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Circles M & N (centered at points M and N respectively) are tangent to each other and to AC & BC. If the radius of the circle with center N is 5/(1+sqrt2) m, what is the radius of M, in meters?

 

 Feb 14, 2020
 #1
avatar+109561 
+2

Let the radius of M  =   m

 

We can from a right triangle

 

Two  legs = (m - 5/ [ 1 + sqrt (2) ]

 

And the  hypotenuse =  m + 5 / [ 1 + sqrt (2) ]

 

So.....by the pythagorean Theorem we have that

 

(m - 5/ [ 1 + sqrt (2) ] )^2   +  ( m - 5/ [1 + sqrt (2) ] )^2   = ( m  + 5 / [1 + sqrt (2) ] )^2

 

Solving this  for the larger value of m  we get that   m   =  5 ( 1 + sqrt (2) ) ≈  12.07 m

 

 

 

cool cool cool

 Feb 14, 2020
 #2
avatar+531 
+1

NO = 5/(1+sqrt(2)) = 2.071068294

MO = r / sin(45) = 2.92893287

MN = MO - NO =0.857864576

Proportion:           MN  :  NO  =  MT  :  TR

Radius of circle "M" is:      TR = 12.071068 m   indecision

 Feb 16, 2020
edited by Dragan  Feb 16, 2020
edited by Dragan  Feb 17, 2020
 #3
avatar+108725 
+1

I like your drawings Dragan    wink

Melody  Feb 16, 2020
 #4
avatar+531 
+1

Thanks, Melody! That's how we did 50+ years ago.  laugh

Dragan  Feb 16, 2020
edited by Dragan  Feb 16, 2020
edited by Dragan  Feb 16, 2020
 #5
avatar+108725 
+1

Yes, I know   wink

 

Still Geogebra is really fun to use. It is a free download. 

 

Maybe you would like to try it sometime..      laugh

 

I like your neat hand-drawn pics though.     cool

Melody  Feb 16, 2020
 #6
avatar+108725 
+1

Hey Dragan,

 

Why is your pic no longer displaying???   sad

Melody  Feb 16, 2020
edited by Melody  Feb 16, 2020
 #7
avatar+531 
+1

Thanks for GeoGebra!  smiley

Dragan  Feb 18, 2020
edited by Dragan  Feb 18, 2020
 #8
avatar+108725 
0

I found GeoGebra a little hard at first.

But persist because you will get the hang of it and it is a lot of fun to use.

Melody  Feb 18, 2020

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