What is the area of the tilted square?

[@vishwasmalhotra3714]

The 3 triangles in right side are similar. You can easily find edges . Then sum and then square .

I did the same and it worked very well, but the solution presented in the video is really beautiful and more elegant 🙂

[@ballin1006]

True

[@MrAndeweg]

How do you know they are similar? I don't get it. Do you mean they all share the same angles for the remaining 180° of the triangle?

[@red-baitingswine8816]

@@MrAndeweg You have to assume that all the vertices of the 3 triangles are on the side of the square (that was not given). Then all 3 top angles are the same (angles of the same line with the vertical), and all have a right angle, therefore all have 2 angles the same, therefore similar.

[@keara.ls3]

Yeah, but that’s a whole lot more steps.

[@sutapadey5274]

Evolution of the Pythagorean theorem 1. Pythagorean Theorem 2. Gougu Theorem 3. "Your Favourite" Theorem 4. "My Favourite Right Triangle" Theorem 5. *Avoids saying the name* 6. "Here we go again" 7. *Idgaf what it should be called* 8. "The" Theorem

[@azamatalmenov5084]

🤣 lol bro

[@valemontgomery9401]

I remember he tried to make a stink about that!

[@davidreid2708]

Mental arithmetic after noting that the vertical on the squares 2 and 3 coincides. Answer 45 in about 15 seconds.

[@piman9280]

There is no evolution - it's just the Pythagorean theorem. Presh, as much as we all love him, is just too stubborn to say it.

[@orngng]

"The"orem

[@boguslawszostak1784]

Another simple method. (solved "in memory in 30 sec)" The middle of the three triangles has sides 1 and 2, so the square of the hypotenuse is 1 + 2 * 2 = 5. The height of the top of the "three squares" above the base line is 3 + 2 + 1 = 6, so it is 3 times greater then height of middle triangel. So the searched area is (3 ^2) * 5 = 45

[@martineyles]

3 similar triangles on the right. We know the ratio of the bottom and left of the middle triangle is 1 to 2 and the total height of the triangles is 6. Therefore we can replace the boxes on the right with a right angled triangle with known sides of 6 and 3. We can then draw two squares on the sides with known areas of 36 and 9. The area of the triangle on the hypotenuse is known to be the sum of these, 45.

[@Tiqerboy]

I just it when I solve a problem exactly as Presh does. Easy, very straightforward solution, did this one in my head. Nice way to start the morning.

[@johnsavard7583]

This one was relatively easy. Focus on the fact that there's a square of size 2 on top of a square of size 3, and those two squares' left edges are aligned. So you know the slope of the sides of the square: a rise of 2 over a run of 1. So as the total rise is 6, you now know by the Pythatorean theorem its area is 9 plus 36, or 45.

[@andrewzulu]

Neat solution, Presh! I did use the 1.5, though. The horiz. length between the RHS of the 1.5 square and the bottom of the blue square being equal to 3 by similar triangles. Then, the length of the blue square is 2*sqrt[1.5^2 + 3^2]. Double this to get area of the blue triangle, and there you have it!

[@As_A________Commenter]

I did the exact same thing, well done mate

[@matttipton5403]

You assumed that the point on the 1.5 square touching the blue square is in the exact middle of the blue square. This information was not given, it happened to be true by luck. What if the problem were changed such that the supporting square on the left was 2x2. The position of the left square is unknown so it would make no difference to the solution.

[@vacdota5516]

argh, stop assuming geometrical illustration

[@anandk9220]

Answer : 45 square units Reason : Tilted Square edge = 1 : 2 : 3 Length of middle part of this edge = √5 (by Pythagoras Theorem) So, by using ratios, Total length of square edge = 3 √5 Hence, Area = side^2 = 45 square units EDIT : I think the above solution is the EASIEST WAY to solve this problem. In case there's any other best way out, kindly share in my comment replies.

[@sand747]

that's what i did

[@jimmjamme3067]

I set the origin at the top left corner of the 2 square, y=mx, -5=2x for the corner of the big triangle, so its length is 3, then use pythagoras to get 3^2 + 6^2 = h^2 = area

[@AkashRaj-nq7qm]

kzbin.info/www/bejne/d6WllH2Om7qfepY

[@rasulrahimov744]

it was very easy compared to other problems in this channel i made it almost instantly

[@zyc4nthropy728]

The video length is pi. Nice touch.

[@EgeSaribas05]

Exactly

[@Chasemuss]

First one I've gotten in a few years. I typically don't try them, but due to watching so many videos, I figured this out easily

[@G8tr1522]

i'm pretty sure you can solve this without knowing the side length "1" and "1.5" if you know the intersecting point of the 1.5 length square bisects the big square.

[@smitsoni6078]

Yes I used that method.

[@THS92]

Nice solution. I proceeded a little differently: ​I split the side of the blue square in a ratio of 1: 2: 3 ⇒ x = x₁ + 2x₁ + 3x₁, Middle triangle is x₂² = 2² + 1² ⇒ x₂ = √5 and its follow that x = 3 × √5 ⇒ S = (3 × √5)² = 9 × 5 = 45 :-)

[@AkashRaj-nq7qm]

kzbin.info/www/bejne/d6WllH2Om7qfepY

[@MUJAHID96414]

It is hard to write

[@Sam_on_YouTube]

Solved in my head in about a minute. Got 45. It goes up 2 and over 1 between the corner of the square of length 2 and 3. Scale that up and it goes up by 6 and over by 3 from the 2 corners. So the hypotenuse squared of that triangle is 3^2+6^2=45. That's the area.

[@richardbloemenkamp8532]

This one was indeed quite simple. After the thumbnail I just clicked the end of the video to see if indeed 45 was the correct answer.

[@red-baitingswine8816]

@@richardbloemenkamp8532 Took me longer. Why? Because it was not given that all 3 vertices are on the side of the square. But when you assume that it verifies all the other givens.

[@AkashRaj-nq7qm]

kzbin.info/www/bejne/d6WllH2Om7qfepY

[@louche3501]

Love how the vid is 3 mins and 14 seconds

[@Rahulm-sy9xy]

Assume the bottom right corner of the square of size 3 is 0,0. We know that the side of the square is passing through the corners of the squares of sizes 3 and 2, which are (-3,3) and (-2,5). The line equation can then be solved as y=2x+9. The point where this line meets the square of size 1 has a y value of 6 and other corner of the square on this side has y value 0. Substitute them in the line to get the points (-3/2,6) and (-9/2,0). The distance between these points which is the length of the side of the square is √45, so the area of the square is 45.

[@jimmykitty]

Seems like this problem is much harder 😕 But easy! I could solve it 😊

[@Akrushta]

Hmm btw Trigonometry would be handy in such cases I assume.....

[@jimmykitty]

@@Akrushta Oh Really! Can you pls explain me how? 😊

[@Akrushta]

@@jimmykitty nono I mean, same types of questions are asked In SMO but angles were also concerned, in this also once by similarities, you've proved the ratio is atan(1/2) then we get to know by pytha that the addition of hypotenuse formed by the squares 1u + 2u equals the hypo formed by 3u, then by similarities and BPT, the two triangles at the bottom are congruent, so the ratio of atan(1/2) remains same so the hypo becomes √(9+2.25) = √11.25 which is half of the side of square so area = (2√11.25)² = 45, you can also use cotangents and cosecants but this question was made simple by reducing the angle part....😊 So no need

[@jimmykitty]

@@Akrushta Thanks a Zettabyte for the elegant explanation 😊

[@MUJAHID96414]

Me too 😊

[@Kriz10]

Thank you Mind Your Decisions . After constantly watching your videos I was Able to solve this problem. Much love ❤️❤️

[@AkashRaj-nq7qm]

kzbin.info/www/bejne/d6WllH2Om7qfepY

[@arcaneminded]

I found the hypotenuse of the '2' block (sqrt(5)) and then said the top block is 1/2 * sqrt(5) that and the bottom is 3/2 * sqrt(5). Sum them all up and you have L = 3 * sqrt(5). So the area is 9 * 5 = 45. Edit: Looks like this was the most common approach from what I can tell.

[@guestuser2373]

I did the same.

[@gokulbansal5520]

I didn't get how you took the base of the triangle to be half of 6(the triangle which's hypotenuse was the side of the square).

[@kvsh5172]

Because all three small triangles on the right have all equal form. The length from the bottom right of the 2cm box to the blue box is known (2+1). So the ratio of the triangle is 2 height : 1 length. Of you use the same ratio for the big triangle you get 6 height : 3 length.

[@anshumanagrawal346]

Similar triangles

[@ayoutubecommenter7494]

1. They are similar triangles 2. You cant count it, the 1 cm triangle has 0,5 base, the 2 has 2 (mentioned in the video) and the 3 has 1,5 (bacause again, it's similar) so it's 0,5+1+1,5=3

[@dharma5227]

they were similar triangles and in geometry such triangles have same angles and ratios of sides

[@orangenostril]

It was easier for me to see it as finding the slope of the larger line (2/1), so since it goes up 6 it goes over 3

[@titantv007]

Perfect contents I can never learn from schools. It just sharpen my brains.

[@Aiden-xn6wo]

Solved it in 30 seconds, nice puzzle.

[@viswas2389]

Realy

[@XariksBeatbox]

Same!!

[@AkashRaj-nq7qm]

kzbin.info/www/bejne/d6WllH2Om7qfepY

[@rhariprakashxics3676]

Its easy to find using congruence triangles.As the blocks with side 1,2 and 3 makes similar triangles triangles with the blue square.Then by congruence we can find the hypotenuse side of all the three triangles and when we sum up that we get the side of the square as 3√5 and thus area of square is a^2 = 45 sq.units

[@adameager7114]

I see a lot of people saying that they used the 1.5x1.5 square to solve this one, but until you work it out as presented, can you be sure that that square's corner is actually at the midpoint of the larger square"s edge? Thanks for this easy one, it made me feel smart!

[@tombratcher6938]

I agree. I looked at this and thought, that 1.5 thing tells us nothing, because you could just have a slightly bigger square further to the left instead

[@asphaltpilgrim]

I may be wrong, but if we are turning the square on it's corner the base will rise up (1.5) as much as the perpendicular moves right (1.5) so we can get the 3x1.5 triangle first that way?

[@adameager7114]

@@asphaltpilgrim That works, but first you have to show that the vertical edge's midpoint indeed moves over 1.5, which is only shown by working out the triangles on the right.

[@chinareds54]

Yeah I think the 1.5 might be there to set up a more complicated problem (which this was adapted from). For example, we could draw a line from the intersection point to the top right corner and make a triangle, and ask what the area of that triangle would be. (Solving Presh's problem would lead us to the fact that the 1.5 hits the blue square at the midpoint, therefore the newly constructed triangle would be 1/4 the area of the square which is 45/4).

[@roichir7699]

Easy. You have two similar right triangles there. On the right side the hypothenuse is the sidelength of the square with one side being 6, the height of the boxes combined. The other one is the triangle on the left where one side is 1.5 at the middle of the side of the square. Because it is a square and it stands on a flat ground you can get the other side of the triangle on the right as 3 by turning the left triangle by 90 degrees and doubling it. Then just Phytagoras and you are done.

[@ziadalkhory4019]

I finally managed to solve one of the problems in your channel after 3 yrs of subscribing 😂😂😂

[@AkashRaj-nq7qm]

kzbin.info/www/bejne/d6WllH2Om7qfepY

[@sethmizrachi8337]

I got the same answer but a bit more complicated method. I took the middle triangle with lengths 1 and 2 and got the hypotenuse of sqrt(5) then I scaled it up to the size of the tilted square and I got 3sqrt(5) squared which gets the same answer

[@neelakandan.v.m2991]

I also got the answer but in even more complicated way. I found that the small triangle made by the square with side 3 and the small triangle made by the square with side 1.5 are similar and both their hypotenuse form the big square's half of the side length.So( (3)^2+(1.5)^2 )×2 and then square it again.

[@antianti4331]

I went even more complicated way Calculated hypotenuse of each small triangle (sqrt(1.25), sqrt(5), sqrt(11.25)), then took out of the sqrt the relevant multiplier, to obtain sqrt(5) term in each case, so x1=1.25/5 = 0.25, x3=11.25/5 = 2.25, sqrt(0.25)=0.5, sqrt(2.25) = 1.5, then sum sqrt(5)×(0.5+1+1.5)= 3sqrt(5), which is exactly your answer.

[@RISE.....]

I bet I got most complicated method ( integration by diagonal)

[@tiltingwindmill]

@@antianti4331 yup - this is how I did it.

[@marcello3621]

Look at the side of the square that intersects the vertices of squares of lengths 1, 2 and 3. Consider in special the point which intersects the square of length 2. The distance of this point and the square of length 1 is *x*. The another distance is *y*. Note that x + 1 + y = 2 ---->>> *x + y = 1*. Let this one for later. Then i used triangle similarity to find the values of x and y: 3/1 + x = 2/1 3 = 2x + 2 2x = 1 x = 1/2 y = 1/2 To find the length of blue square, you need to create an triangle which one of the sides has length in terms of the square's length. Assuming sq. side is *L*: (1 + 2 + 3)² + 3² = L² (Pythagoras theorem) L² = area = 45

[@silkthread422]

Nice, I stumble across your videos here and there and was never able to solve any problem. This is the first one I managed to solve. Bit proud of the little mass between my ears.

[@AkashRaj-nq7qm]

kzbin.info/www/bejne/d6WllH2Om7qfepY

[@chinareds54]

I used coordinate geometry ( which wasn't really needed but whatever), set the origin point to the bottom right corner, so the intersection points on the bottom right side with the first 2 squares were (-3,3) and (-2,5), making the slope 2. The three triangles have a combined height of 6, so if you drop a vertical down, it creates a right triangle with base 3 and height 6, with the side of blue square being the hypotenuse. Area = s^2 = 3^2+6^2 = 45.

[@vidyadharjoshi5714]

The 3 right side triangles are similar. Middle right triangle height = 2 and base = 1. So top base = 0.5 and bottom base = 1.5. Consider the big right triangle with base = 1.5+1+0.5 = 3 and height = 1+2+3 = 6. So the hypotenuse = sqrt (9 + 36). So area of triangle = 45

[@drivingforgiraffes]

If we assume the objects on the right are squares that are touching the blue square, that square (1) is touching the blue square on the blue square's corner, that the blue square is resting one corner on the ground, and that all measurements in the diagram are correct, then area of the blue square is 45 units. If ANY of those assumptions are incorrect, then more measurements would be needed to determine the area of the blue square.

[@hazaaahmadhazaa2770]

I used the information which you said was useless which is 1.5 , I used the triangle between the square with side length 3 and the blue square the height of that triangle is 3 same as the side of the square and the base is 1.5 same as the length of that useless square due to the inclination so i calculate the hypotenus which =(1.5x1.5)+(3x3) =2.25+9 =11.25 =SQUARE ROOT of 11.25 this will be half of the length of the blue square ( because its a mid point of the heights of 1+2+3)! so to get the length of blue square we multiply the SQUARE ROOT of 11.25 by 2 so area of blue square will be ( SQUARE ROOT of 11.25 x 2 ) x( SQUARE ROOT of 11.25 x 2 ) = 45

[@charlesdang2557]

An exercise in picking out the most relevant information. The 1.5 length square is a distraction. Thanks for an easier one to start the week.

[@red-baitingswine8816]

But you had to assume all 3 vertices are on the side of the square (not given).

[@charlesdang2557]

@@red-baitingswine8816 he said it was supported on the right side by the 3 squares. that means the vertices are touching the side of the unknown square.

[@red-baitingswine8816]

@@charlesdang2557 At the risk of splitting hairs, the one statement doesn't unequivocally imply the other. But I admit I used your premise to solve the problem (then saw that all the givens agree with that - including the 1.5 square). In a sense you are correct and maybe smarter than me.

[@charlesdang2557]

@@red-baitingswine8816 nah, I am just a middling among giants who frequent this channel. it's an assumption from physics and engineering problems. if it is supporting, then it is touching... let's both be happy to solve the easier monday problems haha

[@red-baitingswine8816]

@@charlesdang2557 : )

[@batchrocketproject4720]

Solved in head from thumbnail: the gradient of the right edge of the square is 2/1 (2x2 cube, offset by 3-2). Vertical height to top right corner of tilted square is 6 (sum the stacked squares). Make right triangle with tilted square side as hypotenuse giving 2 legs of 6 (height) and 3 (by gradient 2/1) => blue square side √45, square for area = 45. And THAT'S the answer. Thanks Presh.

[@varun-xu8gv]

We can also solve this using trigonometry, at 1:55 consider the highlighted right angled triangle here if we assume that the angle between the base of triangle and side of square is x then tanx= 2/1. Now consider the triangle formed at the bottom, the angle subtended by it's base is corresponding to angle x hence it is equal to x so in this triangle too tanx should be 2/1, perpendicular in this triangle is 3 and assume the length of base is l then 2/1 = 3/l therefore l= 3/2. Now apply Pythagoras theorem to find the hypotenuse of this triangle which is one part of this square's side. It will come out to be 3√5/2. In the highlighted triangle too do the same and the second part of the square's side comes out to be √5. Apply corresponding angle in the small triangle formed at the top, let it's base be of length y since the angle subtended by it's base is also equal to x so 1/y = 2/1 which gives us the base as 1/2. Apply Pythagoras theorem to find the hypotenuse of this triangle which gives us the third part of the side, it comes out to be √5/2. Add all of these to get the side of square which will be 3√5 and hence the area will be 45.

[@nicholaskehler9169]

The angle of the square relative to the stack of 3 blocks 1/2 or 1 over for 2 up the total stack is 6 so the triangle formed has sides 3 and 6 with the square of the hypotenuse being the answer. As such 3^2+6^2=9+36=45

[@deathhawk300]

The 1.5 does not need to be used. There’s a right triangle where we know the legs, the one created by the 3 and 2 squares. The height is 2 and the base is 1. Then a larger triangle is formed which is similar due to the shaded square resting at all points on (the same) angle. The height that larger triangle is 6 and since it’s similar to the 2-1 triangle, that makes the base 3. Now it’s 3^2 + 6^2 = area. Since the side of the squares are the same no need to square root then square. 9+36=45.

[@ΚΑΝΑΒΟΣΜΑΝΩΛΗΣ]

In 14 secs (sqrt1.25+sqrt5+sqrt11.25)^2=45!!! Thanks for giving us nice puzzles

[@masterbullets3105]

This is the first video in mind your decisions that I have found answer before the video end 👍

[@MrMartinae06]

Similar triangles always make me happy in the morning!

[@xdragon2k]

This is the first time I actually pause to try and solve and actually get the correct answer.

[@realnazarene5379]

Easy. The blue square has a side whose length is the hypotenuse of a right triangle whose sides are 6 and 3, i.e. square root of 45. Therefore the are of the blue square is 45.

[@jalengibson5313]

So I did this with the 1.5 and 3 squares. You have the base 3 and height 1.5 for both of the triangles between the blue square & the 1.5 square and the 3 square and the blue square. The hypotenuse of that small triangle is half the length of the blue square's side. Use the Pythagorean theorem and then double the answer and then square that answer.

[@pacoman37]

Same, took me like 5 min, then spent another 10 trying to figure out what I missed and why there were two other squares

[@peterkwan1448]

The length between the top left corner of the box with side length 3 and the bottom left corner of the box with side length 2 = 1. Since the three triangles on the right-hand side are similar, the side length of the square = √(2²+1²)+√(1²+(1/2)²)+√(3²+(3/2)²) = √5 (1+1/2+3/2) = 3√5 Area = (3√5)² = 45

[@ikeetkroketjes8431]

suppose the hypotenuse of the smallest triangle (with a side 1 ) is x. there is 3 equal triangles, one is the one with side 1, one is 2 times as big and one is 3 times as big. so the side length of the square is 6xthe triangle with side 2 has a side of 1 as well by substracting, so its hypotenuse is sqrt5. its also 2x so 6x=3sqrt5. the area of the square is that squared so A=45

[@AkashRaj-nq7qm]

kzbin.info/www/bejne/d6WllH2Om7qfepY

[@anon6514]

Middle triangle hypoteneuse = root5 bottom is 3/2 that. and top is 1/2 that. Sum = 3root5 Squared = 45

[@KCatch22]

I liked that I was able to solve this one in my head simply by looking at the thumbnail.

[@CristiCroicu]

This was really fun. Nice problem. More of these please 👌

[@AkashRaj-nq7qm]

kzbin.info/www/bejne/d6WllH2Om7qfepY

[@average_student4925]

Using 'Z' Rule, the length of the side touching the corner of large square and length 3 square, will be 1.5 units. Find the hypotenuse using Pythagoras theorem. These will be ((3√5)/2, √5, √5/2) units.. The sum of these will give you the length of a side of the large square, which is 3√5 units. The Area of a square is 's²' which is (3√5)² = *45 units²*

[@lucafwn]

For once, I got the right answer, and in the exact way it was presented. It is very satisfying.

[@AkashRaj-nq7qm]

kzbin.info/www/bejne/d6WllH2Om7qfepY

[@saraidou6722]

The point similar to the lower point raised by 1.5 is now raised by 7.5 (in the upper side) therfore with pytagor asq+ (a/2)sq =7.5sq so: a = root 45

[@xz1891]

The 1.5 square IS MEANINGFUL. Use "1 line 3 rt angles" model, u instantly know that the lower two triangles are congruent, and their rt angle sides are 1.5 and 3, respectively, and the two square's touch pts with big square are mid pts. U got same answer, same speed.

[@wwoods66]

It _looks_ meaningful, since you can see it's congruent to the triangle beside the bottom square in the stack. But you could replace that square on the left with _any other_ size square; it never comes into the solution to the problem of the blue square's area.

[@xz1891]

@@wwoods66 tru, but there is reason why 1.5 one is used, bcz it's at mid pt, ez to prove

[@wwoods66]

@@xz1891 Easy to prove, but a distraction from the question asked.

[@xz1891]

@Bill Woods disagree. the same can be said to, say, square 2

[@wwoods66]

@@xz1891 But you _do_ need the squares in the stack, to know that the height up to the corner is 3+2+1=6.

[@eurgen674]

Look at the square whose side is 2 and 3, there is a triangle which sides are 1,2 and square root(5). So the side of big square is 3*sqart(5), and final area of big square is 45.

[@paulcassidy8130]

Take the right angled triangle to the left of the 2 unit square. This has sides adjacent to the right angle of 1 and 2. Therefore the hypotenuse is √5. This hypotenuse is 1/3 of the length of the side of the blue square. Therefore its length is 3 * √5 The area of the blue square is therefore (3 * √5) ^2 = 9 x 5 = 45 units squared.

[@AkashRaj-nq7qm]

kzbin.info/www/bejne/d6WllH2Om7qfepY

[@richardwiseman422]

The ratio of the height to the width made by the 3 triangles is 2:1. Since the height=6 then the base=3 then the hypotenuse is the square root of 6^2 and 3^2 or the square root of 45. The area of the square is 45^.5 times 45^.5 or 45.

[@AkashRaj-nq7qm]

kzbin.info/www/bejne/d6WllH2Om7qfepY

[@AkashRaj-nq7qm]

kzbin.info/www/bejne/iJmtomqbgL-Dhqc

[@tom-kz9pb]

The 3 small right triangles on the right have the same incline, so ratio of the opposite to adjacent sides should be the same. The middle one is 1x2, so bottom one is 1.5x3 and top is 0.5x1. The horizontal length of the bottom side of large triangle, when drawing down vertically from upper right corner of square is then 1 + 1.5 + .5 = 3. So the two sides of the large right triangles are 3 and 6, then Pythagorean does the rest.

[@normallife3045]

I was wondering why base is half of 6. Thanks.

[@deezynar]

One of the few that I quickly solved in my head.

[@priyanshu4586]

We can easily solve this puzzle by bottom two triangle These triangle is congruent So, 1.5,3, 3/2(✓5) Now ,side of square ( 3✓5) Area of square (3✓5)*(3✓5)= 45

[@dpage446]

How did you prove the triangles to be congruent?

[@priyanshu4586]

Send me ur email. I'll send you its solution

[@AkashRaj-nq7qm]

kzbin.info/www/bejne/iJmtomqbgL-Dhqc

[@theobolt250]

Simple and elegant.

[@myrecord5551]

I have assumed right inclined side of a square as 3 sections. Let it be x, y, z with intersecting points. y=√(1²+2²) =√5 And now take base of uppermost triangle as 'p' hence using similarity 1/p=2/1 p=1/2 Now in uppermost triangle x=√(1²+(1/2) ²) =√5/2 Now in bottom right triangle base is √(z² - 3²) With similarity 3/√(z² - 3²) = 2/1 z=(3√5)/2 Now sum of 3 sections of a square side x+y+z = (√5/2) + (√5) + ((3√5) /2) = 3√5 Hence area of square = (3√5) ² = 45

[@SK-vg3xh]

Very easy. All three triangles are similar. The base of the middle is 1. Therefore the base of upper and lower triangles are 0.5 and 1.5 respectively. Therefore the side of the sqare should be √(3²+6²)=√45

[@xoxb2]

I got the method, but my maths is terrible - I couldn't remember how to do those similar triangles. So I knew it was 36 plus whatever proportion the bottom bit needed to be! I'm just so pleased that Pythagoras has at last been the answer! I've lived in hope a long time.

[@vladimirrainish841]

45. The side of the tilted square is √(6²+3²) = √45. Square with side 1.5 is not needed at all.

[@jakubkrcma]

Before watching the video, my answer is 45 (and the left supporting square is completely unneeded). The height of the stack is 1+2+3=6. The slope is 1:2 because the difference between 3 and 2 is 1. So, the horizontal distance of the lower right corner of the blue square from the vertical line dropped from the upper right corner is 3. Therefore, the side of the blue square is SQR(45) because 3^2+6^2=45. Therefore, the area is (square root of 45)^2 = 45. Now let's see if I made a crucial error somewhere in that simple thought process.

[@mikumikudice]

The 1.5 square gave me the tip that both the greatest triangle and the triangle formed from the two smaller ones' basis are equal to 1.5 (that you can check by seeing that the middle triangle has a basis equal to 1, and the smallest has a basis equals to 0.5 (1 + 2x = 3). So then you can discover the hypotenuse (9 + 1.5^2 = 11.25) then multiply it by 4 (2^2, because it's two times the triangle side that is already squared). So you get 45

[@kinosmead]

Funnily enough I got the answer by just guessing. I couldn't figure a proof for it, but I took as given that the corner of Square 1.5, and the corner of Square 3, were each individually centerpoints of the sides of the blue square. From there, half of the square's sidelength would be determined as hypotenuse of a triangle 3×1.5. Turns out my given was entirely true.

[@babyyoda5074]

The square 3's corner is midpoint of blue square's side length. It can be proved as follows The rightangled triangle formed by square 3 on blue square is similar to the rightangled triangle formed by extending the sidelength of sq1 to base of sq2. The height of these triangles are 3 each. That means the length of base is also equal. If perpendicular and base are equal then hypotenuse is also equal which is half of sidelength of blue square.

[@rontyson6118]

That's exactly how I got it, too. Using 1.5 with 3. I was tired after 30 minutes of nothing, lol.

[@AkashRaj-nq7qm]

kzbin.info/www/bejne/d6WllH2Om7qfepY

[@이관모-p8s]

Simple and clear answer.

[@XTG3PQR]

Can be done with trigonometry as well. Side of blue square is 6/sinθ. sinθ= 2/√5

[@gustavgnoettgen]

Finally one where I got the same solution and quickly too

[@leftybassist4]

[6÷(sin(90-atan(1/2)))]^2 is what I got in my head. Not the most elegant solution by any stretch, but it does equal 45

[@shaggy6249]

Solved it in 29 seconds, nice puzzle

[@soumya3056]

This one is lot easier than other questions that I have attended

[@SuperMerlin100]

The top left corns of the 2 and 3 boxes give us the slant (2 over 1). The with the one box we know the rise is 6. sq(6^2+(6/2)^2)=sq45=side length of the blue square. So the are of the blue square is 45.

[@AkashRaj-nq7qm]

kzbin.info/www/bejne/d6WllH2Om7qfepY

[@Chrisoikmath_]

If the element that the other line segment is equal to 1.5 units was not given, I could solve this problem!

[@talhajubair2827]

I solved it the exact way you did. I feel very nice

[@akashgarg5770]

That feeling when you can solve problems on this channel.

[@242math]

love this fun geometry problem, excellent presentation

[@AkashRaj-nq7qm]

kzbin.info/www/bejne/d6WllH2Om7qfepY

[@bagel040]

First, I found the hypotenuse of the 2x1 triangle, which is sqrt 5, and used that to find the hypotenuse of both other triangles because they are similar, which summed up to 6sqrt5 / 2, or 3sqrt5, which was the length of one of the sides, so i just squared it to get 45.

[@prathampatel1740]

pretty simple, I just took the sine of one of the bottom angles, and then the sin of the other one, which is 90-(theta), and the I just squared them and adding them equating them to one, getting L^2 = 45, assuming the side length to be L

[@JasonLFB]

아마도 이 문제는 피타고라스의 정리를 모르는 사람들도 풀 수 있게 설계한 문제라서 좌측 하단의 사각형이 있는게 아닐까요? 직각삼각형의 합동, 삼각형의 닮음, 삼각형의 넓이, 사다리꼴의 넓이, 정사각형의 개념만 알면 피타고라스의 정리 없이도 풀리는 문제입니다.

[@unknownwriter1507]

Easiest q ever on this channel

[@key_of_destiny4712]

I literally was trying to figure out the root of 45. It’s amazing the little details your brain will leave out that just make you think harder than you actually have to

[@SQRTime]

Hi . In case you are interested in math competitions, please consider kzbin.info/www/bejne/ombIl2t4g8moetU and other videos in the Olympiad playlist. Hope you enjoy.

[@thewaytruthandlife]

2^2 + 1 = 5 ---> sqrt(5) all ratio's are the same for all small triangles.... so 1 vs 0,5 for the smallest and 3 vs 1,5 for the biggest so conclusive: square side length = SQRT(4+1) + SQRT (1+0,25) + SQRT (9+2,25) So the area = ( SQRT(5) + SQRT(1,25) + SQRT (11,25) ) ^2 to be exact which is 45 exactly.

[@AkashRaj-nq7qm]

kzbin.info/www/bejne/d6WllH2Om7qfepY

[@circumsizedmind]

What if the middle square is not aligned with the bottom one ? Then the side square with size 1.5 males sense

[@ojasdeshpande7296]

There are no circles involved but you made a pi minute video

[@francovenica1531]

Hey, I had a math problem ages ago that I wasn't able to solve: A person sum up all the numebers of 4 digits, so 1000 + 1001 + 1002 + n + ... + 9998 + 9999 = X . The problem is that he forgot the number n, and the idea is that that n = X / 9999 . You gotta find n and X.

[@AlbinNilssonAlcaztrul]

Sheesh. I literally used trigonometry instead of pythogoras for the final step. Will I ever be able to consider myself an engineer when I make everything so complicated? Back to basics..

[@AMANSHARMA-ci2yh]

Use similarity in three ∆ touching big square we get answers easily side length =3√5

[@shaggy6249]

maybe it’s just me but a hand icon pointing at whatever you’re talking about helps me learn faster

[@annson24]

Darn, that's easier to what I did. I took the angle of the right triangle by arctan (2/1), then use the angle to sin()=6/x

[@chetopuffs]

trig is cheating.

[@냥이-t7f]

고마워요 😄 당신 덕분에 수학에 흥미,관심이 생겼어요!

[@titan1235813]

I did it in the exact same way. Nice little puzzle 👌🏻 😀

[@PhillipJohnsonphiljo]

Yay got it right but differently. Used the boxes to get the 2 smallest hypotenuses of the top triangles and then doubled the sum of those to get the side length. (known side lengths were 0.5, 1, x and 1, 2, x for those triangles)

[@dangerkeith3000]

I paused the video at 0:52. I worked it out to be 45

[@denelson83]

I recognized the special 1, 2, root(5) triangle and went from there, arriving at the same answer.

[@RicoGrafity]

1:59 wooow i didnt see that coming

[@lutheraugustine1247]

Teaching me to not over think. Eliminate unnecessary data

[@Neel-sama]

I could do this one!!! I'm so proud

[@arachnidigits8824]

Easy! At first I thought it's a difficult problem, but I could solve it in 2 minutes.

[@SQRTime]

Great job! See if you can attack this one as well kzbin.info/www/bejne/a4aTl5WsZbqcg6c