Three mutually-tangent circles have centers at given distances from each other; find each radius, and find...
$begingroup$
Three circles of different radii are tangent to each other externally. The distance between their centers are $9 cm$, $8 cm$, and $11 cm$.
Find the radius of each circle.
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
edited Oct 2 '18 at 20:44
Blue
48.5k870154
asked Jan 24 '15 at 12:59
richmondrichmond
312
$endgroup$
add a comment |
$begingroup$
Three circles of different radii are tangent to each other externally. The distance between their centers are $9 cm$, $8 cm$, and $11 cm$.
Find the radius of each circle.
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
edited Oct 2 '18 at 20:44
Blue
48.5k870154
asked Jan 24 '15 at 12:59
richmondrichmond
312
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
Three circles of different radii are tangent to each other externally. The distance between their centers are $9 cm$, $8 cm$, and $11 cm$.
Find the radius of each circle.
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
edited Oct 2 '18 at 20:44
Blue
48.5k870154
asked Jan 24 '15 at 12:59
richmondrichmond
312
$endgroup$
Three circles of different radii are tangent to each other externally. The distance between their centers are $9 cm$, $8 cm$, and $11 cm$.
Find the radius of each circle.
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
edited Oct 2 '18 at 20:44
Blue
48.5k870154
asked Jan 24 '15 at 12:59
richmondrichmond
312
edited Oct 2 '18 at 20:44
Blue
48.5k870154
asked Jan 24 '15 at 12:59
richmondrichmond
312
edited Oct 2 '18 at 20:44
Blue
48.5k870154
edited Oct 2 '18 at 20:44
Blue
48.5k870154
edited Oct 2 '18 at 20:44
Blue
48.5k870154
48.5k870154
asked Jan 24 '15 at 12:59
richmondrichmond
312
asked Jan 24 '15 at 12:59
richmondrichmond
312
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
add a comment |
$begingroup$
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
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
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}
$$
which is $3.05537320587455$.
answered Nov 21 '16 at 6:11
robjohn♦robjohn
268k27308633
$endgroup$
add a comment |
$begingroup$
Mathematica solving for 3 variables.
Clear[a, b, c];
Solve[9 - a == b && 8 - b == c && 11 - c == a, {a, b, c}]
(* {{a -> 6, b -> 3, c -> 5}} *)
answered Sep 9 '16 at 10:57
Fred KlineFred Kline
51921039
$endgroup$
add a comment |
$begingroup$
here is one way to do this:
(a) construct a triangle $ABC$ with sides $8, 9$ and $11.$
(b) find the incenter $I$ of $ABC$
(c) the common value $AI = BI = CI$ is the radius you want.
the same can be done algebraically by finding
(a) two angles using the cosine rule
(b) use herons formula to find the area of $ABC$
(c) in-radius = $dfrac{area}{semi perimeter}$
(d) $AI = dfrac{r}{tan A/2}$
answered Jan 25 '15 at 16:21
abelabel
26.6k12048
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Solving for the radii, we get ${3,5,6}$.
For the area we get
$$
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}
$$
which is $3.05537320587455$.
answered Nov 21 '16 at 6:11
robjohn♦robjohn
268k27308633
$endgroup$
add a comment |
$begingroup$
Solving for the radii, we get ${3,5,6}$.
For the area we get
$$
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}
$$
which is $3.05537320587455$.
answered Nov 21 '16 at 6:11
robjohn♦robjohn
268k27308633
$endgroup$
add a comment |
$begingroup$
Solving for the radii, we get ${3,5,6}$.
For the area we get
$$
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}
$$
which is $3.05537320587455$.
answered Nov 21 '16 at 6:11
robjohn♦robjohn
268k27308633
$endgroup$
Solving for the radii, we get ${3,5,6}$.
For the area we get
$$
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 $3.05537320587455$.
answered Nov 21 '16 at 6:11
robjohn♦robjohn
268k27308633
answered Nov 21 '16 at 6:11
robjohn♦robjohn
268k27308633
answered Nov 21 '16 at 6:11
robjohn♦robjohn
268k27308633
answered Nov 21 '16 at 6:11
robjohn♦robjohn
268k27308633
268k27308633
add a comment |
add a comment |
$begingroup$
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}} *)
answered Sep 9 '16 at 10:57
Fred KlineFred Kline
51921039
$endgroup$
add a comment |
$begingroup$
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}} *)
answered Sep 9 '16 at 10:57
Fred KlineFred Kline
51921039
$endgroup$
add a comment |
$begingroup$
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}} *)
answered Sep 9 '16 at 10:57
Fred KlineFred Kline
51921039
$endgroup$
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}} *)
answered Sep 9 '16 at 10:57
Fred KlineFred Kline
51921039
answered Sep 9 '16 at 10:57
Fred KlineFred Kline
51921039
answered Sep 9 '16 at 10:57
Fred KlineFred Kline
51921039
answered Sep 9 '16 at 10:57
Fred KlineFred Kline
51921039
51921039
add a comment |
add a comment |
$begingroup$
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) $AI = dfrac{r}{tan A/2}$
answered Jan 25 '15 at 16:21
abelabel
26.6k12048
$endgroup$
add a comment |
$begingroup$
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) $AI = dfrac{r}{tan A/2}$
answered Jan 25 '15 at 16:21
abelabel
26.6k12048
$endgroup$
add a comment |
$begingroup$
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) $AI = dfrac{r}{tan A/2}$
answered Jan 25 '15 at 16:21
abelabel
26.6k12048
$endgroup$
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) $AI = dfrac{r}{tan A/2}$
answered Jan 25 '15 at 16:21
abelabel
26.6k12048
answered Jan 25 '15 at 16:21
abelabel
26.6k12048
answered Jan 25 '15 at 16:21
abelabel
26.6k12048
answered Jan 25 '15 at 16:21
abelabel
26.6k12048
26.6k12048
add a comment |
add a comment |
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$begingroup$
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