Calculate the length X between two tangent circles | Important Geometry skills explained

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Calculate the length X between two tangent circles | Important Geometry skills explained

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[theoyanto]

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 🤓👍🏻

[ybodoN]

Generalization: the length of a line segment tangent to two tangent circles of radius _R_ and _r_ is _2√(Rr)._

[philipkudrna]

Pythagoras: 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!

[VishalGupta-ohmb]

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

[albertofernandez]

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

[gdmathguy]

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

[santiagoarosam]

X es la altura de un triangulo equilatero de lado (7+21)=28
Gracias y un saludo.

[alster]

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)

[williamwingo]

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. 🤠

[alanhilder]

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

[Kame]

∆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

[tomtke]

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 <triangle hypotenuse)
V = 14 <triangle vertical side>

therefore:
classic: a^2 + b^2 = c^2
for a right triangle

so...
X = sqrt(H^2 - V^2)
= sqrt(28^2 - 14^2)
= sqrt(784 - 196)
= sqrt(588)
= 24.25 units

Verify -->
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 ✔️✔️✔️

[carstenlarsen]

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

[calvinmasters]

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.

[tipeon]

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)

[MrPaulc]

I didn't spot 14root3 but calculated a decent approximation from 2root147 instead because root144 is 12 :) . Same near-answer though.

[RobG]

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.

[BasuraMethmina]

This is my first comment sir but i have watched your many videos. Your geometry videos are very interesting. Please bring videos to us everyday❤

[yoops]

After having got the triangle, divide by 14 to have small numbers, then multiply the result by 14.

[aymanabdellatief]

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.