Thanks dear for such an elegant feedback. Thank you so much for taking the time to leave this comment. You are awesome 👍 Take care dear and stay blessed😃
With what I have learned from you in previous videos, I am happy to report that I got the correct answer. It took me a lot longer than 16 minutes though (more like an hour and fifteen). Thx much for these wonderful video lessons!
He explains everything very thoroughly and is easy to follow. My question is this: Why were all of my math class problems very much like this, but the exam questions were always more like: "Given two sheep flying, one red, the other headed west, how much does a pound of asphalt cost given that the cow is ten years old?"
Thank you professor for so many fun and challenging problems. I like to think about them in advance and if I get stuck, watch your straight forward, and easy to follow solution. My skill level is increasing with each one and I am beginning to marvel at your creativity in finding such interesting problems.
Thank you so much for your nice and clear explanation! I think it a bit faster if we skip calculating the yellow part. Area of the green = 1/2x [Area of the quarter circle ABC - (3/4 area of the blue circle+area of the square EOFB)]=1.22 sq unit
Challenging, but I'm happy my skills are improving because I got the right strategy even though I got my arithmetic wrong... Thanks again this was an excellent example
You're welcome Ikfat 😊 Thanks dear for such an elegant feedback. Thank you so much for taking the time to leave this comment. You are awesome 👍 Take care dear and stay blessed😃
You don't need to find the area of the yellow part specifically. Get 3/4 area of circle + area of square. Take it away from the quadrant area and divide by 2 to get 1 green area.
True, but finding the yellow area is easy, and finding the values of all three shaded areas makes my OCD happy... And as someone said on another video, watch time is gold on TH-cam, so it's better to be as detailed and as thorough as possible in these videos. As impatient as I am, if I were making this video it would have been about half the length, but the way he does them, a first- or second-grader can easily follow them due to him breaking it down so thoroughly. My version would be a bit higher difficulty due to more assumptions.
Here's another method, perhaps a little simpler, but I try to avoid value judgements: The diagram is symmetrical by reflection along the line DB; so the area of the green region DCF is equal to the corresponding region AED. This means the desired area DCF is equal to the area of the big quadrant ABC, minus the area of square EBFO, minus 3/4 of the area of the small circle--since the portion of the square inside the circle takes up 1/4 of it--all divided by two. The radius of the circle is given as sqrt(2). Then the area of the square EBFO is (sqrt(2))^2, or 2 square units; and by Pythagoras, the diagonal OB is sqrt(sqrt(2)^2 + sqrt(2)^2) = sqrt(2 + 2) = sqrt(4) = 2 linear units. The radius of the big quadrant ABC is 2 + sqrt(2); so the area is (pi/4) * (2 + sqrt(2))^2); Three-fourths of the area of the circle is (3/4) * pi * (sqrt(2)^2) = (3*pi/4) * 2 = 3*pi/2. So the whole expression is A = (pi/4) * ((2 + sqrt(2))^2) - 2 - 3*pi/2, all divided by 2. Dividing through by 2, this becomes A = (pi/4)/2 * ((2 + sqrt(2))^2) - 2/2 - (3*pi/2)/2; carrying out the division: = (pi/8) * (2 + sqrt(2))^2) - 1 - 3*pi/4; multiplying out (2 + sqrt(2))^2: = (pi/8) * (4 + 4*sqrt(2) + 2) - 1 - 3*pi/4; collecting terms: = (pi/8) * (6 + 4*sqrt(2)) -1 -3*pi/4; multiplying out the leftmost two terms: = 6*pi/8 + (4*pi*sqrt(2))/8 - 1 - 3*pi/4; simplifying: = 3*pi/4 + (pi*sqrt(2))/2 - 1 - 3*pi/4; the 3*pi/4 term subtracts out; and Voila! A = ((pi*sqrt(2))/2) - 1 Thank you, ladies and gentlemen; I'll be here all week.
So nice of you Govinda! You are awesome 👍 I'm glad you liked it! Please keep sharing premath channel with your family and friends. Take care dear and stay blessed😃
Thanks Philip dear for such an elegant feedback. Thank you so much for taking the time to leave this comment. You are awesome 👍 Take care dear and stay blessed😃
You're welcome Fedor 😊 Thanks dear for such an elegant feedback. Thank you so much for taking the time to leave this comment. You are awesome 👍 Take care dear and stay blessed😃
As always, I try working these problems out before watching the video. What tripped me here was not recognizing that BD goes through O, which would have helped calculate the radius of the larger quarter circle.
At 8:28, it seems to me that the solution can be had by taking 3/4 of the area of the blue circle, adding the area of the square to that and taking the area of the large circle quadrant, subtracting the aforementioned area and dividing that result by 2! No need to calculate the yellow area (z)!
Right on my friend! Thanks dear for such an elegant feedback. Thank you so much for taking the time to leave this comment. You are awesome 👍 Take care dear and stay blessed😃
So nice of you Faisal dear! You are awesome 👍 I'm glad you liked it! Eventually, I'll cover Probability problems as well. Please keep sharing premath channel with your family and friends. Take care dear and stay blessed😃
To put the problem in perspective,It is like this.The symmetry is not difficult to see.From the fig.R=r(1+sq.root of 2).From here on,find the small yellow and green areas using formulae for area of a sector,area of triangle,area of semi circle of radius 'r'.Since the two circles touch each other,their centres and the point of contact are cllinear.
Area of big circle minus area of circle gives the are of two green shades and one yellow shade. Yellow shade area is obtained by subtracting area of the square minus area sector within the circle thus we can find area two green shades. Hence the area of required green shade can be find out.
If a final decimal answer isn't required, you can simplify the answer further to x = (π/√2) - 1 by dividing both the numerator and denominator of the fraction by √2.
Practice. If you can follow along, you understand it. Next time you see a similar problem you'll remember the logic and will be able to apply it. This puts you ahead of 95% of the population. Personally I still can't name a single Cardashian but am I bothered?
Nice! But if you take 1/8 of the big circle, you must not calculate „z“, it is much more easier: r = √2 R = √2+2 Green Area = (πR^2/8) - (3/4)π - 1 = (π√2 - 2)/2 🙂
That was my remark 6 month ago already. 1/8 of the big cirkel, minus 3/8 of the small cirkel (you have 3/4???), And minus half the little square (so the remaining triangle)
@@LarzB Many thanks, Larz B, congratulations (didn’t see it…). You’re right, the part of the circle (r = √2) is (3/8). But: πr² leads to 2π - therefore (3/8)2π = (3/4)π. 😊 - I corrected my mistake above...
Hello, The video should have mentioned π=3.14 at the beginning, not at the time, 16:08. Also, the value of r=√2 did not appear until after opening the site! This problem could have ∞ answers. Regards,
It is nit picking but the final simplification should have used the rule that √a/a = 1/√a. Obviously not needed if you are going to evaluate using approximations for π an √2 but it would be needed if we were asked to give the answer in the simplest form ( as is often requested on exam papers)
Not a proof, but. If two circles are tangent at the same point i.e. tangent to each other, their radius is perpendicular to that tangent. Said tangent is shared by each circle, each circle's center is collinear with the other radius and point of the tangent.
interesting videos like all the rest, but it looks to me like x would be the area of the two symmetrical sections together... so the area of the one green section would be x divided by 2.
OE = 1.414. So OB = sqrt ( OEsq + OFsq ) = 2 So Radius of quarter circle = 2+1.414 = 3.414 So Area of quarter circle = 9.154, Area ADB = Area BDC = 0.5* 9.154 = 4.577. Area BEF = Area of sq OEBF - 1/4 Area of inner circle = 1.414*1.414 - 0.25*pi*2 = 2 - 1.57 = 0.43, Half of this = 0.215 Area BDC = Area of CDF + half Area of inner circle + half Area EFB. Area CDF = Area BDC - half Area of Inner Circle - half Area EFB = 4.577 - 3.14 - 0.215 = 1.22
Very clear and methodical, but it bothers me that you use more significant figures for the square root of 2 than you use for pi. Why not be consistent?
Sure dear! Dear Aditya, I'm wrapping up Semester final exams for my students this week. Next week, I'll have more time to work on your project. Take care dear and stay blessed😃
Little bit too complicated. Just calculate the area of 1/8 of the big circle with R= 2+sqrt2 and substract half the square (1) and substract 3/8 of the inner circle with r.
Splendid video bro. keep it up. Thank you for helping students.
Thanks dear for such an elegant feedback. Thank you so much for taking the time to leave this comment. You are awesome 👍 Take care dear and stay blessed😃
@@PreMath Thank you for pinning my comment
Thank you for helping ME! (I’m 40)
Very good. I'm learning a lot from your videos. Thanks
With what I have learned from you in previous videos, I am happy to report that I got the correct answer. It took me a lot longer than 16 minutes though (more like an hour and fifteen). Thx much for these wonderful video lessons!
He explains everything very thoroughly and is easy to follow. My question is this: Why were all of my math class problems very much like this, but the exam questions were always more like: "Given two sheep flying, one red, the other headed west, how much does a pound of asphalt cost given that the cow is ten years old?"
Thank you professor for so many fun and challenging problems. I like to think about them in advance and if I get stuck, watch your straight forward, and easy to follow solution. My skill level is increasing with each one and I am beginning to marvel at your creativity in finding such interesting problems.
Thank you so much for your nice and clear explanation!
I think it a bit faster if we skip calculating the yellow part.
Area of the green = 1/2x [Area of the quarter circle ABC - (3/4 area of the blue circle+area of the square EOFB)]=1.22 sq unit
Challenging, but I'm happy my skills are improving because I got the right strategy even though I got my arithmetic wrong... Thanks again this was an excellent example
تمرين جميل جيد. شرح واضح مرتب. شكرا جزيلا لكم والله يحفظكم ويرعاكم ويحميكم. تحياتنا لكم من غزة فلسطين .
Wow !!!! Explanation was elegant and perfect 👌 I could solve this problem easily 🙂 Thanks again
You're welcome Ikfat 😊
Thanks dear for such an elegant feedback. Thank you so much for taking the time to leave this comment. You are awesome 👍 Take care dear and stay blessed😃
You don't need to find the area of the yellow part specifically. Get 3/4 area of circle + area of square. Take it away from the quadrant area and divide by 2 to get 1 green area.
Класс! Логично!🌺
True
very much easier
Good
True, but finding the yellow area is easy, and finding the values of all three shaded areas makes my OCD happy...
And as someone said on another video, watch time is gold on TH-cam, so it's better to be as detailed and as thorough as possible in these videos. As impatient as I am, if I were making this video it would have been about half the length, but the way he does them, a first- or second-grader can easily follow them due to him breaking it down so thoroughly. My version would be a bit higher difficulty due to more assumptions.
Got it. But only because I've watched so many of your videos. You are a great teacher.
Keep up the great work. Thanks to you my speed and intuition to solve these questions has increased. I have just started following your videos.
Ty sir,highly lmpressed by ur teaching methods
I hope in future we see some exciting things
This problem is tough but your solving method is very good.Any one understand simply.
Here's another method, perhaps a little simpler, but I try to avoid value judgements:
The diagram is symmetrical by reflection along the line DB; so the area of the green region DCF is equal to the corresponding region AED. This means the desired area DCF is equal to the area of the big quadrant ABC, minus the area of square EBFO, minus 3/4 of the area of the small circle--since the portion of the square inside the circle takes up 1/4 of it--all divided by two.
The radius of the circle is given as sqrt(2). Then the area of the square EBFO is (sqrt(2))^2, or 2 square units;
and by Pythagoras, the diagonal OB is sqrt(sqrt(2)^2 + sqrt(2)^2) = sqrt(2 + 2) = sqrt(4) = 2 linear units.
The radius of the big quadrant ABC is 2 + sqrt(2); so the area is (pi/4) * (2 + sqrt(2))^2);
Three-fourths of the area of the circle is (3/4) * pi * (sqrt(2)^2) = (3*pi/4) * 2 = 3*pi/2.
So the whole expression is
A = (pi/4) * ((2 + sqrt(2))^2) - 2 - 3*pi/2, all divided by 2.
Dividing through by 2, this becomes
A = (pi/4)/2 * ((2 + sqrt(2))^2) - 2/2 - (3*pi/2)/2; carrying out the division:
= (pi/8) * (2 + sqrt(2))^2) - 1 - 3*pi/4; multiplying out (2 + sqrt(2))^2:
= (pi/8) * (4 + 4*sqrt(2) + 2) - 1 - 3*pi/4; collecting terms:
= (pi/8) * (6 + 4*sqrt(2)) -1 -3*pi/4; multiplying out the leftmost two terms:
= 6*pi/8 + (4*pi*sqrt(2))/8 - 1 - 3*pi/4; simplifying:
= 3*pi/4 + (pi*sqrt(2))/2 - 1 - 3*pi/4; the 3*pi/4 term subtracts out; and Voila!
A = ((pi*sqrt(2))/2) - 1
Thank you, ladies and gentlemen; I'll be here all week.
I actually like your idea of just subtracting the area of the square and 3/4 of the inscribed circle. Makes a lot of practical sense.
difficult problem, but you explain it so easy.
Thank you ,
From, West-Bengal🌹🌹🌹🌹🌹
So nice of you Govinda! You are awesome 👍 I'm glad you liked it! Please keep sharing premath channel with your family and friends. Take care dear and stay blessed😃
Great problem, great explanation, good combination of algebra and geometry!
Thanks Philip dear for such an elegant feedback. Thank you so much for taking the time to leave this comment. You are awesome 👍 Take care dear and stay blessed😃
I LOVE this exercise! Great problem!!
Very good and informative video
Ur explanation is very clear to understand. Thank u Sir 😍
11:32 Alternate Solution:
Area = 1/2 Quadrant ABC - Sector DOF - 1/2 Square OFBE
Area = 1/8 Big Circle - (Θ/360) Blue Circle - 1/2 Square OFBE
Area = 1/8 π(√2 + OB)² - (135/360) π(√2)² - 1/2 (√2)²
Area = 1/8 π(√2 + 2)² - (135/360) π(2) - 1
Area = (3+2√2)π/4 - 3π/4 - 1
Area = √2π/2 - 1
Area = 1.2214
Orderly and easy to follow. Thanks
I liked that a lot! Thanks dear teacher 😊
You're welcome Fedor 😊
Thanks dear for such an elegant feedback. Thank you so much for taking the time to leave this comment. You are awesome 👍 Take care dear and stay blessed😃
As always, I try working these problems out before watching the video. What tripped me here was not recognizing that BD goes through O, which would have helped calculate the radius of the larger quarter circle.
At 8:28, it seems to me that the solution can be had by taking 3/4 of the area of the blue circle, adding the area of the square to that and taking the area of the large circle quadrant, subtracting the aforementioned area and dividing that result by 2! No need to calculate the yellow area (z)!
Great video 👍👍
6:08 could use the square diagonal formula(d=s√2) instead of Pythagorean theorem. That would be the easy way.
Professor loves pythagorean theorem. Shortcut could be c=sqrt2×qrt2 since it is an isosceles right triangle. Hypotenuse = leg × sqrt2
Right on my friend!
Thanks dear for such an elegant feedback. Thank you so much for taking the time to leave this comment. You are awesome 👍 Take care dear and stay blessed😃
Thank you bro, keep it up❤❤👍👍
Could you please make some probability problems sir?
So nice of you Faisal dear! You are awesome 👍 I'm glad you liked it!
Eventually, I'll cover Probability problems as well.
Please keep sharing premath channel with your family and friends. Take care dear and stay blessed😃
Excellent !
To put the problem in perspective,It is like this.The symmetry is not difficult to see.From the fig.R=r(1+sq.root of 2).From here on,find the small yellow and green areas using formulae for area of a sector,area of triangle,area of semi circle of radius 'r'.Since the two circles touch each other,their centres and the point of contact are cllinear.
Area of big circle minus area of circle gives the are of two green shades and one yellow shade. Yellow shade area is obtained by subtracting area of the square minus area sector within the circle thus we can find area two green shades. Hence the area of required green shade can be find out.
я тоже так решал
Bohoth prasanna Hei!
If a final decimal answer isn't required, you can simplify the answer further to x = (π/√2) - 1 by dividing both the numerator and denominator of the fraction by √2.
this turned me on thanks bro
Why BD is a line going through the center of the little circle to the touching point of the little and big circle?
once I got the yellow as 0.43, the blue as 6.28 and the quadrant as 9.15 just subtract and divide by 2 to get 1.22, learning a lot bro
So nice of you dear friend! You are awesome 👍 I'm glad you liked it! Take care dear and stay blessed😃
Area of the required region= Half the quarter of circle - (Half the area of square with side √2 + 45/360 × Area of the inscribed circle) = 1.221
Nice story! I got it correctly! :)
This is blowing my mind. The scary part is that I can follow along because it’s so logical.
Practice. If you can follow along, you understand it. Next time you see a similar problem you'll remember the logic and will be able to apply it. This puts you ahead of 95% of the population. Personally I still can't name a single Cardashian but am I bothered?
Nice! But if you take 1/8 of the big circle, you must not calculate „z“, it is much more easier:
r = √2
R = √2+2
Green Area = (πR^2/8) - (3/4)π - 1 = (π√2 - 2)/2 🙂
That was my remark 6 month ago already. 1/8 of the big cirkel, minus 3/8 of the small cirkel (you have 3/4???), And minus half the little square (so the remaining triangle)
@@LarzB
Many thanks, Larz B, congratulations (didn’t see it…). You’re right, the part of the circle (r = √2) is (3/8). But: πr² leads to 2π - therefore (3/8)2π = (3/4)π. 😊 - I corrected my mistake above...
@@murdock5537 small mistake, happens to the best. The way of thinking was right
Hello,
The video should have mentioned π=3.14 at the beginning, not at the time, 16:08. Also, the value of r=√2 did not appear until after opening the site! This problem could have ∞ answers.
Regards,
How to prove that the 2 green parts are equal according to symmetry?
Many thanks!
Wonderful!
Thank a lot brother😀
It is nit picking but the final simplification should have used the rule that √a/a = 1/√a. Obviously not needed if you are going to evaluate using approximations for π an √2 but it would be needed if we were asked to give the answer in the simplest form ( as is often requested on exam papers)
How to prove that the radius of the larger circle R does in fact go through O, the center of the smaller circle?
Not a proof, but. If two circles are tangent at the same point i.e. tangent to each other, their radius is perpendicular to that tangent. Said tangent is shared by each circle, each circle's center is collinear with the other radius and point of the tangent.
interesting videos like all the rest, but it looks to me like x would be the area of the two symmetrical sections together... so the area of the one green section would be x divided by 2.
Thank you sir
You are awesome 👍 Take care dear and stay blessed😃
This only works if F=E. Why two different letters that might represent different values?
Those are 3 points on small circle
Sometimes humans like complicating things when there's a simpler solution.
from algeria big thanks
Yellow area not really required to be find we can simply get the area of upper green part
.5*{Area of quad-(area of 3/4 of ⭕️ +area of oebf◾️)}
OE = 1.414. So OB = sqrt ( OEsq + OFsq ) = 2 So Radius of quarter circle = 2+1.414 = 3.414
So Area of quarter circle = 9.154,
Area ADB = Area BDC = 0.5* 9.154 = 4.577.
Area BEF = Area of sq OEBF - 1/4 Area of inner circle = 1.414*1.414 - 0.25*pi*2 = 2 - 1.57 = 0.43, Half of this = 0.215
Area BDC = Area of CDF + half Area of inner circle + half Area EFB.
Area CDF = Area BDC - half Area of Inner Circle - half Area EFB = 4.577 - 3.14 - 0.215 = 1.22
Very interesting problem
You are awesome Jasbir 👍 I'm glad you liked it! Please keep sharing premath channel with your family and friends. Take care dear and stay blessed😃
Pl summerise the theory at end of the class.
Don't need yellow area. Green regionx2= quadrant - square - three quarters of circle.
Very clear and methodical, but it bothers me that you use more significant figures for the square root of 2 than you use for pi. Why not be consistent?
hay quá cảm ơn người bạn nhé.
Why bo +od =bod is a straight line please proof it
Right triangle
Pl. Check it
#Pythagoras #PythagoreanTheorem #RightTriangle
Try to solve it with integral calculus
Sir it is now over 1 week, please make a video on my question.
Sure dear!
Dear Aditya, I'm wrapping up Semester final exams for my students this week. Next week, I'll have more time to work on your project.
Take care dear and stay blessed😃
Not bragging or nuffin’, but I got this one right!
How the hell…?
like the video!
R=2+2=4, R=2+√2=?
1/2{ㅠR2* 1/4 - 3/4*ㅠr2 - r2) ???
2 + square root of 2 (1.4(1)) = 3.4(1)
R = 3.4
OB = 2
#Radius
Little bit too complicated. Just calculate the area of 1/8 of the big circle with R= 2+sqrt2 and substract half the square (1) and substract 3/8 of the inner circle with r.
I find this method too - much more nicer...
Square root of 2 = 1.4(1)
S=√2π/2-1≈1,23
Radius
Area = [0.25π(2+√2)² - 0.75π√2² - √2²]÷2
y = 2 pi
Watch the magic of Trigonometrical Ratios in this Video:-th-cam.com/video/dK6bDFSsM1I/w-d-xo.html
Square root of 2; square root of 2; 2
AB=√2*2 so A=(π*r*r)/4
A = 2 pi ÷ 4
Area = 2 pi ÷ 4
c = 2
2 + square root of 2
Square root of 4 = 2
First
You are awesome Naeem 👍 Take care dear and stay blessed😃
@@PreMath you have unique questions with unique solution...I like your way of teaching
@@naeemkhan-tm9hx
You are the best Naeem dear 😀
Pythagorean Theorem
Pythagoras
1.4; 1.4; 2
When you know the answer, that video is seams so long. But anyway five star for this seller from aliexpress)))
Z=(racine 2x racine de2)-(r au carré×pi÷4)=0,43
radius = Square root of 2 = 1.4(1)
r = 1.4(1) = Square root of 2
radius = Square root of 2 = 1.4(1)