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

@dormeur20092010

4 жыл бұрын

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

4 жыл бұрын

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

@NoTLucas

4 жыл бұрын

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

@sgsnake2x

4 жыл бұрын

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

@AKagNA

4 жыл бұрын

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

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

@ShubhamKumar-vp6ro

2 жыл бұрын

Even didn't get a blow of it.

@plasmakitten4261

2 жыл бұрын

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

2 жыл бұрын

"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

5 ай бұрын

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

5 ай бұрын

@@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).

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

@MrCScottie

4 жыл бұрын

It’s a challenge

@caniggiaful

4 жыл бұрын

My mind jumped right away to calculus!

@dlevi67

4 жыл бұрын

Nothing else remains.

@DoReMeDesign

4 жыл бұрын

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

@PMA65537

4 жыл бұрын

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

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.

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

@vitortutors6030

4 жыл бұрын

Yeah man

@vitortutors6030

4 жыл бұрын

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

@ericli8792

4 жыл бұрын

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

@joaovictormendes2293

4 жыл бұрын

@@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

4 жыл бұрын

@@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.

"No trigonometry allowed" *Procedes to use trigonometry*

@thestaringenclave

4 жыл бұрын

Exactly.

@Bignic2008

4 жыл бұрын

He did say that part was just for fun.

@DoReMeDesign

4 жыл бұрын

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

@squidprince2456

4 жыл бұрын

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

@royk7903

4 жыл бұрын

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))

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

Part 4: solve the area of the square

@eurovisioncyan9550

4 жыл бұрын

399.99999999999999999999...

@sabyasachirimpa

4 жыл бұрын

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

@AjayKumar-il3qn

4 жыл бұрын

Area of square = r^2

@sabyasachirimpa

4 жыл бұрын

@@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

4 жыл бұрын

@@sabyasachirimpa r^2 does not involve integration

0:41 That turkey guy has a weird Naim

@justs_

4 жыл бұрын

Take my like and get out

@clary3860

4 жыл бұрын

Do you think its funny?

@madhusudan6772

4 жыл бұрын

@@clary3860 of course it is.

@EgorRandomize

4 жыл бұрын

Haha I got it

@Rohith.

4 жыл бұрын

It's naim not naim

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

2 жыл бұрын

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

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

@Its__Good

4 жыл бұрын

What?

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

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

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

Жыл бұрын

I found it easier this way.

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.

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.

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

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

@arielfuxman8868

3 жыл бұрын

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

@syed3344

Жыл бұрын

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

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

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

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

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

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!!!!

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

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.

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

4 жыл бұрын

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

Great thought Mr.Presh.

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

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

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.

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

@SKYPRODUCTION2021

4 жыл бұрын

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

Nice and smart solution. Congrats

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

It would be interesting to solve it through calculus too!

@TheLostEngineer19

4 жыл бұрын

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

@marianoravec1236

4 жыл бұрын

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

@Cheater3k

4 жыл бұрын

@@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

4 жыл бұрын

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

@saahilmittal5791

4 жыл бұрын

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

Great demonstration!!!

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

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

I would be curious though to see the calculus solution

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

Thank you so much Presh Talwalkar!

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

@animeedits3507

3 жыл бұрын

I did same but without watching this

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

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

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

"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

4 жыл бұрын

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

4 жыл бұрын

I found without trigonometry

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

@IStMl

4 жыл бұрын

Not trigo, only geometry that’s even worse

@ddays4393

4 жыл бұрын

How did you solve with integration

@pluto8404

4 жыл бұрын

@@ddays4393 horseshoe method.

@ddays4393

4 жыл бұрын

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

@joaovictormendes2293

4 жыл бұрын

@@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

Good video thanks !

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 KZread creator

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

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

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

@mattmartin7028

3 жыл бұрын

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

Thanks for the video!!

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

@t.minojan7029

3 жыл бұрын

same too u

@syed3344

Жыл бұрын

cap

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

@jedinxf7

3 жыл бұрын

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

Beautiful method!!

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.

I had formulated this question and solved it by integral calculus

@hanahlavata128

4 жыл бұрын

How did you formulated

@tamajongmichaelnkeh1978

4 жыл бұрын

@@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.

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

Wow, nice! Thank you!

Love the solution. It's neat.

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

@tbarnes36

4 жыл бұрын

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

@Dr.Waffles

4 жыл бұрын

2027????

@Dr.Waffles

4 жыл бұрын

@Siddharth Doshi i feel like we are getting wooooshed

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

@gabrielnettoferreira267

4 жыл бұрын

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

4 жыл бұрын

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

@maicom802

4 жыл бұрын

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

@gabrielnettoferreira267

4 жыл бұрын

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

4 жыл бұрын

Teoricamente dá pra resolver essa questão só com ensino médio

Nicely done

Well done Dr. Perkins...

"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

3 жыл бұрын

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

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

@akaRicoSanchez

4 жыл бұрын

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.

Nice. Both geometry, trigonometry and calculus are elegant solutions.

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.

With integrals is a lot easier

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

@asktnegi5723

4 жыл бұрын

Sherlock?

@thiccaxe

4 жыл бұрын

Holmes?

@asktnegi5723

4 жыл бұрын

@@thiccaxe yep

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

Like how the several mathematicians are acknowledged. Pretty fascinating. Takes real youthful serene focus.

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

@tarushkumar1032

4 жыл бұрын

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

@yurenchu

4 жыл бұрын

@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

4 жыл бұрын

@yuri renner Thanks a lot !

@yurenchu

4 жыл бұрын

@Palak Agarwal, You're welcome!

@mikumikuareka

4 жыл бұрын

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.

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

@purushottamsinghbhadoria5727

2 жыл бұрын

Where?

@mr.legend.6741

2 жыл бұрын

@@purushottamsinghbhadoria5727 In the school course.

Thank you Presh.

I did the same solution approx 8 years back and is still available in my mail

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

@joeeeee256

4 жыл бұрын

We can indeed :)

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

@CheaterCodes

4 жыл бұрын

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

Wonderful! 👏👏👏

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

This is a question *of class 10 cbse India*

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

@marcoantonioloureiro5883

4 жыл бұрын

No, its area from equilateral triangle.

@AntonMakesStuff

4 жыл бұрын

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

@juyifan7933

4 жыл бұрын

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

@bluerizlagirl

4 жыл бұрын

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

4 жыл бұрын

Trigonometry is just generalized Pythagoras

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.

Great! This made my day!

중학교때 풀던 기억이 ㅠㅠ

@user-nana833

3 жыл бұрын

ㅋㅋ 우리나라 중학교 과정

@user-nana833

3 жыл бұрын

똑같은 그림이 고딩때도 나오긴 함ㅠㅜ

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

Sehr schön! Nice! Muito bom!

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

2 жыл бұрын

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

@mayankraj9249

2 жыл бұрын

@@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

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

@therambunctiousrobloxian9323

4 жыл бұрын

and you can't use gogol tranlast

@seeker1857

4 жыл бұрын

Yes because there is nothing called CALCULAS

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.

Excellent sir

nice job, pt

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

@sumchamgosynkdive686

4 жыл бұрын

Me too 🤣

@itsahmd295

4 жыл бұрын

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

3 жыл бұрын

im in 10 and i didnt even try

Can you do this with integration?

@ChepitowSoHer

4 жыл бұрын

I have the same question.

@nocallerid1060

4 жыл бұрын

Yess

@ChepitowSoHer

4 жыл бұрын

@@nocallerid1060 how?

@nocallerid1060

4 жыл бұрын

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

4 жыл бұрын

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

Great way to solve

Great Solution

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

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

@vaibhavsharma6210

4 жыл бұрын

My teacher also

That was brilliant. It's a very important problem to be known by math lovers.

simply billiant! :-) Thanks a lot!