Finding the radius of the smallest circle that can circumscribe an equilateral triangle
Clash Royale CLAN TAG#URR8PPP
up vote
0
down vote
favorite
Q:A puzzle board is in the form of an equilateral triangle that has an area of $7sqrt3$ if the board is placed on a circular table, what should be the min area of the table so that the whole board fits inside the table.
A: $frac883$
I get that the side of the triangle is $2sqrt7$ and also that in an equilateral triangle the median, perpendicular bisector, altitude and angle bisector are the same. I'm however still stuck with how to get the radius without resorting to sin/cos etc. Please help!
P.S.:Please dont downvote the question, else i'm gonna have restrictions placed on my profile :(
geometry circle triangle
asked Oct 2 '15 at 14:40
bia
498
add a comment |Â
up vote
0
down vote
favorite
Q:A puzzle board is in the form of an equilateral triangle that has an area of $7sqrt3$ if the board is placed on a circular table, what should be the min area of the table so that the whole board fits inside the table.
A: $frac883$
I get that the side of the triangle is $2sqrt7$ and also that in an equilateral triangle the median, perpendicular bisector, altitude and angle bisector are the same. I'm however still stuck with how to get the radius without resorting to sin/cos etc. Please help!
P.S.:Please dont downvote the question, else i'm gonna have restrictions placed on my profile :(
geometry circle triangle
asked Oct 2 '15 at 14:40
bia
498
Hint: the center of the equilateral triangle is at a special place regarding the median of the sides, in term of length...
– Martigan
Oct 2 '15 at 14:43
To a given equilateral triangle, there is one and only one circle that circumscribes it. Therefore, there does not exist the "smallest" circle.
– Mick
Oct 2 '15 at 16:13
The area you’ve got looks like an approximation. It seems like there should be a factor of $pi$ in there somewhere.
– amd
Oct 2 '15 at 20:33
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Q:A puzzle board is in the form of an equilateral triangle that has an area of $7sqrt3$ if the board is placed on a circular table, what should be the min area of the table so that the whole board fits inside the table.
A: $frac883$
I get that the side of the triangle is $2sqrt7$ and also that in an equilateral triangle the median, perpendicular bisector, altitude and angle bisector are the same. I'm however still stuck with how to get the radius without resorting to sin/cos etc. Please help!
P.S.:Please dont downvote the question, else i'm gonna have restrictions placed on my profile :(
geometry circle triangle
asked Oct 2 '15 at 14:40
bia
498
Q:A puzzle board is in the form of an equilateral triangle that has an area of $7sqrt3$ if the board is placed on a circular table, what should be the min area of the table so that the whole board fits inside the table.
A: $frac883$
I get that the side of the triangle is $2sqrt7$ and also that in an equilateral triangle the median, perpendicular bisector, altitude and angle bisector are the same. I'm however still stuck with how to get the radius without resorting to sin/cos etc. Please help!
P.S.:Please dont downvote the question, else i'm gonna have restrictions placed on my profile :(
geometry circle triangle
geometry circle triangle
asked Oct 2 '15 at 14:40
bia
498
asked Oct 2 '15 at 14:40
bia
498
edited Oct 2 '15 at 14:45
asked Oct 2 '15 at 14:40
bia
498
asked Oct 2 '15 at 14:40
bia
498
asked Oct 2 '15 at 14:40
bia
498
498
Hint: the center of the equilateral triangle is at a special place regarding the median of the sides, in term of length...
– Martigan
Oct 2 '15 at 14:43
To a given equilateral triangle, there is one and only one circle that circumscribes it. Therefore, there does not exist the "smallest" circle.
– Mick
Oct 2 '15 at 16:13
The area you’ve got looks like an approximation. It seems like there should be a factor of $pi$ in there somewhere.
– amd
Oct 2 '15 at 20:33
add a comment |Â
Hint: the center of the equilateral triangle is at a special place regarding the median of the sides, in term of length...
– Martigan
Oct 2 '15 at 14:43
To a given equilateral triangle, there is one and only one circle that circumscribes it. Therefore, there does not exist the "smallest" circle.
– Mick
Oct 2 '15 at 16:13
The area you’ve got looks like an approximation. It seems like there should be a factor of $pi$ in there somewhere.
– amd
Oct 2 '15 at 20:33
Hint: the center of the equilateral triangle is at a special place regarding the median of the sides, in term of length...
– Martigan
Oct 2 '15 at 14:43
Hint: the center of the equilateral triangle is at a special place regarding the median of the sides, in term of length...
– Martigan
Oct 2 '15 at 14:43
To a given equilateral triangle, there is one and only one circle that circumscribes it. Therefore, there does not exist the "smallest" circle.
– Mick
Oct 2 '15 at 16:13
To a given equilateral triangle, there is one and only one circle that circumscribes it. Therefore, there does not exist the "smallest" circle.
– Mick
Oct 2 '15 at 16:13
The area you’ve got looks like an approximation. It seems like there should be a factor of $pi$ in there somewhere.
– amd
Oct 2 '15 at 20:33
The area you’ve got looks like an approximation. It seems like there should be a factor of $pi$ in there somewhere.
– amd
Oct 2 '15 at 20:33
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
0
down vote
Hint: Use similar triangles to determine where along the bisectors their intersection lies. Sines and cosines will still be there, of course, since they’re ratios of sides of a right triangle, but you won’t be using the functions explicitly.
answered Oct 2 '15 at 20:39
amd
26.8k21046
add a comment |Â
up vote
0
down vote
In an equilateral triangle the orthocenter, centroid, circumcenter and incenter coincide. The center of the circle is the centroid and height coincides with the median. The radius of the circumcircle is equal to two thirds the height.
Use the property in last sentence to find out the radius of circle that completely circumscribes the triangle using the triangle's height.
Height of triangle = root (21)
Radius of circle = 2/3 rd of root (21)
Area of circle = 88/3
answered Jun 19 '16 at 12:37
Avi
1
add a comment |Â
up vote
0
down vote
Consider this image:
The distance from the center of the triangle (a.k.a. apothem) to the midpoint of one of its side is $h/3$, where $h$ is the height of the triangle. So the radius of the circle r is the distance from the center of the triangle to a vertex, and this distance is $2h/3$.
Finding the side of the triangle (L)
We know that the area of the triangle is $7sqrt3$, so we do:
$Area = frac12(base)(height) = frac12Lleft(fracLsqrt32right) = 7sqrt3$
Solving for L, we get $L = 2sqrt7$.
Finding the radius of the circle (r)
As you already calculated, we can use the pythagorean theorem to find the height of the triangle:
$L^2 = left(fracL2right)^2 + h^2$
$h = sqrtL^2 - fracL^24 = fracLsqrt32$
The radius of the circle is then:
$r = frac23h = frac23fracLsqrt32 = fracLsqrt33 = frac2sqrt7sqrt33 = frac2sqrt213$
Finding the area of the circle
$Area_circle = pi r^2 = pi left(frac2sqrt213right)^2 = pi frac(4)(21)9 = frac28pi3 approx frac883$
answered Sep 12 at 3:12
HugoTeixeira
258211
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Hint: Use similar triangles to determine where along the bisectors their intersection lies. Sines and cosines will still be there, of course, since they’re ratios of sides of a right triangle, but you won’t be using the functions explicitly.
answered Oct 2 '15 at 20:39
amd
26.8k21046
add a comment |Â
up vote
0
down vote
Hint: Use similar triangles to determine where along the bisectors their intersection lies. Sines and cosines will still be there, of course, since they’re ratios of sides of a right triangle, but you won’t be using the functions explicitly.
answered Oct 2 '15 at 20:39
amd
26.8k21046
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Hint: Use similar triangles to determine where along the bisectors their intersection lies. Sines and cosines will still be there, of course, since they’re ratios of sides of a right triangle, but you won’t be using the functions explicitly.
answered Oct 2 '15 at 20:39
amd
26.8k21046
Hint: Use similar triangles to determine where along the bisectors their intersection lies. Sines and cosines will still be there, of course, since they’re ratios of sides of a right triangle, but you won’t be using the functions explicitly.
answered Oct 2 '15 at 20:39
amd
26.8k21046
answered Oct 2 '15 at 20:39
amd
26.8k21046
answered Oct 2 '15 at 20:39
amd
26.8k21046
answered Oct 2 '15 at 20:39
amd
26.8k21046
26.8k21046
add a comment |Â
add a comment |Â
up vote
0
down vote
In an equilateral triangle the orthocenter, centroid, circumcenter and incenter coincide. The center of the circle is the centroid and height coincides with the median. The radius of the circumcircle is equal to two thirds the height.
Use the property in last sentence to find out the radius of circle that completely circumscribes the triangle using the triangle's height.
Height of triangle = root (21)
Radius of circle = 2/3 rd of root (21)
Area of circle = 88/3
answered Jun 19 '16 at 12:37
Avi
1
add a comment |Â
up vote
0
down vote
In an equilateral triangle the orthocenter, centroid, circumcenter and incenter coincide. The center of the circle is the centroid and height coincides with the median. The radius of the circumcircle is equal to two thirds the height.
Use the property in last sentence to find out the radius of circle that completely circumscribes the triangle using the triangle's height.
Height of triangle = root (21)
Radius of circle = 2/3 rd of root (21)
Area of circle = 88/3
answered Jun 19 '16 at 12:37
Avi
1
add a comment |Â
up vote
0
down vote
up vote
0
down vote
In an equilateral triangle the orthocenter, centroid, circumcenter and incenter coincide. The center of the circle is the centroid and height coincides with the median. The radius of the circumcircle is equal to two thirds the height.
Use the property in last sentence to find out the radius of circle that completely circumscribes the triangle using the triangle's height.
Height of triangle = root (21)
Radius of circle = 2/3 rd of root (21)
Area of circle = 88/3
answered Jun 19 '16 at 12:37
Avi
1
In an equilateral triangle the orthocenter, centroid, circumcenter and incenter coincide. The center of the circle is the centroid and height coincides with the median. The radius of the circumcircle is equal to two thirds the height.
Use the property in last sentence to find out the radius of circle that completely circumscribes the triangle using the triangle's height.
Height of triangle = root (21)
Radius of circle = 2/3 rd of root (21)
Area of circle = 88/3
answered Jun 19 '16 at 12:37
Avi
1
answered Jun 19 '16 at 12:37
Avi
1
answered Jun 19 '16 at 12:37
Avi
1
answered Jun 19 '16 at 12:37
Avi
1
1
add a comment |Â
add a comment |Â
up vote
0
down vote
Consider this image:
The distance from the center of the triangle (a.k.a. apothem) to the midpoint of one of its side is $h/3$, where $h$ is the height of the triangle. So the radius of the circle r is the distance from the center of the triangle to a vertex, and this distance is $2h/3$.
Finding the side of the triangle (L)
We know that the area of the triangle is $7sqrt3$, so we do:
$Area = frac12(base)(height) = frac12Lleft(fracLsqrt32right) = 7sqrt3$
Solving for L, we get $L = 2sqrt7$.
Finding the radius of the circle (r)
As you already calculated, we can use the pythagorean theorem to find the height of the triangle:
$L^2 = left(fracL2right)^2 + h^2$
$h = sqrtL^2 - fracL^24 = fracLsqrt32$
The radius of the circle is then:
$r = frac23h = frac23fracLsqrt32 = fracLsqrt33 = frac2sqrt7sqrt33 = frac2sqrt213$
Finding the area of the circle
$Area_circle = pi r^2 = pi left(frac2sqrt213right)^2 = pi frac(4)(21)9 = frac28pi3 approx frac883$
answered Sep 12 at 3:12
HugoTeixeira
258211
add a comment |Â
up vote
0
down vote
Consider this image:
The distance from the center of the triangle (a.k.a. apothem) to the midpoint of one of its side is $h/3$, where $h$ is the height of the triangle. So the radius of the circle r is the distance from the center of the triangle to a vertex, and this distance is $2h/3$.
Finding the side of the triangle (L)
We know that the area of the triangle is $7sqrt3$, so we do:
$Area = frac12(base)(height) = frac12Lleft(fracLsqrt32right) = 7sqrt3$
Solving for L, we get $L = 2sqrt7$.
Finding the radius of the circle (r)
As you already calculated, we can use the pythagorean theorem to find the height of the triangle:
$L^2 = left(fracL2right)^2 + h^2$
$h = sqrtL^2 - fracL^24 = fracLsqrt32$
The radius of the circle is then:
$r = frac23h = frac23fracLsqrt32 = fracLsqrt33 = frac2sqrt7sqrt33 = frac2sqrt213$
Finding the area of the circle
$Area_circle = pi r^2 = pi left(frac2sqrt213right)^2 = pi frac(4)(21)9 = frac28pi3 approx frac883$
answered Sep 12 at 3:12
HugoTeixeira
258211
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Consider this image:
The distance from the center of the triangle (a.k.a. apothem) to the midpoint of one of its side is $h/3$, where $h$ is the height of the triangle. So the radius of the circle r is the distance from the center of the triangle to a vertex, and this distance is $2h/3$.
Finding the side of the triangle (L)
We know that the area of the triangle is $7sqrt3$, so we do:
$Area = frac12(base)(height) = frac12Lleft(fracLsqrt32right) = 7sqrt3$
Solving for L, we get $L = 2sqrt7$.
Finding the radius of the circle (r)
As you already calculated, we can use the pythagorean theorem to find the height of the triangle:
$L^2 = left(fracL2right)^2 + h^2$
$h = sqrtL^2 - fracL^24 = fracLsqrt32$
The radius of the circle is then:
$r = frac23h = frac23fracLsqrt32 = fracLsqrt33 = frac2sqrt7sqrt33 = frac2sqrt213$
Finding the area of the circle
$Area_circle = pi r^2 = pi left(frac2sqrt213right)^2 = pi frac(4)(21)9 = frac28pi3 approx frac883$
answered Sep 12 at 3:12
HugoTeixeira
258211
Consider this image:
The distance from the center of the triangle (a.k.a. apothem) to the midpoint of one of its side is $h/3$, where $h$ is the height of the triangle. So the radius of the circle r is the distance from the center of the triangle to a vertex, and this distance is $2h/3$.
Finding the side of the triangle (L)
We know that the area of the triangle is $7sqrt3$, so we do:
$Area = frac12(base)(height) = frac12Lleft(fracLsqrt32right) = 7sqrt3$
Solving for L, we get $L = 2sqrt7$.
Finding the radius of the circle (r)
As you already calculated, we can use the pythagorean theorem to find the height of the triangle:
$L^2 = left(fracL2right)^2 + h^2$
$h = sqrtL^2 - fracL^24 = fracLsqrt32$
The radius of the circle is then:
$r = frac23h = frac23fracLsqrt32 = fracLsqrt33 = frac2sqrt7sqrt33 = frac2sqrt213$
Finding the area of the circle
$Area_circle = pi r^2 = pi left(frac2sqrt213right)^2 = pi frac(4)(21)9 = frac28pi3 approx frac883$
answered Sep 12 at 3:12
HugoTeixeira
258211
answered Sep 12 at 3:12
HugoTeixeira
258211
answered Sep 12 at 3:12
HugoTeixeira
258211
answered Sep 12 at 3:12
HugoTeixeira
258211
258211
add a comment |Â
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f1461168%2ffinding-the-radius-of-the-smallest-circle-that-can-circumscribe-an-equilateral-t%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Hint: the center of the equilateral triangle is at a special place regarding the median of the sides, in term of length...
– Martigan
Oct 2 '15 at 14:43
To a given equilateral triangle, there is one and only one circle that circumscribes it. Therefore, there does not exist the "smallest" circle.
– Mick
Oct 2 '15 at 16:13
The area you’ve got looks like an approximation. It seems like there should be a factor of $pi$ in there somewhere.
– amd
Oct 2 '15 at 20:33