I love your videos! The problems seem so complex at first, especially with so few given variables, but you explain and work through them using simple theorums and formulas to solve them in ways I wouldn't have thought of! You've made me look at some of these complex problems in completely new ways! Thank you for being such an awesome and inspirational teacher!
you explain very well, right now, with the help of a translator, I am writing in English, so I’m from Russia without even understanding English, everything is clear and I wanted to say that at 3:51 a minute it was possible to draw the diameter of a large circle horizontally and it turns out the diameter is 8 and side 9 turns out the radius of a small one circle is equal to 1 or draw the radius of the large circle and the radius of the small circle to the right side and it turns out 4+r and at the bottom this part will be equal to 5 it turns out 5-4=1
As usual, a useful mind exerciser. Thank you, sir! But let me make another friendly Eritrean suggestion to your otherwise impeccable illustrative explanation. 1) when you place the hash marks on a side of a polygon, I suggest that you use different number of hash marks to identify that the sides do not have the same dimensions (unless they do). Ex., if a triangle ABC has 3 different side measurements, I would use one hash mark, 2 hash marks & 3 Hash marks respectively to clearly identify that the sides ARE NOT congruent with each other, 2) Using the triangle ABC for illustration, as you know, small letter side measurements (a, b, c) are normally used OPPOSITE and corresponding to those capital letters (A, B, C) that indicate the angle identifiers. Maybe you were in a hurry, but, I noticed, you did not follow this helpful method when you were solving the final Triangle using the Pythagorean Theorem. [ 7:50 min to 11.10 min]
At a quick glance, If Rg is the radius of the green circle, Rg is 8/2 = 4. The center of the green circle is x=4 ,y=4 from the bottom left corner. The blue circle radius is Rb. The center of the blue circle is x = 9-rb, y= 8-rb. Draw a line from the center of each circle, this passes through the common tangent. Forming a right angled triangle with this line and the vertical and horizontal. Using pythagorous (9-2Rb-4)^2 + (8-4-2Rb)^2 = (Rb+Rg)^2,. This simplifies to (5-2Rb)^2 + (4-2Rb)^2=(Rb+4) ^2. Then 25-20 Rb+4Rb^2 + 16 -16Rb+4Rb^2 = Rb^2 +8 Rb + 16. Collecting like terms, 25 -44 Rb =0. Rb = 25/44 . The green circle area , Pi * radius ^2 = Pi * 16 and the Blue circle area = Pi * (25/44)^2. To check this , drawing a square around the blue circle the diagnal length is sqrt(2) *2 Rb and passes through the center of the blue circle and the common tangent between circles. I hope this is correct.
I have basically worked out the same method as you ( I have become more adept from watching your videos !) , I am doing the algebra in the comments section without sketching aids or looking through your video. On checking I do want ( (9 - Rb-4)^2 + (8-4-Rb)^2 = (rb+Rg)^2. Then 25 - 10Rb + Rb^2 + 16 -8Rb + Rb^2 = Rb^2 + 8Rb + 16. This gives 25 -26 Rb = 0 and Rb = 25/26. So the Blue circle radius is 25/26 and area PI * (25/26) ^ 2 =PI * 0.92 and the green circle area is PI * 16. I have a slight difference here. Although I did not use this, The common tangent does not coincide with the diagnal formed from a square around the blue circle so I need to be careful reading off explanation sketches. The square diagnal will be 45 degrees from horizontal and vertical and pass through the center of the blue circle. Thank you again, when i have time I do enjoy working through your premath.
Great almost had it just didn't see the 5-r, but I'm still pleased that I deducted the rest of it... More practice needed, I often seem to be waiting for the penny to drop, ah well, if you don't fail now and then you never get anywhere, so bring it on !😃
After watching and enjoying many of your presentations, I have a suggestion for you. Rather than jumping into the solution, present a strategy first allowing the viewer to pause and try to solve the problem. Thanks for all the fun.
I did green circle R=4 (half the rectangle height) so area of 16pi sq un. Blue circle was fiddlier. I ended up with joining the circle centres for the hypotenuse with side length 4+r. The other two sides were 4-r and 5-r. (4-r)^2+(5-r)^2=(4+r)^2 produced r=1 with the quadratic formula. so and area of pi sq un Decimal equivalents approx 50.27 and 3.14 sq un respectively. With the quadratic formula I also had the spurious answer of 25 = r. Unusually, the retained answer was the minus answer - it's usually the other way. Just checked the video and see you used the same right triangle as me.
Wouldn’’t the simpler way have been to draw a line tangent to the other side of the green circle which bisects the blue circle, which only leaves 1 unit available(because we know that the green circle has a diameter of 8 units, so 8-9=1, which is the radius of the blue circle. We still get the Area of the green circle as 16pi, and the area of the blue circle as pi. It’s much simpler than mucking about with the pythagorean theorem.
@@GiuseppeGarufiKeeran you can clearly see that it does, but I guess it's better to be 100% sure and do the math, although I can clearly see the answer staring me in the face.
So nice of you Federico! 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😃
С таким радиусом касательный перпендикуляр к большому кругу должен проходить прямо через центр. Возможно, получилось бы как-то доказать это, не зная самого радиуса.
I'm a little confused, is the answer 17π or 16π and π as 2 separate answers, one for each circle. What would a math teacher accept for the answer? I've been out of school for a long time...
Dear friend, you have a very valid question. Well, to be safe, I'd first calculate the areas individually. Then put the answer as a total as well. As a math professor, I don't penalize my students for these kind of tiny things. Benefit of the doubt always goes to my student. Bottom line: the answer is supposed to be 17Pi. Thanks for asking. You are awesome 👍 Take care dear and stay blessed😃
It's ambiguous. I went to 17π, but when he stopped at 16π and π, I went back to look at the wording of the problem. The title says "Calculate the Area of the Green & Blue Circles" but the card inside the video says "Calculate the areas [plural] of Green and Blue circles". So the first is asking for one number (the total), and the other is asking for two numbers. :-)
If the area of the rectangle is 72 square units (8 x 9), and the area of the large circle is approximately 50.24 square units (16 x 3.14) ... why can we not find the area of the small circle by simply subtracting the area of the large circle from the area of the rectangle? It seems intuitive at first glance but it doesn't work out on paper. What am I missing?
Area of the rectangle is 8x9. 72. Area of the large circle is pi r sq.So pi x 4x4 which is 50.265. The area of the blue circle is rectangle minus large circle. 21.74. They don't dish out CSE Grade 1s for nothing you know.
The first circle is easy, it's 16 PI units^2, the smaller one takes a little bit of work. We can draw a line from A to B. Then we can draw a perpendicular line that bisects that line that is tangent to both circles... Then we can draw a horizontal line from A to the right until it touches the outer perimeter of the rectangle. We know that A to the tangent is 4 since that's the radius of the larger circle. We know that from the edge of the large circle to the edge of the rectangle is 1 since 9-8 is one because the diameter of the larger circle is 8. Now we can draw a vertical line on the right side of the large circle that is tangent to it. We know that this line is parallel with the outer edge of the rectangle. Now, if and only if this tangent line happens to intersect the center of the smaller circle, we are then done. If it does that means that the smaller circle has a radius of one thus giving us an area of PI units^2. If it doesn't then it will require more geometry with parallel lines, perpendicular bisectors, completing the rectangle or square, possibly using the Pythagorean Theorem or Trigonometric functions, etc... until we have enough information to determine either is radius, or its diameter. Then once we know that, calculating it's area is very trivial.
If we trace a diameter of the small circle paralel to the side of length 8 we find is tangent to the greater circle. Therefore r has to be one et voilá!
Area of Green circle = 16*pi. Area of Blue Circle = pi. (Form a triangle from centres so that (4 + r)Sq = (4-r)sq + (5-r)sq. Solve for r will give you 1)
I have an Rb^2 term missing in my algebra. Putting this in Rb^2 -26 Rb +25 = 0 and (Rb -1) * (Rb-25) = 0 and a Radius of 25 is not possible so Rb = 1 ! We now have the same radii ! Green circle area 16 Pi and the blue circle area Pi.
Skipped a step there. You would need to argue that the segment AB passes through the point of tangency of the two circles before you can assume that it is equal to 4 + r.
Muy larga explicación. El área del circulo grande es evidente que el radio es 4 ya que si trazamos una paralela al lado vertical tangente al circulo, entonces el radio es igual a la mitad del lado. Y asi la distancia entre esta paralela y el lado verticasl del rectángulo es igual a 9 - 8 que es igual a 1 que es el radio del circulo menor. Por lo tanto las áreas son respectivamente PI x R al cuadrado y PI x r al cuadrado. La explicación que se muestra en el video es extremadamente larga por eso le doy dedo abajo.
one of the best teachers .
Thanks my friend. You are the best!
Take care dear and stay blessed😃
The best, not "one of the best"
I love your videos! The problems seem so complex at first, especially with so few given variables, but you explain and work through them using simple theorums and formulas to solve them in ways I wouldn't have thought of! You've made me look at some of these complex problems in completely new ways! Thank you for being such an awesome and inspirational teacher!
Wow, what a thorough and clear explanation, sir! You are an amazing teacher, PreMath! I wish you a million subscribers!
تمرين جميل جيد. شرح واضح مرتب . شكرا جزيلا لكم والله يحفظكم ويرعاكم ويحميكم . تحياتنا لكم من غزة فلسطين .
you explain very well, right now, with the help of a translator, I am writing in English, so I’m from Russia without even understanding English, everything is clear and I wanted to say that at 3:51 a minute it was possible to draw the diameter of a large circle horizontally and it turns out the diameter is 8 and side 9 turns out the radius of a small one circle is equal to 1 or draw the radius of the large circle and the radius of the small circle to the right side and it turns out 4+r and at the bottom this part will be equal to 5 it turns out 5-4=1
As usual, a useful mind exerciser. Thank you, sir! But let me make another friendly Eritrean suggestion to your otherwise impeccable illustrative explanation.
1) when you place the hash marks on a side of a polygon, I suggest that you use different number of hash marks to identify that the sides do not have the same dimensions (unless they do). Ex., if a triangle ABC has 3 different side measurements, I would use one hash mark, 2 hash marks & 3 Hash marks respectively to clearly identify that the sides ARE NOT congruent with each other,
2) Using the triangle ABC for illustration, as you know, small letter side measurements (a, b, c) are normally used OPPOSITE and corresponding to those capital letters (A, B, C) that indicate the angle identifiers. Maybe you were in a hurry, but, I noticed, you did not follow this helpful method when you were solving the final Triangle using the Pythagorean Theorem. [ 7:50 min to 11.10 min]
Reminded Us of Pi r squared, Didn't spend time on that assoc/distrib algebra thing. If I remembered it all, wouldn't have watched at all.
At a quick glance, If Rg is the radius of the green circle, Rg is 8/2 = 4. The center of the green circle is x=4 ,y=4 from the bottom left corner. The blue circle radius is Rb. The center of the blue circle is x = 9-rb, y= 8-rb. Draw a line from the center of each circle, this passes through the common tangent. Forming a right angled triangle with this line and the vertical and horizontal. Using pythagorous (9-2Rb-4)^2 + (8-4-2Rb)^2 = (Rb+Rg)^2,. This simplifies to (5-2Rb)^2 + (4-2Rb)^2=(Rb+4) ^2. Then 25-20 Rb+4Rb^2 + 16 -16Rb+4Rb^2 = Rb^2 +8 Rb + 16. Collecting like terms, 25 -44 Rb =0. Rb = 25/44 . The green circle area , Pi * radius ^2 = Pi * 16 and the Blue circle area = Pi * (25/44)^2. To check this , drawing a square around the blue circle the diagnal length is sqrt(2) *2 Rb and passes through the center of the blue circle and the common tangent between circles. I hope this is correct.
Because of your previous problems; I can solve this!
Thank you!!!
Great sir.you are ane of the best teacher in our society.
Simply explanation. Amazing.
I have basically worked out the same method as you ( I have become more adept from watching your videos !) , I am doing the algebra in the comments section without sketching aids or looking through your video. On checking I do want ( (9 - Rb-4)^2 + (8-4-Rb)^2 = (rb+Rg)^2. Then 25 - 10Rb + Rb^2 + 16 -8Rb + Rb^2 = Rb^2 + 8Rb + 16. This gives 25 -26 Rb = 0 and Rb = 25/26. So the Blue circle radius is 25/26 and area PI * (25/26) ^ 2 =PI * 0.92 and the green circle area is PI * 16. I have a slight difference here. Although I did not use this, The common tangent does not coincide with the diagnal formed from a square around the blue circle so I need to be careful reading off explanation sketches. The square diagnal will be 45 degrees from horizontal and vertical and pass through the center of the blue circle. Thank you again, when i have time I do enjoy working through your premath.
You are a very brilliant mathematician. India
Cool solution! Very interesting)
Thanks! Good refresher!
Great almost had it just didn't see the 5-r, but I'm still pleased that I deducted the rest of it... More practice needed, I often seem to be waiting for the penny to drop, ah well, if you don't fail now and then you never get anywhere, so bring it on !😃
Very well explained 👍
Thanks a lot Ramani. Thank you so much for your continued love and support. Take care dear and stay blessed😃 You are awesome.
احسنتم وبارك الله فيكم وعليكم والله يحفظكم ويرعاكم ويحميكم
Very nice explanation to students.Once we get the sides of ∆ABC take r=1.The sides are 3,4,5 Pythagorean triplet.
Green circle - 16π
Blue circle - π
S=17π≈53,43
After watching and enjoying many of your presentations, I have a suggestion for you. Rather than jumping into the solution, present a strategy first allowing the viewer to pause and try to solve the problem. Thanks for all the fun.
I did green circle R=4 (half the rectangle height) so area of 16pi sq un.
Blue circle was fiddlier. I ended up with joining the circle centres for the hypotenuse with side length 4+r. The other two sides were 4-r and 5-r. (4-r)^2+(5-r)^2=(4+r)^2 produced r=1 with the quadratic formula. so and area of pi sq un
Decimal equivalents approx 50.27 and 3.14 sq un respectively. With the quadratic formula I also had the spurious answer of 25 = r. Unusually, the retained answer was the minus answer - it's usually the other way.
Just checked the video and see you used the same right triangle as me.
Nyc one ... thanks for sharing...
Thank you too.
Take care dear and stay blessed😃
Good teacher.
Thank you so much for explain🙏🌝
Do you keep a book of formula at hand. Can you recommend such a book. Thank you for your uniquely enlightening content.
Good video!
Your voice modulation excites to listen. Goodman. What if phythogeres is born???.
Wouldn’’t the simpler way have been to draw a line tangent to the other side of the green circle which bisects the blue circle, which only leaves 1 unit available(because we know that the green circle has a diameter of 8 units, so 8-9=1, which is the radius of the blue circle. We still get the Area of the green circle as 16pi, and the area of the blue circle as pi. It’s much simpler than mucking about with the pythagorean theorem.
You can't be sure that the tangent bisects the blue circe in 2 identical half.
@@GiuseppeGarufiKeeran you can clearly see that it does, but I guess it's better to be 100% sure and do the math, although I can clearly see the answer staring me in the face.
"Clearly see" Is not how math works. Not at all. You have to demonstate everything
Thank you for sharing
So nice of you Federico! 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😃
with ease I could solve it. But fantastic problem.
Nice! I've solved it in a different way; got that radius of blue circle is about 0,9954 thus his area is a little bit less than π^2... : )
С таким радиусом касательный перпендикуляр к большому кругу должен проходить прямо через центр. Возможно, получилось бы как-то доказать это, не зная самого радиуса.
I'm a 10th student , but also i can understand
Tq pre math
Hi sir, great video, please tell which software you use to create these videos. How do you record the video? Thank you
Sir I am also interested about it.please share how to create this types beautiful videos.
I'm a little confused, is the answer 17π or 16π and π as 2 separate answers, one for each circle. What would a math teacher accept for the answer? I've been out of school for a long time...
Dear friend, you have a very valid question. Well, to be safe, I'd first calculate the areas individually. Then put the answer as a total as well.
As a math professor, I don't penalize my students for these kind of tiny things. Benefit of the doubt always goes to my student.
Bottom line: the answer is supposed to be 17Pi.
Thanks for asking.
You are awesome 👍 Take care dear and stay blessed😃
Thank you for the quick response. My instincts lead me to the final answer 17π. I'm really enjoying your videos.
@@1paultv22 Take care and stay safe my friend. Please keep sharing my channel with your family and friends.
It's ambiguous. I went to 17π, but when he stopped at 16π and π, I went back to look at the wording of the problem. The title says "Calculate the Area of the Green & Blue Circles" but the card inside the video says "Calculate the areas [plural] of Green and Blue circles". So the first is asking for one number (the total), and the other is asking for two numbers. :-)
@@wwoods66 Thanks Bill for your honest feedback. In the future, it'll have more clarity. You are awesome 👍Take care dear and stay blessed😃
It's much easier to just read the radii off the graph and use A = (pi) r^2 directly; trivial even.
If the area of the rectangle is 72 square units (8 x 9), and the area of the large circle is approximately 50.24 square units (16 x 3.14) ...
why can we not find the area of the small circle by simply subtracting the area of the large circle from the area of the rectangle? It seems intuitive at first glance but it doesn't work out on paper. What am I missing?
Because that would also include the white area
#Just #a #big #fat #wow ! Great.👏👏👏👏
diameter of the Circle inscribed in the rectangle = width of the rectangle
So area = π*(width)^2/4
i now this and Fantastic!
My teacher use your math problem but left out your RUclips link channel 😏😏😏😏
i love this
Thank you sir
Thanks a lot Gowri. You are the best!
Take care dear and stay blessed😃
Area of the rectangle is 8x9. 72. Area of the large circle is pi r sq.So pi x 4x4 which is 50.265. The area of the blue circle is rectangle minus large circle. 21.74.
They don't dish out CSE Grade 1s for nothing you know.
If you translate the height of the rectangle and touch the big circle you obtain an 8x8 square. Obviously the radius of the small circle is 9 - 8
That doesn't always work. Just a coincidence here.
Well done
I retract my comment. It seemed "snarky" after I read it. I see now that the real problem is "proving" that r = 1 (it looked obvious from the graph.)
Sir I think area of small triangle is not correct I solve this by alternate method then I get area =21.54 pie
when he wrote rejected i felt that
Thank for permath
The first circle is easy, it's 16 PI units^2, the smaller one takes a little bit of work. We can draw a line from A to B. Then we can draw a perpendicular line that bisects that line that is tangent to both circles... Then we can draw a horizontal line from A to the right until it touches the outer perimeter of the rectangle. We know that A to the tangent is 4 since that's the radius of the larger circle. We know that from the edge of the large circle to the edge of the rectangle is 1 since 9-8 is one because the diameter of the larger circle is 8. Now we can draw a vertical line on the right side of the large circle that is tangent to it. We know that this line is parallel with the outer edge of the rectangle. Now, if and only if this tangent line happens to intersect the center of the smaller circle, we are then done. If it does that means that the smaller circle has a radius of one thus giving us an area of PI units^2. If it doesn't then it will require more geometry with parallel lines, perpendicular bisectors, completing the rectangle or square, possibly using the Pythagorean Theorem or Trigonometric functions, etc... until we have enough information to determine either is radius, or its diameter. Then once we know that, calculating it's area is very trivial.
If we trace a diameter of the small circle paralel to the side of length 8 we find is tangent to the greater circle. Therefore r has to be one et voilá!
Area of Green circle = 16*pi.
Area of Blue Circle = pi. (Form a triangle from centres so that (4 + r)Sq = (4-r)sq + (5-r)sq. Solve for r will give you 1)
keep it up please
Big and little
Large and small
No negative numbers were involved in the solution, which used subtraction.
I have an Rb^2 term missing in my algebra. Putting this in Rb^2 -26 Rb +25 = 0 and (Rb -1) * (Rb-25) = 0 and a Radius of 25 is not possible so Rb = 1 ! We now have the same radii ! Green circle area 16 Pi and the blue circle area Pi.
#Radius
radius of little circle is 1 because length og rektangle is 9 and diameter of big circle is 8 😁😊
Good
#Pythagoras #PythagoreanTheorem
thanks for good pension
Great
Skipped a step there.
You would need to argue that the segment AB passes through the point of tangency of the two circles before you can assume that it is equal to 4 + r.
No need to prove something which is always the case. You can't draw two tangent circles without the radii forming a straight line.
good
Fantastic!
Pi (d/2)^2
Pythagorean Theorem
Radius = 4
Area of the rectangle subtract the area of big circle is equal to the area of the small circle (72 - 16Pi)
A = 16 pi
i now this
R = 4
Pythagoras
To easy
Muy larga explicación. El área del circulo grande es evidente que el radio es 4 ya que si trazamos una paralela al lado vertical tangente al circulo, entonces el radio es igual a la mitad del lado. Y asi la distancia entre esta paralela y el lado verticasl del rectángulo es igual a 9 - 8 que es igual a 1 que es el radio del circulo menor. Por lo tanto las áreas son respectivamente PI x R al cuadrado y PI x r al cuadrado. La explicación que se muestra en el video es extremadamente larga por eso le doy dedo abajo.
The area of the green circle is 4.5*4.4*3.14=65.585 and the area of square is 72 so,the area72-65.585=6,415
Why this channel is called PreMath? What do you mean, before math?!