Three mutually-tangent circles have centers at given distances from each other; find each radius, and find...












6












$begingroup$


Three circles of different radii are tangent to each other externally. The distance between their centers are $9 cm$, $8 cm$, and $11 cm$.




  1. Find the radius of each circle.


  2. Find the area in between the three circles.




The distance between the centers of two tangent circles should be the sum of their radii. If we call the radii $r_1,r_2,r_3$ then we get a system of equations $$begin{align}r_1+r_2&=9\r_2+r_3&=8\r_3+r_1&=11end{align}$$ From this we can find $r_1,r_2,r_3$.



Let us call $O_1,O_2,O_3$ the centers of the circles. For the area between the centers we could compute it as the difference of the area of the triangle $O_1O_2O_3$ minus the three circular sectors that the circles cover. We have the area of the triangle because we have its sides $9,8,11$ and we could use Heron's formula with them. For the circular sectors we know their radii but we also need their amplitude. Ah, but the amplitude is the same as the angles of the triangle. I guess we can compute the angles, because we have the sides of the triangle and we can use The law of cosines to solve for the angles.




Is this correct?



Would I have to use the $arccos(x)$ function to get the final answer or there is a simpler way?






euclidean-geometry







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$endgroup$












  • $begingroup$
    What did you tried so far ?
    $endgroup$
    – servabat
    Jan 24 '15 at 13:01








  • 9




    $begingroup$
    The first time I read the title I thought it said "Solid menstruation (cycles)".
    $endgroup$
    – Pp..
    Jan 25 '15 at 14:16










  • $begingroup$
    $(r_1,r_2,r_3)=(6,3,5)$, by the way.
    $endgroup$
    – Akiva Weinberger
    Nov 3 '15 at 17:05
















6












$begingroup$


Three circles of different radii are tangent to each other externally. The distance between their centers are $9 cm$, $8 cm$, and $11 cm$.




  1. Find the radius of each circle.


  2. Find the area in between the three circles.




The distance between the centers of two tangent circles should be the sum of their radii. If we call the radii $r_1,r_2,r_3$ then we get a system of equations $$begin{align}r_1+r_2&=9\r_2+r_3&=8\r_3+r_1&=11end{align}$$ From this we can find $r_1,r_2,r_3$.



Let us call $O_1,O_2,O_3$ the centers of the circles. For the area between the centers we could compute it as the difference of the area of the triangle $O_1O_2O_3$ minus the three circular sectors that the circles cover. We have the area of the triangle because we have its sides $9,8,11$ and we could use Heron's formula with them. For the circular sectors we know their radii but we also need their amplitude. Ah, but the amplitude is the same as the angles of the triangle. I guess we can compute the angles, because we have the sides of the triangle and we can use The law of cosines to solve for the angles.




Is this correct?



Would I have to use the $arccos(x)$ function to get the final answer or there is a simpler way?






euclidean-geometry







share|cite|improve this question











$endgroup$












  • $begingroup$
    What did you tried so far ?
    $endgroup$
    – servabat
    Jan 24 '15 at 13:01








  • 9




    $begingroup$
    The first time I read the title I thought it said "Solid menstruation (cycles)".
    $endgroup$
    – Pp..
    Jan 25 '15 at 14:16










  • $begingroup$
    $(r_1,r_2,r_3)=(6,3,5)$, by the way.
    $endgroup$
    – Akiva Weinberger
    Nov 3 '15 at 17:05














6












6








6


1



$begingroup$


Three circles of different radii are tangent to each other externally. The distance between their centers are $9 cm$, $8 cm$, and $11 cm$.




  1. Find the radius of each circle.


  2. Find the area in between the three circles.




The distance between the centers of two tangent circles should be the sum of their radii. If we call the radii $r_1,r_2,r_3$ then we get a system of equations $$begin{align}r_1+r_2&=9\r_2+r_3&=8\r_3+r_1&=11end{align}$$ From this we can find $r_1,r_2,r_3$.



Let us call $O_1,O_2,O_3$ the centers of the circles. For the area between the centers we could compute it as the difference of the area of the triangle $O_1O_2O_3$ minus the three circular sectors that the circles cover. We have the area of the triangle because we have its sides $9,8,11$ and we could use Heron's formula with them. For the circular sectors we know their radii but we also need their amplitude. Ah, but the amplitude is the same as the angles of the triangle. I guess we can compute the angles, because we have the sides of the triangle and we can use The law of cosines to solve for the angles.




Is this correct?



Would I have to use the $arccos(x)$ function to get the final answer or there is a simpler way?






euclidean-geometry







share|cite|improve this question











$endgroup$




Three circles of different radii are tangent to each other externally. The distance between their centers are $9 cm$, $8 cm$, and $11 cm$.




  1. Find the radius of each circle.


  2. Find the area in between the three circles.




The distance between the centers of two tangent circles should be the sum of their radii. If we call the radii $r_1,r_2,r_3$ then we get a system of equations $$begin{align}r_1+r_2&=9\r_2+r_3&=8\r_3+r_1&=11end{align}$$ From this we can find $r_1,r_2,r_3$.



Let us call $O_1,O_2,O_3$ the centers of the circles. For the area between the centers we could compute it as the difference of the area of the triangle $O_1O_2O_3$ minus the three circular sectors that the circles cover. We have the area of the triangle because we have its sides $9,8,11$ and we could use Heron's formula with them. For the circular sectors we know their radii but we also need their amplitude. Ah, but the amplitude is the same as the angles of the triangle. I guess we can compute the angles, because we have the sides of the triangle and we can use The law of cosines to solve for the angles.




Is this correct?



Would I have to use the $arccos(x)$ function to get the final answer or there is a simpler way?






euclidean-geometry




euclidean-geometry






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Oct 2 '18 at 20:44









Blue

49.7k870158




49.7k870158










asked Jan 24 '15 at 12:59









richmondrichmond

312




312












  • $begingroup$
    What did you tried so far ?
    $endgroup$
    – servabat
    Jan 24 '15 at 13:01








  • 9




    $begingroup$
    The first time I read the title I thought it said "Solid menstruation (cycles)".
    $endgroup$
    – Pp..
    Jan 25 '15 at 14:16










  • $begingroup$
    $(r_1,r_2,r_3)=(6,3,5)$, by the way.
    $endgroup$
    – Akiva Weinberger
    Nov 3 '15 at 17:05


















  • $begingroup$
    What did you tried so far ?
    $endgroup$
    – servabat
    Jan 24 '15 at 13:01








  • 9




    $begingroup$
    The first time I read the title I thought it said "Solid menstruation (cycles)".
    $endgroup$
    – Pp..
    Jan 25 '15 at 14:16










  • $begingroup$
    $(r_1,r_2,r_3)=(6,3,5)$, by the way.
    $endgroup$
    – Akiva Weinberger
    Nov 3 '15 at 17:05
















$begingroup$
What did you tried so far ?
$endgroup$
– servabat
Jan 24 '15 at 13:01






$begingroup$
What did you tried so far ?
$endgroup$
– servabat
Jan 24 '15 at 13:01






9




9




$begingroup$
The first time I read the title I thought it said "Solid menstruation (cycles)".
$endgroup$
– Pp..
Jan 25 '15 at 14:16




$begingroup$
The first time I read the title I thought it said "Solid menstruation (cycles)".
$endgroup$
– Pp..
Jan 25 '15 at 14:16












$begingroup$
$(r_1,r_2,r_3)=(6,3,5)$, by the way.
$endgroup$
– Akiva Weinberger
Nov 3 '15 at 17:05




$begingroup$
$(r_1,r_2,r_3)=(6,3,5)$, by the way.
$endgroup$
– Akiva Weinberger
Nov 3 '15 at 17:05










3 Answers
3






active

oldest

votes


















2












$begingroup$

Solving for the radii, we get ${3,5,6}$.



For the area we get
$$
begin{align}
&sqrt{14(14-8)(14-9)(14-11)}\[9pt]
&-frac{(8+9-11)^2}8arccosleft(frac{8^2+9^2-11^2}{2cdot8cdot9}right)\
&-frac{(11+8-9)^2}8arccosleft(frac{11^2+8^2-9^2}{2cdot11cdot8}right)\
&-frac{(9+11-8)^2}8arccosleft(frac{9^2+11^2-8^2}{2cdot9cdot11}right)
end{align}
$$
which is $3.05537320587455$.



enter image description here






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    Mathematica solving for 3 variables.



    Clear[a, b, c];
    
Solve[9 - a == b && 8 - b == c && 11 - c == a, {a, b, c}]
    

    
(* {{a -> 6, b -> 3, c -> 5}} *)





    share|cite|improve this answer









    $endgroup$





















      0












      $begingroup$

      here is one way to do this:



      (a) construct a triangle $ABC$ with sides $8, 9$ and $11.$



      (b) find the incenter $I$ of $ABC$



      (c) the common value $AI = BI = CI$ is the radius you want.



      the same can be done algebraically by finding



      (a) two angles using the cosine rule



      (b) use herons formula to find the area of $ABC$



      (c) in-radius = $dfrac{area}{semi perimeter}$



      (d) $AI = dfrac{r}{tan A/2}$






      share|cite|improve this answer









      $endgroup$














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        3 Answers
        3






        active

        oldest

        votes








        3 Answers
        3






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        2












        $begingroup$

        Solving for the radii, we get ${3,5,6}$.



        For the area we get
        $$
        begin{align}
        &sqrt{14(14-8)(14-9)(14-11)}\[9pt]
        &-frac{(8+9-11)^2}8arccosleft(frac{8^2+9^2-11^2}{2cdot8cdot9}right)\
        &-frac{(11+8-9)^2}8arccosleft(frac{11^2+8^2-9^2}{2cdot11cdot8}right)\
        &-frac{(9+11-8)^2}8arccosleft(frac{9^2+11^2-8^2}{2cdot9cdot11}right)
        end{align}
        $$
        which is $3.05537320587455$.



        enter image description here






        share|cite|improve this answer









        $endgroup$


















          2












          $begingroup$

          Solving for the radii, we get ${3,5,6}$.



          For the area we get
          $$
          begin{align}
          &sqrt{14(14-8)(14-9)(14-11)}\[9pt]
          &-frac{(8+9-11)^2}8arccosleft(frac{8^2+9^2-11^2}{2cdot8cdot9}right)\
          &-frac{(11+8-9)^2}8arccosleft(frac{11^2+8^2-9^2}{2cdot11cdot8}right)\
          &-frac{(9+11-8)^2}8arccosleft(frac{9^2+11^2-8^2}{2cdot9cdot11}right)
          end{align}
          $$
          which is $3.05537320587455$.



          enter image description here






          share|cite|improve this answer









          $endgroup$
















            2












            2








            2





            $begingroup$

            Solving for the radii, we get ${3,5,6}$.



            For the area we get
            $$
            begin{align}
            &sqrt{14(14-8)(14-9)(14-11)}\[9pt]
            &-frac{(8+9-11)^2}8arccosleft(frac{8^2+9^2-11^2}{2cdot8cdot9}right)\
            &-frac{(11+8-9)^2}8arccosleft(frac{11^2+8^2-9^2}{2cdot11cdot8}right)\
            &-frac{(9+11-8)^2}8arccosleft(frac{9^2+11^2-8^2}{2cdot9cdot11}right)
            end{align}
            $$
            which is $3.05537320587455$.



            enter image description here






            share|cite|improve this answer









            $endgroup$



            Solving for the radii, we get ${3,5,6}$.



            For the area we get
            $$
            begin{align}
            &sqrt{14(14-8)(14-9)(14-11)}\[9pt]
            &-frac{(8+9-11)^2}8arccosleft(frac{8^2+9^2-11^2}{2cdot8cdot9}right)\
            &-frac{(11+8-9)^2}8arccosleft(frac{11^2+8^2-9^2}{2cdot11cdot8}right)\
            &-frac{(9+11-8)^2}8arccosleft(frac{9^2+11^2-8^2}{2cdot9cdot11}right)
            end{align}
            $$
            which is $3.05537320587455$.



            enter image description here







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Nov 21 '16 at 6:11









            robjohn♦robjohn

            271k27316643




            271k27316643























                1












                $begingroup$

                Mathematica solving for 3 variables.



                Clear[a, b, c];
                
Solve[9 - a == b && 8 - b == c && 11 - c == a, {a, b, c}]
                

                
(* {{a -> 6, b -> 3, c -> 5}} *)





                share|cite|improve this answer









                $endgroup$


















                  1












                  $begingroup$

                  Mathematica solving for 3 variables.



                  Clear[a, b, c];
                  
Solve[9 - a == b && 8 - b == c && 11 - c == a, {a, b, c}]
                  

                  
(* {{a -> 6, b -> 3, c -> 5}} *)





                  share|cite|improve this answer









                  $endgroup$
















                    1












                    1








                    1





                    $begingroup$

                    Mathematica solving for 3 variables.



                    Clear[a, b, c];
                    
Solve[9 - a == b && 8 - b == c && 11 - c == a, {a, b, c}]
                    

                    
(* {{a -> 6, b -> 3, c -> 5}} *)





                    share|cite|improve this answer









                    $endgroup$



                    Mathematica solving for 3 variables.



                    Clear[a, b, c];
                    
Solve[9 - a == b && 8 - b == c && 11 - c == a, {a, b, c}]
                    

                    
(* {{a -> 6, b -> 3, c -> 5}} *)






                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Sep 9 '16 at 10:57









                    Fred KlineFred Kline

                    51721040




                    51721040























                        0












                        $begingroup$

                        here is one way to do this:



                        (a) construct a triangle $ABC$ with sides $8, 9$ and $11.$



                        (b) find the incenter $I$ of $ABC$



                        (c) the common value $AI = BI = CI$ is the radius you want.



                        the same can be done algebraically by finding



                        (a) two angles using the cosine rule



                        (b) use herons formula to find the area of $ABC$



                        (c) in-radius = $dfrac{area}{semi perimeter}$



                        (d) $AI = dfrac{r}{tan A/2}$






                        share|cite|improve this answer









                        $endgroup$


















                          0












                          $begingroup$

                          here is one way to do this:



                          (a) construct a triangle $ABC$ with sides $8, 9$ and $11.$



                          (b) find the incenter $I$ of $ABC$



                          (c) the common value $AI = BI = CI$ is the radius you want.



                          the same can be done algebraically by finding



                          (a) two angles using the cosine rule



                          (b) use herons formula to find the area of $ABC$



                          (c) in-radius = $dfrac{area}{semi perimeter}$



                          (d) $AI = dfrac{r}{tan A/2}$






                          share|cite|improve this answer









                          $endgroup$
















                            0












                            0








                            0





                            $begingroup$

                            here is one way to do this:



                            (a) construct a triangle $ABC$ with sides $8, 9$ and $11.$



                            (b) find the incenter $I$ of $ABC$



                            (c) the common value $AI = BI = CI$ is the radius you want.



                            the same can be done algebraically by finding



                            (a) two angles using the cosine rule



                            (b) use herons formula to find the area of $ABC$



                            (c) in-radius = $dfrac{area}{semi perimeter}$



                            (d) $AI = dfrac{r}{tan A/2}$






                            share|cite|improve this answer









                            $endgroup$



                            here is one way to do this:



                            (a) construct a triangle $ABC$ with sides $8, 9$ and $11.$



                            (b) find the incenter $I$ of $ABC$



                            (c) the common value $AI = BI = CI$ is the radius you want.



                            the same can be done algebraically by finding



                            (a) two angles using the cosine rule



                            (b) use herons formula to find the area of $ABC$



                            (c) in-radius = $dfrac{area}{semi perimeter}$



                            (d) $AI = dfrac{r}{tan A/2}$







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Jan 25 '15 at 16:21









                            abelabel

                            26.6k12148




                            26.6k12148






























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