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Two small semicircles are inscribed in a larger semicircle, and a circle is drawn such that it is internally tangent to all three semicircles, as seen above. If the diameter of the biggest semicircle is 12, find the radius of the small circle.

 

 Dec 16, 2019
 #1
avatar+109742 
+1

If we  connect the centers  ofthe two larger circles.....we can form  the base of  an isoceles triangle that will be  6 units in length

 

Let the radius of the smaller circle = r

 

And the two  sides of this triangle will be segments joining the centers of the larger circle with the center of the smaller circle ....and the length of these sides  = radius of larger circles  + radius of smaller circle=  3 + r

 

And using the  Pythagorean Theorem, we can find the height of this isosceles triangle as

 

√ [ (3 + r)^2  - 3^2]  =   

√ [ 9 + 6r + r^2 - 9] =

√ [ 6r + r^2 ]

 

So....the height of this triangle  plus the radius of the smaller  circle =   radius of the larger circle....so...

 

√[6r + r^2] + r  = 6     subtract  r from both sides

√ [ 6r + r^2]  = r - 6     square both sides

6r + r^2  = r^2 -12r  + 36      

18 r  = 36      divide both sides by 18

r = 2 = radius of smaller circle

 

 

cool cool cool

 Dec 16, 2019
 #2
avatar+24430 
+2

Two small semicircles are inscribed in a larger semicircle, and a circle is drawn such that it is internally tangent to all three semicircles, as seen above.

If the diameter of the biggest semicircle is 12, find the radius of the small circle.

\(\begin{array}{|rcll|} \hline \mathbf{(6-r)^2 +3^2} &=& \mathbf{(r+3)^2} \\ 36-12r+r^2+9 &=& r^2+6r+9 \quad | \quad -(r^2+9)\\ 36-12r &=& 6r\\ 18r &=& 36 \quad | \quad : 18 \\ \mathbf{r} &=&\mathbf{ 2 } \\ \hline \end{array}\)

 

laugh

 Dec 16, 2019

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