What Is The Area? HARD Geometry Problem

People around the world have asked me this and similar problems. Learn how to solve using only geometry (and a little bit of algebra).
There are many other ways to solve the problem too
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[@Its__Good]

The relief when you give up quickly, then find out that the solution involves a mathematic rule you didn't know.

[@dormeur20092010]

I solved it in my head in few minutes by using only junior high school geometry (you only need Pythagoras theorem and the area of a disc). Sorry buddy :(

[@AKagNA]

@@dormeur20092010 i solved this only by using my pre-school geometry information. Hah!

[@NoTLucas]

@@AKagNA I solved this with my innate knowledge of shapes, :P

[@sgsnake2x]

Alper Avcı technically you can, the rules are there and you already know them. However thinking deeply is the hard bit

[@AKagNA]

@@sgsnake2x i dont think any pre school geometry information would be enough for that.

[@rational-ec4rk]

"Trigonometry or Calculus not allowed" Me: Can't solve using Calculus or Trigonometry either.

[@ShubhamKumar-vp6ro]

Even didn't get a blow of it.

[@plasmakitten4261]

I'm not sure how you would do this with calc 1-2, but with calc 3 you can set up a coordinate system with the lower left corner at the origin, slice the shape in half along the line x=10 or y=10 (they work the same way), and then you have a region with one flat side that you can use to set up a double integral, with an integrand of 1 since this is a 2d surface. Then multiply by 2 at the end since the integral only gives you the half you cut with x or y = 10 and you get the same answer.

[@christianwiese9887]

"don't use trigonometry". what is trigonometry if not the metrics of trigons? what exactly was forbidden? did you mean that sine, cosine and tangens are forbidden?

[@saturnslastring]

I saw the thumbnail and solved it before I watched the video using calculus because I didn't know it was banned and the calculus version is way way easier.

[@saturnslastring]

@@plasmakitten4261 I solved it using what I learned from my first week in calc 2 :P I set up the coordinate system so the origin is at the center of the diagram and took the definite integral with respect to x of the arc (x+10)^2 + (y+10)^2 =400 between the bounds of x=0 and x=10(sqrt3)-10. That gave me the area of 1/4, of the center shape, so I just multiplied it by 4 to get 400pi/3-400(sqrt3)+400 which is the exact correct answer. You could simplify it to 400(pi/3-(sqrt3)+1).

[@Ozymandias83]

“Find the answer without using any of the appropriate methods to do so”

[@MrCScottie]

It’s a challenge

[@caniggiaful]

My mind jumped right away to calculus!

[@dlevi67]

Nothing else remains.

[@DoReMeDesign]

Using just geometry is a perfectly reasonable condition, because finding the simultaneous equations is a far more elegant way of doing it.

[@PMA65537]

area of equilateral triangle using sqrt(3) - how is that not using trig?

[@ArtumTsumia]

The 'fun' part about a problem like this is so long as you know how to calculate area of a square and circle, you can slowly break the area up into it's sections. So you can solve for the square, then the quarter circle, then subtract to get the 'outside' of the circle. Then you can take the area of half the square (the triangle) and proceed to continue to break things down. It's more roundabout than knowing some of the angle formulas going in, but they can be derived from the core area formulas.

[@joshualiley]

"No trigonometry allowed" *Procedes to use trigonometry*

[@thestaringenclave]

Exactly.

[@Bignic2008]

He did say that part was just for fun.

[@DoReMeDesign]

He used trigonometry after the geometric solution. Did you even watch the video?

[@squidprince2456]

Honestly in this case it's difficult to define where the geometry stops and the trigonometry begins

[@royk7903]

He used trigonometry to find the area of the equilateral triangle in the third equation and just didn’t mention it. The area of triangle formula -> 1/2(bc(sinA))

[@evescent7312]

I was like, integrals are a thing.. And then, "no calculus" . ☺

[@vitortutors6030]

Yeah man

[@vitortutors6030]

When I see that, I thoght he'll use integrals but not, and I get a crazy

[@ericli8792]

I am taking ap calc rn, could u explain how you would use an integral to get all the areas?

[@joaovictormendes2293]

@@ericli8792 I build my integral for this way: www.wolframalpha.com/input/?i=+2+integral+10+to+10sqrt%283%29+integral+%28-root%2820%5E2-x%5E2%29%2B20%29+to+root%2820%5E2-x%5E2%29+dydx

[@evescent7312]

@@ericli8792 Mr. Mendes already replied with a solution. So if you're wondering how he got that, refer to this picture I drew in paint (I'm proud of it). imgur.com/a/hgF2PiK You can just find the area(vertical lines) subtract area(horizontal lines) from 10 to 10sqrt(3), multiply by two since you only got half. If you're wondering how the guy got the 10sqrt(3), I included the drawing of a 30-60-90 triangle with a hypothenuse of 20. :) There you go.

[@ashwindeshpande7557]

0:41 That turkey guy has a weird Naim

[@justs_]

Take my like and get out

[@clary3860]

Do you think its funny?

[@madhusudan6772]

@@clary3860 of course it is.

[@EgorRandomize]

Haha I got it

[@Rohith.]

It's naim not naim

[@ryanramjattan8714]

Part 4: solve the area of the square

[@eurovisioncyan9550]

399.99999999999999999999...

[@sabyasachirimpa]

For the dumbos, part 4 will come in 22nd century.

[@AjayKumar-il3qn]

Area of square = r^2

[@sabyasachirimpa]

@@AjayKumar-il3qn No it needs very complex integration formula to find out that, only rocket scientists can find the value of the area of the square.

[@AjayKumar-il3qn]

@@sabyasachirimpa r^2 does not involve integration

[@Mariorox1956]

I probably spent longer doing the trig. version than I would have if I had done it as he did. Knowing that arch segment actually helped me, since I could look at a shape and go "Oh, that's a triangle with two of those arch shapes taken away" I also love how the arcs perfectly trisect themselves.

[@NUGGet-3562]

When trying to solve this, I didn't wanna assume anything. Knowing that the arcs trisect each other probably would've helped.

[@nberz692]

The worst thing, when you know all this formulas and understand what is going on, but you didn't solve it

[@light_0_yagami]

Can't find the third equation but after watching the approach it's good

[@Its__Good]

What?

[@2600mx]

It is nice to generalize area a, b, & c. A slight different approach is just use variable a & b: 2a+b=square - quarter of the circle; a + b = 1/12 of the circle - (1/6 of the circle - the equilateral triangle). This will reduce the equations to two variables and two equation.

[@Wagon_Lord]

When you get your equilateral triangle, you can see quite clearly that the area of the square is [Eq. Triangle+ two 30 degree circular arcs + region a]. This lets you quickly find an expression for a, then you can find an expression for b by using simple algebraic manipulation, and again find the area of c. Nice to see I got the same answers at least.

[@touringmoose]

I found it easier this way.

[@2b2getther]

The hardest part of this is to figured that out yourself in limited time.

[@magnetoninja7386]

A quick way is to look at it as a square and 4 "arc circles". Then by symmetry and 30-60-90 triangles, you notice that the arc is going to have a central angle of 30 degrees, you calculate the "arc circle" by finding the sector minus the triangle (1/2absinC) and then find the square using LoC. Basic techniques and quick :)

[@technodrome]

I look forward to solving this set of questions. I love the circle and square area problems.

Al-Kashi did not discover the Law of Cosines, he just expressed one of Euclid's theorem in terms of cosines. If anything, it's Euclid's Law of Cosines.

[@yurenchu]

It can't be "Euclid's Law of Cosines", because Euclid didn't use any cosines.

[@Trixex]

Had a similar problem in my a level further maths book, I used this method and was surprised with how hard the question was. Then it was when I realized that there was an eazier way to do it

[@maicom802]

This video is so useful. Here in Brazil, precisely in my city, this is a very common problem in the exam to be in medical school. I will take the exam this weekend and I am sure it will be there.

[@jalalalarabi5307]

It was very interesting Mr Presh, as usual couple thanks...

[@SKYPRODUCTION2021]

After 10 ..u no need this type of geometry if u take science..

[@ferwildfire865]

Great video thank you! I have a suggestion for you, why don't you include a MindYourDecionsMeter that marsk from one to ten the difficulty of your challenges? It would ovbiously be your opinion.

[@quantumbracket6995]

you could have also subtracted the third equation from the second getting a+b, then you know that: area of the square - 4(a+b) = c

[@gifropan]

In working out the area of the equilateral triangle haven't we used trigonometry? Because the square root of 3 divided by two is the cosine of 30 degrees. I enjoy your videos immensely.

[@tomgreg2008]

Got it! Had to think about the geometry for quite a bit though. Thanks!

[@sujalarora8947]

Though I was not able to solve in a general way, but solved using calculus. When i did it correctly, it gave me much pleasure. I think it's a great feeling for a guy in 11th class

[@irenetonks]

"we'll use trigonometry for fun" Me:😰😱😱

[@nishantajitsaria2884]

This is probably the first presh talwalkar question I've ever solved 100% on my own, what a problem 😍

[@mubtasimnawarmubtasimnawar3013]

i was thinking so much of sectors that I forgot that joining the radius(r) together forms an equilateral triangle. Thank you soo much for this video and explanation!!!!

[@hungphamvan4624]

This is so awesome.! I spent a day but couldn't figure it out. ❤️❤️

[@arielfuxman8868]

Finally someone who is not afraid to tell the truth. An honest person.

[@syed3344]

i just straight up started doing the question (used calculus) and then realised i wasn't allowed to do so;-;

[@twami86]

Hey Presh I emailed you this problem some time ago! Sadly wasn't on the list :(

[@miennho9614]

It's so amzing.....I am very excited to watch this video. It help me much of knowlegge!

[@adhillA97]

It's really interesting because without looking at the solution I can see a really easy way to work it out in the reverse order, but I have no idea how you would work out the first without already knowing the other two and just subtracting them.

[@Idisagreethisisnotanon]

I spent two hours on this with my friends how did I not figure this out

[@LordSandwichII]

"Geometry only, no trigonometry!" Me: Yay finally! An easy problem! 3 hours later Me: (staring at paper) "Wait how the hell are you supposed to do this without trigonometry?!"

[@akvarius7]

There is a way to solve with the help of the circle. A circle is drawn and a square with the center coinciding with the center of the circle is completed. The side of this square was miraculously equal to the radius of the circle. Chords are drawn on the sides of the square. It turns out something like a cross inscribed in a circle. The required area is equal to the difference between the area of the circle and the area of the cross.

[@someonelol3404]

I found without trigonometry

[@maruthasalamoorthiviswanat153]

Excellent solution sir. I couldn't find the solution using with geometry. Its really superb solution. I remember your Picasso story. Yes you are working for your LIFETIME Sir. Thank you. I am eagerly expecting your next video.

[@jinesh027]

Great, pls post more such interesting and challenging geometry problems 👍👍👍👍

[@andriotik007]

Exercise: Find area without using calculus Presh Talwalkar: ok, lest do some algebra there, there and its done. Easy huh?

[@aram5642]

I would be curious though to see the calculus solution

[@gvilsan35]

Nice and smart solution. Congrats

[@sedatturan602]

Good video thanks !

[@Blaqjaqshellaq]

It would be interesting to solve it through calculus too!

[@TheLostEngineer19]

Can you solve it using calculus please? I'm curious about that.

[@marianoravec1236]

I recommend you to find some informations about double integrals :)

[@Cheater3k]

@@TheLostEngineer19 It's 2 times the integral of (sqrt(r^2 - x^2) - (-sqrt(r^2 - x^2) + r)) from r/2 to sqrt(3)*r/2

[@f.p.5410]

@@marianoravec1236 nah, no need for double integrals, it's a trivial integration domain.

[@saahilmittal5791]

Its much easier with the concept of area under curve, took me 5 min to solve

[@vlad071096]

It can be solved easier using only areas of the square and the quarter-circle.

[@jedinxf7]

didn't watch the video yet but I can't imagine any other way to do it...

[@erik19borgnia]

Wow, nice! Thank you!

[@WoodyC-fv9hz]

Hi Presh, solutions for c,b,a are 126.058, 51.129, and 17.355 (truncated, not rounded). No need for a set of equations here. Focus on the area of the intersection of quarter-circles shaping a dome, both running from the base into a point up top, thus looking like a bishops-hat, comprising of areas "b+c+b+a". Area "b+c+b+a" is worked out via (r^2)/2 * 1.228378 [see footnote]. I already know "b+c+b" ("b+c+b" being the double-sided symmetrically convex lens called the "Fish". The fish is derived via 400 - 2 times the area "a+b+a" ["a+b+a" being the area shaped like a "Hangglider"]). Please note: "Hangglider" is equal to 400 minus "Quartercircle" (Quartercircle being r^2*Pi/4). Area "a" is effortlessly computed via "Bishops-hat" minus "Fish". At last, "Hanglider" minus 2 times the area "a" results in area "b". Finally, "Fish" minus 2 times "b" gives us area "c", solving the problem. Elementary arithmetic. [Footnote:] 1.22837 comes from 2*Pi/3 - Sqrt(3)/2, being a constant term in the area-formula of my second (larger) double-sided symmetrically convex lens (centres apart by distance r). In this case, (r^2)/2*1.22837 represents half of this larger lens, thus representing the "Bishops-hat". Only the upper half of this lens is visible in the sketch, stretching by imagination into the 4th quadrant below and standing up vertically on its point.

[@sogari2187]

i did it completly different but got my solution :'D

[@animeedits3507]

I did same but without watching this

[@RamkrishanYT]

Me: easily integration Him: solution should only be trigonometry ☹️

[@IStMl]

Not trigo, only geometry that’s even worse

[@ddays4393]

How did you solve with integration

[@pluto8404]

@@ddays4393 horseshoe method.

[@ddays4393]

@@pluto8404 ı dont get it when ı search about horseshoe method ı find it's trigonometric functions integral which is intertwined

[@joaovictormendes2293]

@@ddays4393 www.wolframalpha.com/input/?i=+2+integral+10+to+10sqrt%283%29+integral+%28-root%2820%5E2-x%5E2%29%2B20%29+to+root%2820%5E2-x%5E2%29+dydx

[@zixuansweetie]

Great demonstration!!!

[@bpark10001]

There is easier way to get area of one of the little moon-shaped pieces. (Following assumes square side of 1 initially). The left-center intersection point is known to be at Y=1/2 by symmetry. Form triangle from both bottom corners of the square, and the leftmost intersection of the arcs. It is isosceles triangle, 2 sides = 1. Altitude = 1/2, so area = 1/4. Angle (lower right corner of the square) is 30 degrees (drop perpendicular from left arc intersection point to base of square). This is 30-60-90 triangle. The corners of the square are trisected by these triangles. This makes all the isosceles triangles congruent. Area of arc circular segment = 30/360 x pi (1)^2 = pi/12. Area of moon segment = pi/12 - 1/4. Distance points central square from center can be found using 30-60-90 triangle (sqrt(3)/2 - 1/2). This is semi-diagonal of square, so square this and double. 2 * (1 - sqrt(3))^2 to get 1/2 * (2 - sqrt(3)). Add in 4 of the moons, and scale by R^2, to get Presh's answer.

[@bugalaman]

"Don't use trig or calc to figure it out". *proceeds to use trig* You used a triangle and angles at 2:06. If that isn't trigonometry, then I don't know what is.

[@bugalaman]

@Hans von Zettour Trig is literally the study of angles and triangles.

[@danielsgrunge]

I was like damn that's my chance to shine, all the things I studied in coll... "No Calculus" ah ok then

[@vameza1]

Amazing!!! The best part is show another possible solution!

[@pauloalbuquerque6824]

Sehr schön! Nice! Muito bom!

[@kauaamorim1617]

"As always, thanks for watching" -Michael Stevens 1984-2027

[@tbarnes36]

Иө Fun whoa! I doubt it. He will live until he is a scraggly old long bearded man.

[@Dr.Waffles]

2027????

[@Dr.Waffles]

@Siddharth Doshi i feel like we are getting wooooshed

[@tajsinha5268]

Lesson: complicate a simple thing to the limit that no one understands 😂

[@uglygamer2028]

It took me till the 3:20 mark to figure out howe the rest will work out and how to find the answer X’D that visual of the square put into those seperate sections really helped me figure out the meaning of every varrible within the equations, it's like seeing the shapes and knowing how many were subtracting from that amount, and what we are left with we then use as a foot hold to "Math" the next step till we find "a" then "b" then lastly "c" I got pretty hyped for some reason in this video dX props to this channel and this KZbin creator

[@viniciusfernandes2303]

Thanks for the video!!

[@thiagocunhafreitascastro4656]

That’ a common question for admition tests in college in Brazil

[@gabrielnettoferreira267]

Eu diria que eh uma questão nivel ITA pra cima; inclusive, essa questão eh a última no volume 9 do Fundamentos da Matemática Elementar do Gelson Iezzi, grande livro.

[@thiagocunhafreitascastro4656]

Mano, tenho certeza que já fiz uma dessa pra Fuvest ou Unicamp

[@maicom802]

UFMS e UFGD cai todo ano uma dessa....

[@gabrielnettoferreira267]

Calcular area hachurada cai em todo vestibular, mas essa questão em específico eu considero mais difícil. Põe ela numa ufrgs da vida e vamo ver a taxa de acerto sjakakwkw

[@felipec.2854]

Teoricamente dá pra resolver essa questão só com ensino médio

[@tamajongmichaelnkeh1978]

I had formulated this question and solved it by integral calculus

[@hanahlavata128]

How did you formulated

[@tamajongmichaelnkeh1978]

@@hanahlavata128 well, I just posted this same problem in a science forum as a challenge problem. I solved it by getting the equations of the four circles relative to the frame containing axes: left and bottom side, with the origin as the bottom left edge. Then I used integration to get the area enclosed by the graphs of the circles. The rest is just algebraic manipulation.

[@mibsaamahmed]

Wow, this was a fun problem to watch being solved!!

[@justpaulo]

For once I solved this kind of problems! It took me a lot of time, but I finally saw the 3rd equation.

[@joaovictormendes2293]

With integrals is a lot easier

[@najibqunoo7232]

0:33 We have exactly the same question here in germany at 9th grade 😂

[@t.minojan7029]

same too u

[@syed3344]

cap

[@enricolucarelli816]

Wonderful! 👏👏👏

[@akankshagoel3325]

Thanks for the solution I was stuck in this problem for weeks

[@palakagarwal9819]

Can you please show how to solve this using calculus!?

[@tarushkumar1032]

By using integration under curve if you know how to solve them

[@yurenchu]

@Palak Agarwal, Let r be the side of the square. If we draw horizontal and vertical lines through the center of the square, then the requested area c is divided into four triangle-like areas. We can hence calculate the area of c by calculating the area of, say, the top-right one of those four triangle-like areas, and then multiply by 4. Hence, using left side of the square as y-axis and bottom side of square as x-axis, area c is calculated as 4 * ∫ (√(r²-x²) - r/2) dx , from x = r/2 to x = r*sqrt(3)/2 ... substitute x = r*cos(t), dx = -r*sin(t)dt, from t = pi/3 to t = pi/6 ... = 4 * ∫ (√(r²-r²cos²(t)) - r/2) * (-r)*sin(t)dt , from t = pi/3 to t = pi/6 = 4r² * ∫ (√(1-cos²(t)) - 1/2)*sin(t) dt , from t = pi/6 to t = pi/3 = 4r² * ∫ (√(sin²(t)) - 1/2)*sin(t) dt = 4r² * ∫ (sin(t) - 1/2)*sin(t) dt = 4r² * ∫ (sin²(t) - (1/2)*sin(t)) dt ... use the identity cos(2u) = cos²(u) - sin²(u) = 1 - 2sin²(u) ==> sin²(u) = [1-cos(2u)]/2 ... = 4r² * ∫ ( [1-cos(2t)]/2 - (1/2)*sin(t) ) dt = 2r² * ∫ ( 1 - cos(2t) - sin(t) ) dt , from t = pi/6 to t = pi/3 = 2r² * [ t - sin(2t)/2 + cos(t) ], for t = pi/6 to t = pi/3 = 2r² * [ (pi/3 - sin(2*pi/3)/2 + cos(pi/3)) - (pi/6 - sin(2*pi/6)/2 + cos(pi/6)) ] = 2r² * [ (pi/3 - (sqrt(3)/2)/2 + 1/2) - (pi/6 - (sqrt(3)/2)/2 + sqrt(3)/2) ] = 2r² * [ pi/3 - (sqrt(3)/2)/2 + 1/2 - pi/6 + (sqrt(3)/2)/2 - sqrt(3)/2 ] = 2r² * [ pi/6 + 1/2 - sqrt(3)/2 ] = r² * [pi/3 + 1 - sqrt(3)] ... given r = 20 ... = 400 * [pi/3 + 1 - sqrt(3)] = 126.059 (approximately)

[@palakagarwal9819]

@yuri renner Thanks a lot !

[@yurenchu]

@Palak Agarwal, You're welcome!

[@mikumikuareka]

Let's assume that square lays down in quarter I of the Cartesian plane, its sides are parallel or perpendicular to the axes and one of its corners is at (0, 0). If r is a length of a side of a square, there're 4 arches inside of it: y1 = r - sqrt(r^2 - x^2) y2 = sqrt(r^2 - x^2) y3 = sqrt(r^2 - (x - r)^2) y4 = r - sqrt(r^2 - (x - r)^2) Also because the shape is symmetrical around x = r/2, we can ignore one half of it and evaluate only the area of it's another half and double the answer after. I choose to ignore the part where y3 and y4 intersect. So, now we need to learn where y1 and y2 intersect: r - sqrt(r^2 - x^2) = sqrt(r^2 - x^2) 4r^2 - 4x^2 = r^2 4x^2 = 3r^2 x = sqrt(3)/2 * r And the area between these curves that has to be an definite integral of [y2 - y1]. Also let's not forget that we need double of it: s = 2(y2 - y1) = 2sqrt(r^2 - x^2) - 2(r - sqrt(r^2 -x^2)) = 2r - 4sqrt(r^2 - x^2) So now you just have to integrate [2r - 4sqrt(r^2 - x^2)] from [r/2] to [sqrt(3)/2 * r] and after all evaluations, you'll come up with something like: S(r) = (3 + π - 3 sqrt(3)) r^2 / 3 And after substituting r = 20 you'll get: S(20) = 126.058697451 Though I must admit that this integral is extremely difficult to take and it took me a couple of hours to do it and simplify the answer properly. Actually, I didn't have to because there was a correct answer at the end of the video, but I wanted to be sure that this solution really leads to this answer.

[@Moliuan]

2:02 You can solve a by subtracting 2 sectors and 1 triangle from the square. Then solve b then c.

[@akaRicoSanchez]

Yep. That's what I did and it's a lot easier than juggling with the three equations. It's also amusing because basically you are answering the questions in reverse order.

[@Pedritox0953]

Beautiful method!!

[@darnellbaird206]

I used a vent diagram approach to solve. Reduce each portion (quarter circle) to an intersect.

[@darnellbaird206]

I was wrong.

[@AntonMakesStuff]

2:24 isn't this kind of trigonometry tho?

[@marcoantonioloureiro5883]

No, its area from equilateral triangle.

[@AntonMakesStuff]

@@marcoantonioloureiro5883 I know, but it's hard to calculate without trigonometry (I think).

[@juyifan7933]

@@AntonMakesStuff You can calculate the height of equilateral triangle with Pythagoras. You dont need trigonometry.

[@bluerizlagirl]

The formula for the area of an equilateral triangle is well-known, though. It would certainly be on the sheet of formulae given out with any examination paper, which includes areas of shapes and volumes of solids as well as illustrations of alternate, parallel and corresponding angles. Just because you need to use trigonometry to prove it from first principles, doesn't mean using a well-known identity counts as using trigonometry. _I swear I didn't know it was trig, officer! It was written right there on this here crib sheet, innocent as you like ....._

[@LuchoTorres96]

Trigonometry is just generalized Pythagoras

[@holdenh-dawg8772]

*I have a question I don’t know how to solve* Solve for a, sqrt(a)+b=a

[@CheaterCodes]

Substitute a = c^2 Leaves you with c^2-c-b=0 Solve for c, solve for a

[@mayankraj9249]

I'm in Class 8 and I understood the first method. My favorite part of maths is Algebra. Thank you for making me motivated on Geometry and Algebra. I'm going to share this video to my friends and family.

[@HackedPC]

Were you able to solve the problem ? Or just saw the solution

[@mayankraj9249]

@@HackedPC my fav is algebra so obv i've not seen it, yk when he says pause the video and solve it, i didn't get the second method

[@lingleo6355]

Love the solution. It's neat.

[@RekhaGujar83]

This is a question *of class 10 cbse India*

[@mr.legend.6741]

This type of question was there in my 12th class.

[@purushottamsinghbhadoria5727]

Where?

[@mr.legend.6741]

@@purushottamsinghbhadoria5727 In the school course.

[@wilderrivera2166]

You make it so simple and so concise, it is very impressive, I would like to see if you can let us know the software tool you use these math problem.Thank You very much in advance.

[@LuisAStgo]

Great! This made my day!

[@sudoheckbegula]

Hi presh, Which software do you use to make your videos

[@abhishekghosh8304]

I couldn't find the way to make the last equation. That was really hard.

[@HeckaS]

That pure algebra solution was beautiful. I paused and solved using trig before you said not to use trig.

[@anandarunakumar6819]

Did exactly the same way. Glad that i still remember basics..☺

[@mattmartin7028]

Wow congrats, you must be one hell of a smart person! :)

[@kamlesh_darji]

Can we consider the circle inside is 1/4 th of full ??

[@joeeeee256]

We can indeed :)

[@randomrimrock]

I’m just grade 8 and my mind is swollen after watching this 😂

[@sumchamgosynkdive686]

Me too 🤣

[@itsahmd295]

I'm grade 11 and I just couldn't find out the third equation just bec I didn't notice the shape so sad lol

[@binashah3106]

im in 10 and i didnt even try

[@gajavelliprincesses]

Great thought Mr.Presh.

[@LeonardoFazan]

What I managed to do using integrals and coordinates We can see that inside the square we have 4 quarters of a circle, if we ignore 3 of them and focus on just one, we can see that the area that we want is symmetrical if we trace an x and y-axis in the center of the square. By doing this, we can see that in the upper-right sector of our graph is ¼ of the area that we need. So we just need to find it. If the rest of the circle had its center on the (0,0) of the coordinate system it would be very easy to calculate, but it is not. We know that the radius of the circle is 20, and if the quarter circle is fitted perfectly inside the square, it means that the center of the circle is 10 units away from the imaginary (0,0) that we set. The general equation for a circle is x^2 + y^2 = r^2 , but in this case, the center of the circle is displaced by 10 units negatively, 10 on the x-axis, and 10 units negatively in the y-axis. The formula of circle displacement is (x - Dx)^2 + (y - Dy) = r^2 , considering that Dx and Dy are the distance of the displacement. Applying that to our circle we have (x + 10) ^2 + ( y + 10) ^2 = 20 ^2 Knowing that we can know to integrate this equation, but first, we need to know where is the intersection in the y-axis. Doing that is very simple, we just need to substitute y by 0 on the equation and then calculate the x. By doing that we know that the graph intercepts y when x = [10(-1 + sqrt(3))] We can now isolate the Y on the equation, so we can properly integrate it. By doing so we have: y=sqrt(20^2-(x+10)^2)-10 Now we can integrate {sqrt(400-((x+10)^2)) - 10} from 0 to [10(-1 + sqrt(3))] now Which is 100/3 (3 - 3 sqrt(3) + π) = 31.5146 So, the area on the figure on the upper-right sector is 31.515, but it is just ¼ of the total area that the problem requires, so now we can multiply it by 4. By doing it we have 31.515 * 4 = 126.6 Overview: don’t do it using integrals, it is WAY more complicated than it looks (I omitted the integration process of {sqrt(400-((x+10)^2)) - 10}, because it's very boring and difficult to not get lost)

[@saturnslastring]

That's exactly how I did it too. I disagree about your advice to not do it this way. It took me like 5 minutes to get the answer. I'm not sure I could have solved it using geometry any faster.

[@thiccaxe]

“Find the answer without the appropriate methods:” *ITS ELEMENTARY, MY DEAR WATSON*

[@asktnegi5723]

Sherlock?

[@thiccaxe]

Holmes?

[@asktnegi5723]

@@thiccaxe yep

[@BRYDN_NATHAN]

Weird I can use al-kashi law but not calculas.

[@therambunctiousrobloxian9323]

and you can't use gogol tranlast

[@seeker1857]

Yes because there is nothing called CALCULAS

[@luisownerbr]

I've got one of those little rulers, you can like, measure the sides of the funny center square then add a little bit because they're not straight.

[@hjchoi2063]

I solved the problem in a little bit different way with the 2nd approach. The circular sector with the radius of r and the central angle of 30 degrees has the area of π r^2 / 12. Also consider an obtuse triangle of which vertices are two nodes in the radius of the circular sector and the centre of the square. The triangle has the bottom of (√3 - 1)r/2 and the height of r/2, so its area is (√3 - 1) r^2 / 8. If we subtract 2 times the area of the triangle from that of the circular sector, we obtain the one-fourth of the area of the blue region.

[@atharvaverma5013]

Can you do this with integration?

[@ChepitowSoHer]

I have the same question.

[@nocallerid1060]

Yess

[@ChepitowSoHer]

@@nocallerid1060 how?

[@nocallerid1060]

First divide the Area shaded into two equal parts by a vertical line placed at the middle. Then get the equation of the two quarter circle(or the whole circle) then solve for the limits by solving for the point of intersections. Then you can have it.

[@pyazkachori123]

Yes, but the real fun is solving it with geometry, as it requires more thinking

[@Nsomia1567]

중학교때 풀던 기억이 ㅠㅠ

[@user-nana833]

ㅋㅋ 우리나라 중학교 과정

[@user-nana833]

똑같은 그림이 고딩때도 나오긴 함ㅠㅜ

[@viharsarok]

There is no need to use the law of cosines in the second method. The height of the equilateral triangle is √3/2*r so the distance between opposing intersection points is y=[1-2*(1-√3/2)]*r=(√3-1)*r. The area of the square within the shape is thus y^2/2=(2-√3)*r^2.

[@tmcjj2001]

Nicely done

[@gabrielpereiraferreira8646]

This is a commom problom in "colégio naval" a exam for Brazilian army for 14 years old

[@johnsmith4499]

This was the easiest problem I've seen in geometry. Aside from the 'prove a square is a parallelogram' things that you get on standard tests in America.

[@Tiqerboy]

Oh boy, I got into work late, expecting to get my normal work done. And here comes Presh with a difficult geometry puzzle that will no doubt take my entire work day to solve, like it usually does, LOL, and I can't put it down until I get the right answer. Grrrr! If we can't use trig, thinking this can be solved using co-ordinate geometry.

[@piershanson1784]

The trig method can actually be solved without use of trig. You need the pythagorean theorem, and the fact that 2 circular arcs are symmetrical so the intersection between the top-left and bottom left corner centered arcs intersect with a vertical height of r/2.

[@mukteshbodkhe0718]

My teacher also solved it in a far easier way...

[@vaibhavsharma6210]

My teacher also

[@vimleshmeena7275]

I have solved this another Geometry way...😊

[@leif1075]

WAIT A MIBUTE CANT YOU SOLVE FOR PARTS 2A PLUS B BY SUBTRACTING THE AREA OF ONE QUARTER CIRCLE FROM THE AREA OF THE SQUARE?? And 2B PLUS C IS TWICE THE AREA OF ONE QUARTER CIRCLE MINUS THE AREA OF HALF THE SQUARE AKA THE TRIANGLE SO WHY AM I GETTING A DIFFERENT ANSWER??!?? PRESH I THINK YOU MADE A MISTAKES..DIDNT ANYONE ELSE DO IT THIS WAY

[@vimleshmeena7275]

But I Do this .... Let see. One Part of a+b = R²(3√3-Π)/12 Total part of 4(a+b)= 4R²(3√3-Π)/12 Total area of square = R² Now Shaded part area = R²- R²(3√3-Π)/3 😊😊

[@genie9845]

@@leif1075 You too? I'm having the same problem

[@ramjeegupta8355]

@@vimleshmeena7275 how you calculate a+b please explain it

[@leif1075]

@@genie9845 you got the same thing?

[@stephenstruble5064]

Yes! I got exactly the same answer. Mine is written in the form 400-400sqrt(3)+(400pi)/3 though. Which is approximately 126.0587.

[@JWentu]

I solved it exactly as in the first method. Nice problem!