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Radius of a circle touching a rectangle both of which are inside a square
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Given this configuration :
We're given that the rectangle is of the dimensions 20 cm by 10 cm, and we have to find the radius of the circle.
If we somehow know the distance between the circle and the corner of the square then we can easily find the radius. (It's equal to $ sqrt2times R-R$)
I really can't understand how to solve it. Any help appreciated.
geometry circle rectangles
asked 1 hour ago
Mooncrater
216110
add a comment |Â
up vote
4
down vote
favorite
Given this configuration :
We're given that the rectangle is of the dimensions 20 cm by 10 cm, and we have to find the radius of the circle.
If we somehow know the distance between the circle and the corner of the square then we can easily find the radius. (It's equal to $ sqrt2times R-R$)
I really can't understand how to solve it. Any help appreciated.
geometry circle rectangles
asked 1 hour ago
Mooncrater
216110
This is fascinating, provided it is written correctly. At first it seems there is not enough information, but at the moment, I feel confident that there is. I will mess around with this.
– The Count
1 hour ago
1
The information you have is $R(1-cosalpha) = 2a$ and $R(1-sinalpha) = a$, where $a = 10$. From there it follows that $R^2 = (R-2a)^2+(R-a)^2$.
– amsmath
1 hour ago
1
@amsmath juuuusssssttt beat me to it. Nice!
– The Count
1 hour ago
If you set the upper left corner to be the origin $O(0,0)$, then the center of this circle has coordinates $(R,-R)$, so the circle is given by the equation $(x-R)^2+(y+R)^2=R^2$. Now note that the circle contains the point $(20,-10)$, and calculate $R$ from this.
– SMM
1 hour ago
@SMM I deleted my comment because I saw that you had another approach (which I actually like more).
– amsmath
1 hour ago
add a comment |Â
up vote
4
down vote
favorite
up vote
4
down vote
favorite
Given this configuration :
We're given that the rectangle is of the dimensions 20 cm by 10 cm, and we have to find the radius of the circle.
If we somehow know the distance between the circle and the corner of the square then we can easily find the radius. (It's equal to $ sqrt2times R-R$)
I really can't understand how to solve it. Any help appreciated.
geometry circle rectangles
asked 1 hour ago
Mooncrater
216110
Given this configuration :
We're given that the rectangle is of the dimensions 20 cm by 10 cm, and we have to find the radius of the circle.
If we somehow know the distance between the circle and the corner of the square then we can easily find the radius. (It's equal to $ sqrt2times R-R$)
I really can't understand how to solve it. Any help appreciated.
geometry circle rectangles
geometry circle rectangles
asked 1 hour ago
Mooncrater
216110
asked 1 hour ago
Mooncrater
216110
asked 1 hour ago
Mooncrater
216110
asked 1 hour ago
Mooncrater
216110
asked 1 hour ago
Mooncrater
216110
216110
This is fascinating, provided it is written correctly. At first it seems there is not enough information, but at the moment, I feel confident that there is. I will mess around with this.
– The Count
1 hour ago
1
The information you have is $R(1-cosalpha) = 2a$ and $R(1-sinalpha) = a$, where $a = 10$. From there it follows that $R^2 = (R-2a)^2+(R-a)^2$.
– amsmath
1 hour ago
1
@amsmath juuuusssssttt beat me to it. Nice!
– The Count
1 hour ago
If you set the upper left corner to be the origin $O(0,0)$, then the center of this circle has coordinates $(R,-R)$, so the circle is given by the equation $(x-R)^2+(y+R)^2=R^2$. Now note that the circle contains the point $(20,-10)$, and calculate $R$ from this.
– SMM
1 hour ago
@SMM I deleted my comment because I saw that you had another approach (which I actually like more).
– amsmath
1 hour ago
add a comment |Â
This is fascinating, provided it is written correctly. At first it seems there is not enough information, but at the moment, I feel confident that there is. I will mess around with this.
– The Count
1 hour ago
1
The information you have is $R(1-cosalpha) = 2a$ and $R(1-sinalpha) = a$, where $a = 10$. From there it follows that $R^2 = (R-2a)^2+(R-a)^2$.
– amsmath
1 hour ago
1
@amsmath juuuusssssttt beat me to it. Nice!
– The Count
1 hour ago
If you set the upper left corner to be the origin $O(0,0)$, then the center of this circle has coordinates $(R,-R)$, so the circle is given by the equation $(x-R)^2+(y+R)^2=R^2$. Now note that the circle contains the point $(20,-10)$, and calculate $R$ from this.
– SMM
1 hour ago
@SMM I deleted my comment because I saw that you had another approach (which I actually like more).
– amsmath
1 hour ago
This is fascinating, provided it is written correctly. At first it seems there is not enough information, but at the moment, I feel confident that there is. I will mess around with this.
– The Count
1 hour ago
This is fascinating, provided it is written correctly. At first it seems there is not enough information, but at the moment, I feel confident that there is. I will mess around with this.
– The Count
1 hour ago
1
1
The information you have is $R(1-cosalpha) = 2a$ and $R(1-sinalpha) = a$, where $a = 10$. From there it follows that $R^2 = (R-2a)^2+(R-a)^2$.
– amsmath
1 hour ago
The information you have is $R(1-cosalpha) = 2a$ and $R(1-sinalpha) = a$, where $a = 10$. From there it follows that $R^2 = (R-2a)^2+(R-a)^2$.
– amsmath
1 hour ago
1
1
@amsmath juuuusssssttt beat me to it. Nice!
– The Count
1 hour ago
@amsmath juuuusssssttt beat me to it. Nice!
– The Count
1 hour ago
If you set the upper left corner to be the origin $O(0,0)$, then the center of this circle has coordinates $(R,-R)$, so the circle is given by the equation $(x-R)^2+(y+R)^2=R^2$. Now note that the circle contains the point $(20,-10)$, and calculate $R$ from this.
– SMM
1 hour ago
If you set the upper left corner to be the origin $O(0,0)$, then the center of this circle has coordinates $(R,-R)$, so the circle is given by the equation $(x-R)^2+(y+R)^2=R^2$. Now note that the circle contains the point $(20,-10)$, and calculate $R$ from this.
– SMM
1 hour ago
@SMM I deleted my comment because I saw that you had another approach (which I actually like more).
– amsmath
1 hour ago
@SMM I deleted my comment because I saw that you had another approach (which I actually like more).
– amsmath
1 hour ago
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
3
down vote
It is just using the pythagorean theorem:
answered 59 mins ago
Seyed
5,76231221
add a comment |Â
up vote
1
down vote
Place the center of the cricle at $O.$
Let the radius be $R$
The corner of the square is $(R,R)$ I have made a small alteration to the picture to create fewer negative numbers.
Offsetting by the rectange, the corer of the rectangle is $(R-20, R-10)$
And the distance from this point equals the $R.$
That should put you on your way to the solution.
answered 1 hour ago
Doug M
39.7k31750
add a comment |Â
up vote
1
down vote
$(x, y) = (R-20, R-10)$ as a point on the circle $y = sqrtR^2 - x^2$
$R - 10 = sqrtR^2 - (R-20)^2$
$(R- 10)^2 = R^2 - (R-20)^2$
$R^2 - 20R + 100 = R^2 - (R^2 - 40R + 400)$
$R^2 - 60R + 500 = 0$
$(R - 50)(R-10) = 0$
$R = 50$ is the only sensible option.
answered 1 hour ago
Phil H
2,0722311
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
It is just using the pythagorean theorem:
answered 59 mins ago
Seyed
5,76231221
add a comment |Â
up vote
3
down vote
It is just using the pythagorean theorem:
answered 59 mins ago
Seyed
5,76231221
add a comment |Â
up vote
3
down vote
up vote
3
down vote
It is just using the pythagorean theorem:
answered 59 mins ago
Seyed
5,76231221
It is just using the pythagorean theorem:
answered 59 mins ago
Seyed
5,76231221
edited 20 mins ago
answered 59 mins ago
Seyed
5,76231221
answered 59 mins ago
Seyed
5,76231221
answered 59 mins ago
Seyed
5,76231221
5,76231221
add a comment |Â
add a comment |Â
up vote
1
down vote
Place the center of the cricle at $O.$
Let the radius be $R$
The corner of the square is $(R,R)$ I have made a small alteration to the picture to create fewer negative numbers.
Offsetting by the rectange, the corer of the rectangle is $(R-20, R-10)$
And the distance from this point equals the $R.$
That should put you on your way to the solution.
answered 1 hour ago
Doug M
39.7k31750
add a comment |Â
up vote
1
down vote
Place the center of the cricle at $O.$
Let the radius be $R$
The corner of the square is $(R,R)$ I have made a small alteration to the picture to create fewer negative numbers.
Offsetting by the rectange, the corer of the rectangle is $(R-20, R-10)$
And the distance from this point equals the $R.$
That should put you on your way to the solution.
answered 1 hour ago
Doug M
39.7k31750
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Place the center of the cricle at $O.$
Let the radius be $R$
The corner of the square is $(R,R)$ I have made a small alteration to the picture to create fewer negative numbers.
Offsetting by the rectange, the corer of the rectangle is $(R-20, R-10)$
And the distance from this point equals the $R.$
That should put you on your way to the solution.
answered 1 hour ago
Doug M
39.7k31750
Place the center of the cricle at $O.$
Let the radius be $R$
The corner of the square is $(R,R)$ I have made a small alteration to the picture to create fewer negative numbers.
Offsetting by the rectange, the corer of the rectangle is $(R-20, R-10)$
And the distance from this point equals the $R.$
That should put you on your way to the solution.
answered 1 hour ago
Doug M
39.7k31750
answered 1 hour ago
Doug M
39.7k31750
answered 1 hour ago
Doug M
39.7k31750
answered 1 hour ago
Doug M
39.7k31750
39.7k31750
add a comment |Â
add a comment |Â
up vote
1
down vote
$(x, y) = (R-20, R-10)$ as a point on the circle $y = sqrtR^2 - x^2$
$R - 10 = sqrtR^2 - (R-20)^2$
$(R- 10)^2 = R^2 - (R-20)^2$
$R^2 - 20R + 100 = R^2 - (R^2 - 40R + 400)$
$R^2 - 60R + 500 = 0$
$(R - 50)(R-10) = 0$
$R = 50$ is the only sensible option.
answered 1 hour ago
Phil H
2,0722311
add a comment |Â
up vote
1
down vote
$(x, y) = (R-20, R-10)$ as a point on the circle $y = sqrtR^2 - x^2$
$R - 10 = sqrtR^2 - (R-20)^2$
$(R- 10)^2 = R^2 - (R-20)^2$
$R^2 - 20R + 100 = R^2 - (R^2 - 40R + 400)$
$R^2 - 60R + 500 = 0$
$(R - 50)(R-10) = 0$
$R = 50$ is the only sensible option.
answered 1 hour ago
Phil H
2,0722311
add a comment |Â
up vote
1
down vote
up vote
1
down vote
$(x, y) = (R-20, R-10)$ as a point on the circle $y = sqrtR^2 - x^2$
$R - 10 = sqrtR^2 - (R-20)^2$
$(R- 10)^2 = R^2 - (R-20)^2$
$R^2 - 20R + 100 = R^2 - (R^2 - 40R + 400)$
$R^2 - 60R + 500 = 0$
$(R - 50)(R-10) = 0$
$R = 50$ is the only sensible option.
answered 1 hour ago
Phil H
2,0722311
$(x, y) = (R-20, R-10)$ as a point on the circle $y = sqrtR^2 - x^2$
$R - 10 = sqrtR^2 - (R-20)^2$
$(R- 10)^2 = R^2 - (R-20)^2$
$R^2 - 20R + 100 = R^2 - (R^2 - 40R + 400)$
$R^2 - 60R + 500 = 0$
$(R - 50)(R-10) = 0$
$R = 50$ is the only sensible option.
answered 1 hour ago
Phil H
2,0722311
answered 1 hour ago
Phil H
2,0722311
answered 1 hour ago
Phil H
2,0722311
answered 1 hour ago
Phil H
2,0722311
2,0722311
add a comment |Â
add a comment |Â
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This is fascinating, provided it is written correctly. At first it seems there is not enough information, but at the moment, I feel confident that there is. I will mess around with this.
– The Count
1 hour ago
1
The information you have is $R(1-cosalpha) = 2a$ and $R(1-sinalpha) = a$, where $a = 10$. From there it follows that $R^2 = (R-2a)^2+(R-a)^2$.
– amsmath
1 hour ago
1
@amsmath juuuusssssttt beat me to it. Nice!
– The Count
1 hour ago
If you set the upper left corner to be the origin $O(0,0)$, then the center of this circle has coordinates $(R,-R)$, so the circle is given by the equation $(x-R)^2+(y+R)^2=R^2$. Now note that the circle contains the point $(20,-10)$, and calculate $R$ from this.
– SMM
1 hour ago
@SMM I deleted my comment because I saw that you had another approach (which I actually like more).
– amsmath
1 hour ago