Finding the radius of the smallest circle that can circumscribe an equilateral triangle
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Q:A puzzle board is in the form of an equilateral triangle that has an area of $7sqrt{3}$ 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: $frac{88}{3}$
I get that the side of the triangle is $2sqrt{7}$ 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.
geometry circle triangle
asked Oct 2 '15 at 14:40
biabia
498
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add a comment |
$begingroup$
Q:A puzzle board is in the form of an equilateral triangle that has an area of $7sqrt{3}$ 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: $frac{88}{3}$
I get that the side of the triangle is $2sqrt{7}$ 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.
geometry circle triangle
asked Oct 2 '15 at 14:40
biabia
498
$endgroup$
$begingroup$
Hint: the center of the equilateral triangle is at a special place regarding the median of the sides, in term of length...
$endgroup$
– Martigan
Oct 2 '15 at 14:43
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To a given equilateral triangle, there is one and only one circle that circumscribes it. Therefore, there does not exist the "smallest" circle.
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– Mick
Oct 2 '15 at 16:13
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The area you’ve got looks like an approximation. It seems like there should be a factor of $pi$ in there somewhere.
$endgroup$
– amd
Oct 2 '15 at 20:33
add a comment |
$begingroup$
Q:A puzzle board is in the form of an equilateral triangle that has an area of $7sqrt{3}$ 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: $frac{88}{3}$
I get that the side of the triangle is $2sqrt{7}$ 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.
geometry circle triangle
asked Oct 2 '15 at 14:40
biabia
498
$endgroup$
Q:A puzzle board is in the form of an equilateral triangle that has an area of $7sqrt{3}$ 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: $frac{88}{3}$
I get that the side of the triangle is $2sqrt{7}$ 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.
geometry circle triangle
geometry circle triangle
asked Oct 2 '15 at 14:40
biabia
498
asked Oct 2 '15 at 14:40
biabia
498
edited Jan 3 at 16:44
bia
asked Oct 2 '15 at 14:40
biabia
498
asked Oct 2 '15 at 14:40
biabia
498
asked Oct 2 '15 at 14:40
biabia
498
498
$begingroup$
Hint: the center of the equilateral triangle is at a special place regarding the median of the sides, in term of length...
$endgroup$
– Martigan
Oct 2 '15 at 14:43
$begingroup$
To a given equilateral triangle, there is one and only one circle that circumscribes it. Therefore, there does not exist the "smallest" circle.
$endgroup$
– Mick
Oct 2 '15 at 16:13
$begingroup$
The area you’ve got looks like an approximation. It seems like there should be a factor of $pi$ in there somewhere.
$endgroup$
– amd
Oct 2 '15 at 20:33
add a comment |
$begingroup$
Hint: the center of the equilateral triangle is at a special place regarding the median of the sides, in term of length...
$endgroup$
– Martigan
Oct 2 '15 at 14:43
$begingroup$
To a given equilateral triangle, there is one and only one circle that circumscribes it. Therefore, there does not exist the "smallest" circle.
$endgroup$
– Mick
Oct 2 '15 at 16:13
$begingroup$
The area you’ve got looks like an approximation. It seems like there should be a factor of $pi$ in there somewhere.
$endgroup$
– amd
Oct 2 '15 at 20:33
$begingroup$
Hint: the center of the equilateral triangle is at a special place regarding the median of the sides, in term of length...
$endgroup$
– Martigan
Oct 2 '15 at 14:43
$begingroup$
Hint: the center of the equilateral triangle is at a special place regarding the median of the sides, in term of length...
$endgroup$
– Martigan
Oct 2 '15 at 14:43
$begingroup$
To a given equilateral triangle, there is one and only one circle that circumscribes it. Therefore, there does not exist the "smallest" circle.
$endgroup$
– Mick
Oct 2 '15 at 16:13
$begingroup$
To a given equilateral triangle, there is one and only one circle that circumscribes it. Therefore, there does not exist the "smallest" circle.
$endgroup$
– Mick
Oct 2 '15 at 16:13
$begingroup$
The area you’ve got looks like an approximation. It seems like there should be a factor of $pi$ in there somewhere.
$endgroup$
– amd
Oct 2 '15 at 20:33
$begingroup$
The area you’ve got looks like an approximation. It seems like there should be a factor of $pi$ in there somewhere.
$endgroup$
– amd
Oct 2 '15 at 20:33
add a comment |
3 Answers
3
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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
amdamd
30k21050
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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
AviAvi
1
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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 $7sqrt{3}$, so we do:
$Area = frac{1}{2}(base)(height) = frac{1}{2}Lleft(frac{Lsqrt{3}}{2}right) = 7sqrt{3}$
Solving for L, we get $L = 2sqrt{7}$.
Finding the radius of the circle (r)
As you already calculated, we can use the pythagorean theorem to find the height of the triangle:
$L^2 = left(frac{L}{2}right)^2 + h^2$
$h = sqrt{L^2 - frac{L^2}{4}} = frac{Lsqrt{3}}{2}$
The radius of the circle is then:
$r = frac{2}{3}h = frac{2}{3}frac{Lsqrt{3}}{2} = frac{Lsqrt{3}}{3} = frac{2sqrt{7}sqrt{3}}{3} = frac{2sqrt{21}}{3}$
Finding the area of the circle
$Area_{circle} = pi r^2 = pi left(frac{2sqrt{21}}{3}right)^2 = pi frac{(4)(21)}{9} = frac{28pi}{3} approx frac{88}{3}$
answered Sep 12 '18 at 3:12
HugoTeixeiraHugoTeixeira
3281313
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
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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
amdamd
30k21050
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add a comment |
$begingroup$
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
amdamd
30k21050
$endgroup$
add a comment |
$begingroup$
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
amdamd
30k21050
$endgroup$
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
amdamd
30k21050
answered Oct 2 '15 at 20:39
amdamd
30k21050
answered Oct 2 '15 at 20:39
amdamd
30k21050
answered Oct 2 '15 at 20:39
amdamd
30k21050
30k21050
add a comment |
add a comment |
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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
AviAvi
1
$endgroup$
add a comment |
$begingroup$
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
AviAvi
1
$endgroup$
add a comment |
$begingroup$
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
AviAvi
1
$endgroup$
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
AviAvi
1
answered Jun 19 '16 at 12:37
AviAvi
1
answered Jun 19 '16 at 12:37
AviAvi
1
answered Jun 19 '16 at 12:37
AviAvi
1
1
add a comment |
add a comment |
$begingroup$
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 $7sqrt{3}$, so we do:
$Area = frac{1}{2}(base)(height) = frac{1}{2}Lleft(frac{Lsqrt{3}}{2}right) = 7sqrt{3}$
Solving for L, we get $L = 2sqrt{7}$.
Finding the radius of the circle (r)
As you already calculated, we can use the pythagorean theorem to find the height of the triangle:
$L^2 = left(frac{L}{2}right)^2 + h^2$
$h = sqrt{L^2 - frac{L^2}{4}} = frac{Lsqrt{3}}{2}$
The radius of the circle is then:
$r = frac{2}{3}h = frac{2}{3}frac{Lsqrt{3}}{2} = frac{Lsqrt{3}}{3} = frac{2sqrt{7}sqrt{3}}{3} = frac{2sqrt{21}}{3}$
Finding the area of the circle
$Area_{circle} = pi r^2 = pi left(frac{2sqrt{21}}{3}right)^2 = pi frac{(4)(21)}{9} = frac{28pi}{3} approx frac{88}{3}$
answered Sep 12 '18 at 3:12
HugoTeixeiraHugoTeixeira
3281313
$endgroup$
add a comment |
$begingroup$
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 $7sqrt{3}$, so we do:
$Area = frac{1}{2}(base)(height) = frac{1}{2}Lleft(frac{Lsqrt{3}}{2}right) = 7sqrt{3}$
Solving for L, we get $L = 2sqrt{7}$.
Finding the radius of the circle (r)
As you already calculated, we can use the pythagorean theorem to find the height of the triangle:
$L^2 = left(frac{L}{2}right)^2 + h^2$
$h = sqrt{L^2 - frac{L^2}{4}} = frac{Lsqrt{3}}{2}$
The radius of the circle is then:
$r = frac{2}{3}h = frac{2}{3}frac{Lsqrt{3}}{2} = frac{Lsqrt{3}}{3} = frac{2sqrt{7}sqrt{3}}{3} = frac{2sqrt{21}}{3}$
Finding the area of the circle
$Area_{circle} = pi r^2 = pi left(frac{2sqrt{21}}{3}right)^2 = pi frac{(4)(21)}{9} = frac{28pi}{3} approx frac{88}{3}$
answered Sep 12 '18 at 3:12
HugoTeixeiraHugoTeixeira
3281313
$endgroup$
add a comment |
$begingroup$
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 $7sqrt{3}$, so we do:
$Area = frac{1}{2}(base)(height) = frac{1}{2}Lleft(frac{Lsqrt{3}}{2}right) = 7sqrt{3}$
Solving for L, we get $L = 2sqrt{7}$.
Finding the radius of the circle (r)
As you already calculated, we can use the pythagorean theorem to find the height of the triangle:
$L^2 = left(frac{L}{2}right)^2 + h^2$
$h = sqrt{L^2 - frac{L^2}{4}} = frac{Lsqrt{3}}{2}$
The radius of the circle is then:
$r = frac{2}{3}h = frac{2}{3}frac{Lsqrt{3}}{2} = frac{Lsqrt{3}}{3} = frac{2sqrt{7}sqrt{3}}{3} = frac{2sqrt{21}}{3}$
Finding the area of the circle
$Area_{circle} = pi r^2 = pi left(frac{2sqrt{21}}{3}right)^2 = pi frac{(4)(21)}{9} = frac{28pi}{3} approx frac{88}{3}$
answered Sep 12 '18 at 3:12
HugoTeixeiraHugoTeixeira
3281313
$endgroup$
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 $7sqrt{3}$, so we do:
$Area = frac{1}{2}(base)(height) = frac{1}{2}Lleft(frac{Lsqrt{3}}{2}right) = 7sqrt{3}$
Solving for L, we get $L = 2sqrt{7}$.
Finding the radius of the circle (r)
As you already calculated, we can use the pythagorean theorem to find the height of the triangle:
$L^2 = left(frac{L}{2}right)^2 + h^2$
$h = sqrt{L^2 - frac{L^2}{4}} = frac{Lsqrt{3}}{2}$
The radius of the circle is then:
$r = frac{2}{3}h = frac{2}{3}frac{Lsqrt{3}}{2} = frac{Lsqrt{3}}{3} = frac{2sqrt{7}sqrt{3}}{3} = frac{2sqrt{21}}{3}$
Finding the area of the circle
$Area_{circle} = pi r^2 = pi left(frac{2sqrt{21}}{3}right)^2 = pi frac{(4)(21)}{9} = frac{28pi}{3} approx frac{88}{3}$
answered Sep 12 '18 at 3:12
HugoTeixeiraHugoTeixeira
3281313
answered Sep 12 '18 at 3:12
HugoTeixeiraHugoTeixeira
3281313
answered Sep 12 '18 at 3:12
HugoTeixeiraHugoTeixeira
3281313
answered Sep 12 '18 at 3:12
HugoTeixeiraHugoTeixeira
3281313
3281313
add a comment |
add a comment |
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$begingroup$
Hint: the center of the equilateral triangle is at a special place regarding the median of the sides, in term of length...
$endgroup$
– Martigan
Oct 2 '15 at 14:43
$begingroup$
To a given equilateral triangle, there is one and only one circle that circumscribes it. Therefore, there does not exist the "smallest" circle.
$endgroup$
– Mick
Oct 2 '15 at 16:13
$begingroup$
The area you’ve got looks like an approximation. It seems like there should be a factor of $pi$ in there somewhere.
$endgroup$
– amd
Oct 2 '15 at 20:33