 4, and the bottom one is 8 by 12, that way each side of the square adds up to 12. Taking the given areas away from 144 gives 60.

[@mwtclan5249]

If you make one side length x, and then the fractions of the sides as (x-y), if you make the smaller part y, then it makes the equations much simpler

[@Qermaq]

I started with s as the side. a is the short bit on the top of the left side, and b is the short bit on the right of the top side. From the areas, we see that b = 36/s, and a = s - 96/s. Since a(s - b) = 36, we can plug those right in to this one to get (s - 96/s)(s - 36/s) = 36. This leads to a quadratic in s^2, which is what we're after. s^4 - 168s^2 + 3456 = 0. The positive value of s^2 is 144.

[@Qermaq]

In fact this leads to a general solution. If the area of the triangle with only one vertex at a corner of the square (top left in this diagram) is F, and the other two corner triangles have areas G and H, and F + G + H = K, then the area of the square is simply K + sqrt(K^2 - 4GH), and the area of the center region is just sqrt(K^2 - 4GH).

[@romypotash7114]

KZbin showed me this video, and then a video from 6 month prior to it, (trinagle red eara) wich is basically the same question that you solved in a different way... so i think this is the answer to a simpler solution

[@here_be_dragons9184]

You should have kept R as the square side that makes the equations simpler. Rz=96 Ry=36 (R-z)(R-x)=36 Replace in the last equation: (R-96/R)(R-36/R)=36 (R^2-96)(R^2-36)=36*R^2 Since R^2 is the square surface, let's call it S, and is actually what we need, you end with a polynom in S... (S-96)(S-36)=36S S^2+(-96-36)S+(96*36)=36S S^2+(-96-36-36)S+(96*36)=0 S^2-168S+3456=0 Or something like that... Solve that S=24 or S=144 24 is too small, so it's 144 which happens to be square 12. Than 144-(48+18+18)=60

[@meatdepartment8901]

who cares

[@Aashuthosh]

I care!!! Loved your explanation.... I did also think about this and since the numbers were 18 and 48 i also thought we could probably could have used trial and error.... You just get sides of triangles which equal 18 and 48 see which one fits and there you go

[@ViktorLoR_Mainu]

You just watched a video explaining the same fucking problem ​@@meatdepartment8901

[@weilim8942]

Do we need to learn this for gcses

[@samuelenriquealvarezhernan8088]

This channel makes me more intelligent

[@meetarranjjan]

You can take squares side as A and then solve . Directly you will get A^2 as 84 +_ 60 hence red area as 60 because rest is 84

[@brianmacmillan4478]

the answer is x = 1.25. I don't feel like explaining but the distance from x axis and y axis is 3.75 to form the square in the problem. hypotenuese doesn't change in scale so easy to check...good work kid good work

[@brianmacmillan4478]

Ah that makes a 2x by x rectangle. Essentially that but then altered to make 2 x by x/2 rectangles side by side.

[@ivanokmunoz]

I did it without graphics I took the 48 m2 triangle and added a equal triangle to make a rectangle. This rectangle is 96m2, so the only possible sides would be 8 and 12. So 12 x 12 = 144 m2 the total area of the square, minus 48, 18 and 18 is 60 m2 for the red triangle.

[@mhmdmansour9507]

there is much easier way to find the area of the square by taking the GCF of 96 & 36 which is 12

[@terrencepereira169]

I would consider only lower triangle with 48 area. To get 48 as an answer, we need base* height=96. 12 and 8 make 96, which fit the scaling properly. So height= 8, base =12. Then assume it's a square, area=144 then subtract the 3 areas to get 60

[@edheldude]

That's carpenter math but not a proof.

[@terrencepereira169]

@@edheldude Its depends on what you are looking for. I assumed its a time bound exam and there are Marks only for final answer. If you call using common sense as carpenter Maths....that's what it is

[@edheldude]

@@terrencepereira169 I used to study physics and maths so you need to show proof without intuitive leaps.

[@terrencepereira169]

@@edheldude let Andy decide that. It's okay to not have common Intuition. I understand your incompetance

[@edheldude]

@@terrencepereira169 Dude, I'm talking abouts maths and how it's done logically to show proof. Carpenter math works well for solving day-to-day individual cases but you show a logical general proof that solves the specific case. That's how it's done.

[@Midnightspirit500]

honestly I think this is one of those problems where you have to rely on your area to side length ratio intuition because i figured it out quickly from the 48 and the area of a triangle isnt particularly hard to calculate

[@Klosterhasi]

guessing whole numbers going from 1 to 10 for all w,x,y,z gives us a pretty fast brute force solution with about 10*10*10 = 1000 possibilites to try as w = (w+z) - z on the right hand side. this can be done in excel. or any programming language of your choice.

[@Klosterhasi]

and i just did it, took me however long these comments are apart, using excel formula and brute force guessing all the numbers values: 4, 9, 3, 8 for w,x,y,z the side lengths which makes the whole square 12 x 12 = 144 minus the 3 triangles of 18 18 48 makes the red triangle 60 units Edit: 26 minutes cuz i needed to figure out syntax but could be faster now

[@JoshSteiner14]

I had a slightly quicker way to do this by setting the long edge to x, then the shorter edges to y & x-y and z & x-z. It starts you off with only three equations but I still needed to solve via graphing so not much quicker

[@zilvinas5130]

There's a formula that can be used to calculate the area of a triangle while having lengths of all sides. It's called Heron's Formula.

[@sportsloverbaseball]

We don’t have the lengths of the sides, though. We have the areas of the 3 triangles.

[@Aditya_196]

Take S to be side of square then the triangle that has area = 18 and one side S take other side to be 3x ... And u can see for the triangle having area 48 has to have sides S and 8x then u have the other triangle with area 18 have dimensions S - 3x and S - 8x ... U can write this then S² - 3x*S - 8x*S + 24x² = 18 .. u know 3x*S and 8x*S and also that 1/2 * 3x* S = 18 and solve easily afterwards

[@charimonfanboy]

Is it just me or can this be a lot easier. It's a square so... the upper right triangle has sides x and a the lower triangle has sides x and b xa=36 xb=96 factors of 36 are... 1*36, 2*18, 3*12, 4*9, 6*6 factors of 96 are... 1*96, 2*46, 3*33, 4*24, 6*16, 8*12 x is the longest side for both triangles, so for 36 x must be one of... 36, 18, 12, 9, 6 and for 96 x must be one of... 96, 46, 33, 24, 16, 12 and only 12 appears on both sides so the x=12 12*12=144 triangles add up to 84 so the inner triangle area must be 60

[@danielbrooke-gandhi]

The problem doesn't say that any of the sides are integers.

[@charimonfanboy]

@@danielbrooke-gandhi That is a fair point, would have been more complicated if they weren't integers

[@lasalleman6792]

Explanation appears elegant, but to me it was as clear as mud. But I did get the same answer as to the sides and x, y and z. As well as an area of about 59. Mostly through some intuition.

[@MarieAnne.]

I'll leave w, x, y, z as shown. Let s = side length of square. From top-right triangle, we get ys/2 = 18 → y = 36/s → x = s − 36/s From bottom triangle, we get zs/2 = 48 → z = 96/s → w = s − 96/s From top-left triangle, we get xw/2 = 18 (s − 36/s) (s − 96/s) = 36 s² − 96 − 36 + 3456/s² = 36 s² − 168 = −3456/s² s⁴ − 168s² = −3456 s⁴ − 168s² + 84² = 84² − 3456 (s² − 84)² = 3600 s² − 84 = ± 60 s² = 84 ± 60 Since s² is area of square, it must be greater than sum of areas of triangles shown s² = 84 + 60 = 144 Area of red triangle = Area of square − Area of 3 unshaded triangles = 144 − (18 + 18 + 48) = 144 − 84 = 60

[@Andymandyrocks]

Since we know all triangles are Pythagoras triangles.. We can try hit and trial method by using HCF technique.. Since one factor of large and small right triangle is same and other factor of large and small left triangle should add upto large factor.. We find HCF of 36 and 96 is 12.. Now does the other multipliers of 36 and 96 add upto 12.. 9*3 and 12*8.. Well it does.. Hence the side of the square is 12...rest is just subtraction

[@sam1984-w8i]

Area of Bottom ∆ = 48 = .5 * z * (x+y) Area of Top left ∆ = 18 = .5 * w * x Area of Top right ∆ = 18 = .5 * y * (w+z) z * (x+y) = 96 w * x = 36 y * (w+z) = 36 we know x+y=w+z since square z * (x+y) = 96 y * (w+z) = 36 96/z=36/y z/y=96/36 we have to find a common number between 96 and 36 that could satisfy the above equations. z/y=(12*8)/(12*3)---------> the numbers involved were whole numbers hence easier to decipher a common factor. if we take (x+y)=12 then z = 8 and w = 4 x = 36 /4 = 9 and y = 3 Area of inner ∆ = (12)square = 144 - 18 -18 -48= 144 -84 = 60 sq. m

[@tomdekler9280]

This is a quadratic equation with a single variable to consider. The only sad part is that it's an annoyingly big quadratic equation. Take side length s. You can express all the other lengths in this fashion. Let's use the XYWZ you marked. 0.5*y*s = 18 y = 36/s 0.5*z*s = 48 z = 96/s x = s - 36/s w = s - 96/s Then, for the top left triangle, you get xw/2 = 18 xw - 36 = 0 substitute (s - 96/s) * (s - 36/s) - 36 = 0 Working this out gives you s⁴ - 168 s² + 3456 = 0, which is a quadratic equation. Substitute s² for variable u (area of the square) to get u² - 168u + 3456 = 0 (u -24) (u-144) = 0 u = 24 or u = 144 24 can't be the case as the surface of the 48 triangle would be bigger than the area of the square, so the square must be 144. Then it's just 144 - 18 - 18 - 48 = 60, and that's our answer. How exciting.

[@tomdekler9280]

and yes I used the quadratic formula to get 24 and 144 lol, I'm not that clever.

[@pratiktaltia4931]

We can take a smarter approach here by guesstimating the answer since the values are very easy to dissect. Area = 18 i.e base x height = 36 Factors are: 1*36 2*18 3*12 4*9 6*6 The only possible sides are for 1st triangle be 9,4 12,4. Based on this trial and error, arrive at the answer

[@AnonymesWombat]

with graphing catched me

[@fssocrates]

You had a much easier way to solve it. Side of the square = x Bottom fraction of the left side = a Right fraction of the upper side = b ax/2 = 48...(i) bx/2 = 18...(ii) (x-a)(x-b)/2 = 18...(iii) from (i), a = 96/x from (ii), b = 36/x putting these to (iii), (x-96/x)(x-36/x) = 36 solving this, we get, x = 12 So, the area of the square = 144 So, the red area = 144 - (48 + 18 + 18) = 60

[@rupom_1670]

can you please give me the 3d graph website link

[@AndyMath]

www.geogebra.org/3d

[@AndyMath]

www.desmos.com/3d

[@rupom_1670]

@@AndyMaththanks Andy

[@weiwuxianshair]

What about solving with herons formula?

[@nickinterstellar6521]

Those are areas lol

[@yangranay8102]

If it took 23minutes and 22 seconds for a master, how long would it be for the average people?

[@waitwhat6590]

Went with the factorization method and ended up with (x^2-144)(x^2-24)=0 so the value of x is 12 or 2sqrt(6), and then plugged it in x^2-(18+18+48)= Red Area, after a trial and error, found that its 60m^2, yeah?

[@MiniGunTurtle36]

Wouldn’t it be faster once you have the four four-variable equations to put them in a matrix and RREF to find the values?

[@bobkenchidarwin9975]

the requirement for that is that all four systems have to be linear. none of them were unfortunately

[@albertongoogle]

1 minute solution: Since all 3 triangle areas are whole numbers, compare the possible factor pairs. Start with top and right triangles. Both need rectangles with factor pairs of 36. But one factor of top triangle + one factor of right triangle = other factor of right triangle. That gives you 9+3=12. The rest is easy. Sure, i assumed the answer had whole numbers, but with 3 whole number inputs, probably a good assumption. :)

[@octobixer]

Really good trick but doesn’t generalise very well

[@albertongoogle]

@octobixer ya :) but that's the wonderful thing about math - many ways to the same answer!

[@peraserp7569]

we had this problem at our schools first cml competition but the side lengths were given to you

[@CatCatchCold]

Did this with 2 variables, your 'y' and your 'w'. With a bit of simple geometric operation, the required S=48+18-(18-y*w), so these two lengths are the only needed. Since z is obviously 48/18*y, there are two equations: 48/18*y*(w+48/18*y)=96; w*(48/18*y-y)=36. Then there's the answer. However I really can't find a way to solve this without any equation and only 'classic' geometry methods, althogh I felt so close at some point. Frustrating.

[@fabricewoelflin3742]

Hello, in fact I found more simple to look for the value of the side of the square (a) and I found this equation with all the equations you find with the areas of the triangles : a^4-168a^2+3456=0. And the negative values are impossible and one value 2²V6=4.89 impossible as well. So a=12m the side of the square. And after you know the end.

[@ahmedfahadkhan4558]

I would have done it this way which I think is easier. I would have assumed the red area to be x then by adding 48+18+18+x=84+x I would've gotten the area then as it is a square I would have taken the square root on both sides and i would've got the side of the square which is √ 84+x . then I would have taken the area of 18m² which is on the left side (right side of the screen) assumed the above side z and calculated the area by doing (√ 84+x )(z)=18 x 2 and also taking out z which i would have gotten z = 36/ √ 84+x. likewise, i would have done the same for for the 48m² triangle and assuming its smaller side to be y and would have gotten the result y= 96/√ 84+x. then for the next part i would do √ 84+x - z and √ 84+x - y to get the sides of the other 18m² triangle which i would get x-12/√ 84+x and 48+x/√ 84+x (there is no need to take the hypotnease). then by applying bxh/2 (formula for area of triangle) again i would have done ( x-12/√ 84+x)( 48+x/√ 84+x)=18 x 2 after further simplifying i would get (x-12)(48+x)/84+x=36 then again by simplifying i would get 3024 + 36x = 36x -576 + x² then by further simplifying it would become 3600=x² and by taking the square root on both side i would get the answer x=60

[@johnsonyip1127]

to be honest i just multiplied all the areas by 2 and then looked at the total area if i took away one of the 18m areas and if i didnt and ended up with 132m or 168m and went "well one sides probably 12m long but i dont have proof and just eyeballed it, wonder if that true and what he did to prove it."

[@cappycap8144]

how, exciting

[@randomchshorts]

How. Exciting.

[@Felipera_]

I found an easier way that took me exactly 4:43 to find the answer.

[@founder8240]

maybe you can solve it with a simultaneous?

[@matheusribeiro9478]

i did an equation that ended up like this: 3456 = 168L^2 - L^4 This give me that the side of the square is equal to 12 or 2*square root of 6 WHY?????????????

[@CharlesB147]

I apologize, but I completely guessed that the side length of the square was 12, and everything else just ... fit into place. It took me all of ... about 3 minutes to just take a bit of a stab at it, and it just happened to work. So other than an educated guess, I can't really see anything faster. 😅😂

[@NalosEclipso]

WTF THAT'S SO COOL. AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA

[@ChrisThe1]

there's two squares though: 18x8 and 12x12

[@alimassoudi2035]

Hey, I sent you a detailed solution by email (completely analytic and I included the intuition behind). It’s not very tedious if you play with the equations and select the right variables. You end up solving a simple quadratic solution that gives you 2 solutions for the area : 60 and - 60

[@shkhamd]

now solve this problem assuming the square a rectangle.

[@caleb2050]

no "how exciting"?

[@brettstafford9665]

Easy, Heron’s formula a = b = oh wait those are areas

[@naimhasanpappu3901]

You forgot to add 3 2 1 at the beginning of the video.

[@anomanderrake440]

I actually stopped believing in triangles like at all about 3 or 4 years ago

[@cstuart5638]

I solved thinking those numbers meant the side lengths of the triangle. I’m an idiot.

[@HereForTheComments]

Andy Math, will you ever tell us about yourself? Someone in the comments estimated your IQ to be somewhere between 130 and 150. Inquiring minds wanna know what it takes to recognize patterns like you do. Yes, you can solve problems for us and share problem solving strategies with us. But it takes something else, something that can't be taught, to be able to see solutions like this in the first place.

[@meme-ville]

It takes practice and knowledge

[@HereForTheComments]

@@meme-ville I have geometry "knowledge," but it takes "genius" to see the kinds of clues he's pointing out. Without anyone to point them out for you first.

[@meme-ville]

@@HereForTheComments I could say the same thing about chess, but alas a plumber can be better at chess then an astrophysicist, its just about practice and knowledge.

[@HereForTheComments]

@@meme-ville I know plumbers can be smart, but there's a difference between being educated and being a genius. If you're educated, you were taught. Like Andy is teaching us. If you're a genius, things just "make sense" to you because you notice patterns.

[@MG-the-L-E-Min8tor]

Wow 😲,also please don't view this as childish but,FIRST!

[@byrearejohnson7698]

yeah thats super childish

[@yzomariano7130]

​very childish

[@henaytx1799]

You child

[@engrlo]

How not exciting… just kidding!

[@gonzalorubilar634]

i got to x^4 - 168x^2 + 3456 = 0 😹

[@danielbrooke-gandhi]

Substitute r = x^2 and you have a quadratic equation that's solvable via quadratic formula

[@gonzalorubilar634]

@@danielbrooke-gandhi yes, i forgot to say that i solved it and got to the correct answer. I feel so nerd but i enjoy solving this math problems 😹🤓

[@paresh6166]

he looks like Mia Khalifa

[@antonschacht8985]

I’m too young for this

[@Glucenaphene]

Took me like 2 minutes but it was mostly just guessing haha I basically just doubled the area of the two 18m^2 triangles to get 36 and looked for the factors and got 3x12 and 4x9 and the 3+9 equaled 12 so i knew it was s square of 12 then just 144-18-18-48 lol

[@baldability]

yeah same lol i just did 48*2 and saw that it was 12*8, so from there it’s just 144-48-18-18

[@mrbutish]

I know how to solve it on my own, just like watching it get solved because it's satisfying

[@JonasThente-ji5xx]

Good for you

[@devd_rx]

dumb way to solve, just keep eliminating variables and you get your answer, smh, bro tried using a graphing calculator ong

[@darkjudas1]

Graphing was cheating. If you can use external tools we can feed equasions to maple or something

[@gentozik]

I love questions from you bc they are not requiring you to remember 3.5 mega tons of formulas and MagnusKarlsens-grade strategy to realise what you should use for a solution, which is too high to discourage future mathematicians I just can saw one of these, solve it from a thumbnail for couple of minutes ( if I use a calculator ), them watch the answer and be happy

[@fartenko]

How exciting

[@44ThaNatos44]

Thank you so much for your content. It's perfect for a refresh after my lazy brain got rid of the logic for so long. Happy holidays !

[@smsofisami724]

How exciting