Great, I could do this quite easily, but I know there has to be a mix of easy and difficult to suit differing abilities. I'm pretty much now addicted to your daily problem, it's like my daily brain medication, thanks again 🤓👍🏻
thanku premath, i am used to with your videos and solving them, today i juat saw the thumbnail and calculated w/o pen and paper in just 20 seconds...and verified, i was correct
Pythagoras: sqrt((21+7)^2-(21-7)^2)=sqrt(588) or 24,25 cm. Interestingly, as soon as one realizes that the short side is half of the hypothenuse, one could have immediately derived that the remaining side must be 14*sqrt(3) as we are obviously dealing with a 30-60-90 special right triangle!
There is an alternate technique for the rt∆BEA Instead of using Pythagorean, we will use the side properties for a 30-60-90 right triangle. Notice that AB= 2BE, so this means AE= BE√3. Since BE= 14 this would make AE = 14√3 Thus making this our final answer for x since AE || x (AE parallel to x)
two circles r1 = 7 (C.7 : circle of radius 7) r2 = 21 (C.21 : radius 21) two circles touch along the line beween their centers. the smaller radius of C.7 is 7 units above horizontal baseline of both circles while the other, C.21, is 21. The larger circle's center is, therefore: 21 - 7 = 14 units above horizontal thru C.7's center. The approach is to establish a right triangle who's hypotenuse (label: H) is the line between circle centers and passing thru the point where C.7 and C.21 touch.. H = 21 + 7 H = 28 The remaining sides of the right triangle are V (vertical) and X (horizontal distance between circles' centerlines.) The questions problem asks for the distance X. {Note that the diagram shows the X distance to be where C.7 and C.21 circles touch (tangent) to the baseline. Given, now H = 28 hmmm if X = 24.25 V = 14 H = 28 the angle above horizontal for H thru both centers = ?? Tan(angle) = opposite/adjacent Tan(angle) = 14/24.25 angle = arcTan(14/24.25) = arcTan(0.5773) = 30° so then sin(angle) =? V/H H =? V/(sin(30)) H =? 14/(sin(30)) =? 14/0.5 =? 28 previously H = 28 28 == 28 ✔️✔️✔️
When dealing with the pythagorean theorem you can always try to factor out the gcd of the two known sides, which might reveal a primitive triple that you already know or give you much smaller numbers that are easier to deal with. If the radii is e.g. 20 and 5 you can reduce the sides lengths 25 and 15 by their gcd of 5, to get the simplest primitive triple: 3^2 + x^2 = 5^2. You straight away get that x = 4 and X = 20 With 21 and 7, x won't be an integer, but the calculation is still fairly easy: Q = gcd( 21 + 7 , 21 - 7 ) = gcd( 28 , 14 ) = 14 X = √( (28/Q)^2 - (14/Q)^2 ) * Q = √( 2^2 - 1^2 ) * Q = √( 3 ) * 14 = 14√3 You can get Q by doing the gcd twice, if you feel that it is easier that way: K = gcd( 21, 7 ) = 7 21/K = 3 7/K = 1 3+1 = 4 3-1 = 2 L = gcd( 4, 2 ) = 2 Q = K * L = 7 * 2 = 14 With both 21:7 and 20:5 all calculation can be done almost instantly in your head. But it will not be as easy to get that 1775:639 reduces to an ( 8 , x , 17 ) triple times a GCD of 142, i.e x = 15 and X = 2130.
Got the same answer, but here's a simplification: 28 = (7)(4) and 14 = (7)(2); so 28^2 = (7^2)(4^2) = (7^2)(16); and 14^2 = (7^2)(2^2) = (7^2)(4). So the square of the unknown side is (7^2)(16 -- 4) = (7^2)(12) = (7^2)(4)(3) = (7^2)(2^2)(3) = (14^2)(3); and the unknown side itself is √((14^2)(3)) = 14√(3). Saves us multiplying out all the squares.... Cheers. 🤠
1 minit in the head..14x root of 3 bc. triangel--long side = 28 = center to center smal side center left and horisontal to vertical down from center in big cirkel = rest og R (21-7) =14 and a 90 angel at the lower right. means basic x is 28 x root 3../2 = 14x root 3 14 x 1,73205...any math guy has that number by heart
21 + 7 = 28 hypotenuse 21- 7 =14 base 28^2 - 14^2 = x^2 588 =x^2 sqrt 588= 24.287 Answer Draw a line to attach the center of both circles = 28 the base of the circle is the difference between both radii or 21-7 =14 Use the Pythagorean theorem to get x
Yes! I personally make and design. Then I ask my daughter which design and color stands out... Thanks for your feedback! Cheers! You are awesome. Keep smiling👍 Love and prayers from the USA! 😀
Haven't watched yet. X=24.2487 ( square root of 588 ) Horizontal line from A to Vertical line down from B = right angle triangle. Base (D->C)=x Height is =21-7=14 And Hypotenuse =21+7=28 x^2 + height^2= Hypotenuse^2 So x^2= Hypotenuse^2- height^2 x^2 = 28^2-14^2 x^2 = 784-196 x^2=588 x=square root of 588 aprox 24.2487
Interesting history. Pythagoras wasn't the one and only, but he got the cred. The Sumerians had the 3-4-5, so did the Babylonians, Egyptians, Chinese and Maya. Look it up.
Once I could see that the hypotenuse was the sum of the two radii the missing information was revealed, two out of the three sides of the right triangle were found. The value of the hypotenuse was hiding in plain sight! The rest was easy.
You really weren’t kidding when you said the diagram wasn’t drawn 100% true to the scale. That 14 for the height of the triangle look’s about half of the 7 for the line segment below it. Lol. A nice problem nevertheless.
@@PreMath Sir. Doing my best. 95% accuracy with my solution before watching with ordinary difficulty. With olympiad level I still need to watch before solving. I still need to manage different strategies. Amazing channel to watch.
Looking at the image radius is wrong, your talking nonsense is too complicated and just confusing. A mathematician should solve a problem in the simplest way, and not talk nonsense.
Great, I could do this quite easily, but I know there has to be a mix of easy and difficult to suit differing abilities. I'm pretty much now addicted to your daily problem, it's like my daily brain medication, thanks again 🤓👍🏻
Excellent!
Glad to hear that!
Thanks for your continued love and support!
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It’s like my daily headache but I love it
Generalization: the length of a line segment tangent to two tangent circles of radius _R_ and _r_ is _2√(Rr)._
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thanku premath, i am used to with your videos and solving them, today i juat saw the thumbnail and calculated w/o pen and paper in just 20 seconds...and verified, i was correct
Bravo!
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Pythagoras: sqrt((21+7)^2-(21-7)^2)=sqrt(588) or 24,25 cm. Interestingly, as soon as one realizes that the short side is half of the hypothenuse, one could have immediately derived that the remaining side must be 14*sqrt(3) as we are obviously dealing with a 30-60-90 special right triangle!
I'm a bit lacking at geometry but this was a pretty easy one for me. Just need to find out how to use the a²+b²=c² and you got it solved
There is an alternate technique for the rt∆BEA
Instead of using Pythagorean, we will use the side properties for a 30-60-90 right triangle.
Notice that AB= 2BE, so this means AE= BE√3.
Since BE= 14 this would make AE = 14√3 Thus making this our final answer for x since AE || x (AE parallel to x)
There is no Special triangle at all
Thanks for video.Good luck sir!!!!!!!!!
two circles
r1 = 7 (C.7 : circle of radius 7)
r2 = 21 (C.21 : radius 21)
two circles touch along the line beween their centers.
the smaller radius of C.7 is 7 units above horizontal baseline of both circles while the other, C.21, is 21.
The larger circle's center is, therefore:
21 - 7 = 14 units above horizontal thru C.7's center.
The approach is to establish a right triangle who's hypotenuse (label: H) is the line between circle centers and passing thru the point where C.7 and C.21 touch..
H = 21 + 7
H = 28
The remaining sides of the right triangle are V (vertical) and X (horizontal distance between circles' centerlines.)
The questions problem asks for the distance X. {Note that the diagram shows the X distance to be where C.7 and C.21 circles touch (tangent) to the baseline.
Given, now
H = 28
hmmm
if X = 24.25
V = 14
H = 28
the angle above horizontal for H thru both centers = ??
Tan(angle) =
opposite/adjacent
Tan(angle) = 14/24.25
angle = arcTan(14/24.25)
= arcTan(0.5773)
= 30°
so then
sin(angle) =? V/H
H =? V/(sin(30))
H =? 14/(sin(30))
=? 14/0.5
=? 28
previously H = 28
28 == 28 ✔️✔️✔️
When dealing with the pythagorean theorem you can always try to factor out the gcd of the two known sides, which might reveal a primitive triple that you already know or give you much smaller numbers that are easier to deal with.
If the radii is e.g. 20 and 5 you can reduce the sides lengths 25 and 15 by their gcd of 5, to get the simplest primitive triple: 3^2 + x^2 = 5^2.
You straight away get that x = 4 and X = 20
With 21 and 7, x won't be an integer, but the calculation is still fairly easy:
Q = gcd( 21 + 7 , 21 - 7 )
= gcd( 28 , 14 )
= 14
X = √( (28/Q)^2 - (14/Q)^2 ) * Q
= √( 2^2 - 1^2 ) * Q
= √( 3 ) * 14
= 14√3
You can get Q by doing the gcd twice, if you feel that it is easier that way:
K = gcd( 21, 7 ) = 7
21/K = 3
7/K = 1
3+1 = 4
3-1 = 2
L = gcd( 4, 2 ) = 2
Q = K * L = 7 * 2 = 14
With both 21:7 and 20:5 all calculation can be done almost instantly in your head.
But it will not be as easy to get that 1775:639 reduces to an ( 8 , x , 17 ) triple times a GCD of 142, i.e x = 15 and X = 2130.
Many thanks. You made it sound quite logical even to a dummy like me.
Nice...you went easy on us this time.
😮extraordinario profesor, teoremas geométricos combinados con álgebra y con el señor pitágora 😮
Nice, I was able to do this one mentally!
∆ABE is a 30-60-90 triangle because hyp => 2(adj) = 28 => adj = 14
adj = 1
opp = (adj)√3
hyp = 2(adj)
adj = BE = 14
opp = x = AE = 14√3 ≈ 24.24871130596
hyp = AB = 28
Got the same answer, but here's a simplification: 28 = (7)(4) and 14 = (7)(2); so 28^2 = (7^2)(4^2) = (7^2)(16); and 14^2 = (7^2)(2^2) = (7^2)(4).
So the square of the unknown side is (7^2)(16 -- 4) = (7^2)(12) = (7^2)(4)(3) = (7^2)(2^2)(3) = (14^2)(3); and the unknown side itself is √((14^2)(3)) = 14√(3).
Saves us multiplying out all the squares....
Cheers. 🤠
Very well explained👍👍😊
i got this one!
1 minit in the head..14x root of 3
bc. triangel--long side = 28 = center to center
smal side center left and horisontal to vertical down from center in big cirkel = rest og R (21-7) =14
and a 90 angel at the lower right.
means basic x is 28 x root 3../2 = 14x root 3
14 x 1,73205...any math guy has that number by heart
This triangle is half of a equilateral triangle.
So you can use h=(a*√3)/2 with a=28
h=14*√3
h=24,2487 units
LG Gerald
21 + 7 = 28 hypotenuse
21- 7 =14 base
28^2 - 14^2 = x^2
588 =x^2
sqrt 588= 24.287 Answer
Draw a line to attach the center of both circles = 28
the base of the circle is the difference between both radii or 21-7 =14
Use the Pythagorean theorem to get x
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Uh, it's so easy.
Using the Pitagoras theorem:
AB²=(21-7)²+x²
(21+7)²=(21-7)²+x²
28²=14²+x²
(2²•14²)-14²=x²
14²(2²-1)=x²
14²•3=x²
196•3=x²
588=x²
x=√(2²•3•7²)
x=14√3
Thanks for a fun Sunday night problem!❤🥂👍🍺
You are very welcome!
Thanks for your feedback! Cheers!
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thanks so much! very nice problems sir, really smart stuff, do you make these problems yourself?
Yes! I personally make and design. Then I ask my daughter which design and color stands out...
Thanks for your feedback! Cheers!
You are awesome. Keep smiling👍
Love and prayers from the USA! 😀
And in this case, since EB=1/2 AB, ABE is one half of an equilateral triangle with the side of 28 and AE is it's height, so AE=(AB/2)*SQRT(3).
x=14√3≈24,22
I found that scaling down all the lengths by 7 made the math much easier.
I just had to multiply the result by 7 at the end to get x=7×2×sqrt(3)
Since the sin leg is exactly half the hypotenuse, the angle BAE must be 30 deg. Therefore AE = cos 30 deg × 28 = 24.25.
After having got the triangle, divide by 14 to have small numbers, then multiply the result by 14.
très belle démonstration :))
Haven't watched yet. X=24.2487 ( square root of 588 )
Horizontal line from A to Vertical line down from B = right angle triangle.
Base (D->C)=x
Height is =21-7=14
And Hypotenuse =21+7=28
x^2 + height^2= Hypotenuse^2
So
x^2= Hypotenuse^2- height^2
x^2 = 28^2-14^2
x^2 = 784-196
x^2=588
x=square root of 588 aprox 24.2487
X es la altura de un triangulo equilatero de lado (7+21)=28
Gracias y un saludo.
Yay! I solved it, 14 * [sq rt (3)].
I didn't spot 14root3 but calculated a decent approximation from 2root147 instead because root144 is 12 :) . Same near-answer though.
Interesting history.
Pythagoras wasn't the one and only, but he got the cred.
The Sumerians had the 3-4-5, so did the Babylonians, Egyptians, Chinese and Maya. Look it up.
Yes, we shouldn't think the drawing is to scale, for the smaller circle was MUCH larger than one-third the size of the larger, distractingly so.
x=square root of 28^2-14^2=14 root 3=24.25 approximately. 🙂
Excellent!
Thanks for sharing! Cheers!
You are awesome. Keep it up 👍
Love and prayers from the USA! 😀
Once I could see that the hypotenuse was the sum of the two radii the missing information was revealed, two out of the three sides of the right triangle were found. The value of the hypotenuse was hiding in plain sight! The rest was easy.
Sir, the bigger circle should have displayed much bigger in size. Currently, the length of 21 units appears smaller than the length measuring 7 units.
I believe it was not drawn to scale on purpose so that you have to use math and not just measure.
斜辺が4で高さが2の直角三角形の底辺の長さはピタゴラスの定理で一撃だが、
そんな事よりも図にあまりにもリニアリティーが無いのが気になって夜も眠れない。
x = 14/tan(30) from trig.
Sqrt(28^2 - 14^2) = 24.25
24.25
Its simple: X = Sqrt((21+7)^2 - (21-7)^2)
X=14×3^(1/2).
The theory and execution is correct but the drawing is not to scale which is why it is bad example. the 21 r circle should have been much larger.
I solved it and it had been 30.8!
where was it wrong?
원과 길이의 비례는 신경쓰지 않아도 좋아요
x=14sqrt(3)
24.2487
Excellent!
Thanks for sharing! Cheers!
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Pythagoras!
14
Noting the common factor of 7 the answer is 14xsqroot3
x^2+(21-7)^2=(21+7)^2..x=14sqrt3
Excellent!
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You really weren’t kidding when you said the diagram wasn’t drawn 100% true to the scale. That 14 for the height of the triangle look’s about half of the 7 for the line segment below it. Lol. A nice problem nevertheless.
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so easy
28
First view sir
Thank you! Cheers!
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This is a strange perspective, but CE = 7 looks longer than EB = 14. House of Illusion Houdini Attic Phenomenon. 🙂
Disclaimer: Not drawn to the scale!
use sin and cos und you get it within 2 minutes, without knowing about binominal formulas !
Surely it would be exactly 14.
7 + 21 =28
14v3=24.25
X=28 ?
和算で有名な公式が使えるねぇ～
ピタゴラスの方が好き
This one too easy
Excellent!
You are awesome. Keep it up 👍
Love and prayers from the USA! 😀
@@PreMath Sir. Doing my best. 95% accuracy with my solution before watching with ordinary difficulty.
With olympiad level I still need to watch before solving. I still need to manage different strategies. Amazing channel to watch.
@@grandemika Thanks Marco. You are the best!
24.245 unless my covid addled brain is completely unusable
Sir, you spent 8 minutes solving something that shouldn’t take 2 minutes
Gecho meme
Easy cake
Your "not to scale" diagram is disproportionately misleading !
детская задача
This actually wasn't that hard.
too greatly out of scale.
Looking at the image radius is wrong, your talking nonsense is too complicated and just confusing. A mathematician should solve a problem in the simplest way, and not talk nonsense.