The measure of one angle of an octagon is twice that of the other seven angles. What is the measure of each...
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Help would be greatly appreciated.
geometry
asked Feb 2 '13 at 14:05
ShajinatorShajinator
4719
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closed as off-topic by José Carlos Santos, Lee David Chung Lin, egreg, rtybase, Adrian Keister Jan 12 at 15:16
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – José Carlos Santos, Lee David Chung Lin, egreg, rtybase, Adrian Keister
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
Help would be greatly appreciated.
geometry
asked Feb 2 '13 at 14:05
ShajinatorShajinator
4719
$endgroup$
closed as off-topic by José Carlos Santos, Lee David Chung Lin, egreg, rtybase, Adrian Keister Jan 12 at 15:16
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – José Carlos Santos, Lee David Chung Lin, egreg, rtybase, Adrian Keister
If this question can be reworded to fit the rules in the help center, please edit the question.
1
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It seems like you have posted the same question twice.
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– Parth Kohli
Feb 2 '13 at 14:48
2
$begingroup$
It is up to you what you accept, however, my post is not really an answer. It is a comment that can't be posted as a comment. Clayton and amWhy have given actual answers.
$endgroup$
– robjohn♦
Feb 2 '13 at 19:50
add a comment |
$begingroup$
Help would be greatly appreciated.
geometry
asked Feb 2 '13 at 14:05
ShajinatorShajinator
4719
$endgroup$
Help would be greatly appreciated.
geometry
geometry
asked Feb 2 '13 at 14:05
ShajinatorShajinator
4719
asked Feb 2 '13 at 14:05
ShajinatorShajinator
4719
asked Feb 2 '13 at 14:05
ShajinatorShajinator
4719
asked Feb 2 '13 at 14:05
ShajinatorShajinator
4719
asked Feb 2 '13 at 14:05
ShajinatorShajinator
4719
4719
closed as off-topic by José Carlos Santos, Lee David Chung Lin, egreg, rtybase, Adrian Keister Jan 12 at 15:16
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – José Carlos Santos, Lee David Chung Lin, egreg, rtybase, Adrian Keister
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by José Carlos Santos, Lee David Chung Lin, egreg, rtybase, Adrian Keister Jan 12 at 15:16
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – José Carlos Santos, Lee David Chung Lin, egreg, rtybase, Adrian Keister
If this question can be reworded to fit the rules in the help center, please edit the question.
1
$begingroup$
It seems like you have posted the same question twice.
$endgroup$
– Parth Kohli
Feb 2 '13 at 14:48
2
$begingroup$
It is up to you what you accept, however, my post is not really an answer. It is a comment that can't be posted as a comment. Clayton and amWhy have given actual answers.
$endgroup$
– robjohn♦
Feb 2 '13 at 19:50
add a comment |
1
$begingroup$
It seems like you have posted the same question twice.
$endgroup$
– Parth Kohli
Feb 2 '13 at 14:48
2
$begingroup$
It is up to you what you accept, however, my post is not really an answer. It is a comment that can't be posted as a comment. Clayton and amWhy have given actual answers.
$endgroup$
– robjohn♦
Feb 2 '13 at 19:50
1
1
$begingroup$
It seems like you have posted the same question twice.
$endgroup$
– Parth Kohli
Feb 2 '13 at 14:48
$begingroup$
It seems like you have posted the same question twice.
$endgroup$
– Parth Kohli
Feb 2 '13 at 14:48
2
2
$begingroup$
It is up to you what you accept, however, my post is not really an answer. It is a comment that can't be posted as a comment. Clayton and amWhy have given actual answers.
$endgroup$
– robjohn♦
Feb 2 '13 at 19:50
$begingroup$
It is up to you what you accept, however, my post is not really an answer. It is a comment that can't be posted as a comment. Clayton and amWhy have given actual answers.
$endgroup$
– robjohn♦
Feb 2 '13 at 19:50
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Images don't work in comments, so I am posting this even though it is not really an answer.
Using Clayton's answer, and because Valentine's Day is just a couple of weeks away, it seemed appropriate to post this image of the octagon:
$hspace{3.5cm}$
answered Feb 2 '13 at 14:44
robjohn♦robjohn
271k27314643
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$begingroup$
Thanks for the display picture!
$endgroup$
– Parth Kohli
Feb 2 '13 at 14:46
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Prediction #0129384: This will get 100+ upvotes.
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– Parth Kohli
Feb 2 '13 at 14:50
1
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+1 I want THIS heart for my valentine gravatar - (background blue?) ;-) Than it be made so that all the lengths of the sides are congruent? (hint: challenge!)
$endgroup$
– Namaste
Feb 2 '13 at 14:55
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@amWhy: I started out making all the sides congruent, but as the angles are fixed (and therefore all the sides must be parallel to those as drawn), the bottom two sides need to be a different length.
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– robjohn♦
Feb 2 '13 at 15:01
1
$begingroup$
Would the downvoter care to comment? I explained why I was posting a non-answer, although I realize that that doesn't make it an answer.
$endgroup$
– robjohn♦
Feb 2 '13 at 18:14
|
show 1 more comment
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Hint: An octagon has $1080^circ$, and you have the equation $$2theta+theta+cdots+theta=9theta=1080^circ.$$
answered Feb 2 '13 at 14:08
ClaytonClayton
19.6k33288
$endgroup$
add a comment |
$begingroup$
The sum of the interior angles of an octagon is $1080^circ$.
There are eight interior angles, one twice the measures of all others:
$7 theta + (2theta) = 1080^circ$
Now solve for $theta$ (which is the measure of the 7 equal angles),
and then compute $2theta$, the measure of the angle that is twice the size of the other 7 interior angles.
answered Feb 2 '13 at 16:04
NamasteNamaste
1
$endgroup$
add a comment |
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Images don't work in comments, so I am posting this even though it is not really an answer.
Using Clayton's answer, and because Valentine's Day is just a couple of weeks away, it seemed appropriate to post this image of the octagon:
$hspace{3.5cm}$
answered Feb 2 '13 at 14:44
robjohn♦robjohn
271k27314643
$endgroup$
$begingroup$
Thanks for the display picture!
$endgroup$
– Parth Kohli
Feb 2 '13 at 14:46
$begingroup$
Prediction #0129384: This will get 100+ upvotes.
$endgroup$
– Parth Kohli
Feb 2 '13 at 14:50
1
$begingroup$
+1 I want THIS heart for my valentine gravatar - (background blue?) ;-) Than it be made so that all the lengths of the sides are congruent? (hint: challenge!)
$endgroup$
– Namaste
Feb 2 '13 at 14:55
$begingroup$
@amWhy: I started out making all the sides congruent, but as the angles are fixed (and therefore all the sides must be parallel to those as drawn), the bottom two sides need to be a different length.
$endgroup$
– robjohn♦
Feb 2 '13 at 15:01
1
$begingroup$
Would the downvoter care to comment? I explained why I was posting a non-answer, although I realize that that doesn't make it an answer.
$endgroup$
– robjohn♦
Feb 2 '13 at 18:14
|
show 1 more comment
$begingroup$
Images don't work in comments, so I am posting this even though it is not really an answer.
Using Clayton's answer, and because Valentine's Day is just a couple of weeks away, it seemed appropriate to post this image of the octagon:
$hspace{3.5cm}$
answered Feb 2 '13 at 14:44
robjohn♦robjohn
271k27314643
$endgroup$
$begingroup$
Thanks for the display picture!
$endgroup$
– Parth Kohli
Feb 2 '13 at 14:46
$begingroup$
Prediction #0129384: This will get 100+ upvotes.
$endgroup$
– Parth Kohli
Feb 2 '13 at 14:50
1
$begingroup$
+1 I want THIS heart for my valentine gravatar - (background blue?) ;-) Than it be made so that all the lengths of the sides are congruent? (hint: challenge!)
$endgroup$
– Namaste
Feb 2 '13 at 14:55
$begingroup$
@amWhy: I started out making all the sides congruent, but as the angles are fixed (and therefore all the sides must be parallel to those as drawn), the bottom two sides need to be a different length.
$endgroup$
– robjohn♦
Feb 2 '13 at 15:01
1
$begingroup$
Would the downvoter care to comment? I explained why I was posting a non-answer, although I realize that that doesn't make it an answer.
$endgroup$
– robjohn♦
Feb 2 '13 at 18:14
|
show 1 more comment
$begingroup$
Images don't work in comments, so I am posting this even though it is not really an answer.
Using Clayton's answer, and because Valentine's Day is just a couple of weeks away, it seemed appropriate to post this image of the octagon:
$hspace{3.5cm}$
answered Feb 2 '13 at 14:44
robjohn♦robjohn
271k27314643
$endgroup$
Images don't work in comments, so I am posting this even though it is not really an answer.
Using Clayton's answer, and because Valentine's Day is just a couple of weeks away, it seemed appropriate to post this image of the octagon:
$hspace{3.5cm}$
answered Feb 2 '13 at 14:44
robjohn♦robjohn
271k27314643
edited Feb 2 '13 at 15:07
answered Feb 2 '13 at 14:44
robjohn♦robjohn
271k27314643
answered Feb 2 '13 at 14:44
robjohn♦robjohn
271k27314643
answered Feb 2 '13 at 14:44
robjohn♦robjohn
271k27314643
271k27314643
$begingroup$
Thanks for the display picture!
$endgroup$
– Parth Kohli
Feb 2 '13 at 14:46
$begingroup$
Prediction #0129384: This will get 100+ upvotes.
$endgroup$
– Parth Kohli
Feb 2 '13 at 14:50
1
$begingroup$
+1 I want THIS heart for my valentine gravatar - (background blue?) ;-) Than it be made so that all the lengths of the sides are congruent? (hint: challenge!)
$endgroup$
– Namaste
Feb 2 '13 at 14:55
$begingroup$
@amWhy: I started out making all the sides congruent, but as the angles are fixed (and therefore all the sides must be parallel to those as drawn), the bottom two sides need to be a different length.
$endgroup$
– robjohn♦
Feb 2 '13 at 15:01
1
$begingroup$
Would the downvoter care to comment? I explained why I was posting a non-answer, although I realize that that doesn't make it an answer.
$endgroup$
– robjohn♦
Feb 2 '13 at 18:14
|
show 1 more comment
$begingroup$
Thanks for the display picture!
$endgroup$
– Parth Kohli
Feb 2 '13 at 14:46
$begingroup$
Prediction #0129384: This will get 100+ upvotes.
$endgroup$
– Parth Kohli
Feb 2 '13 at 14:50
1
$begingroup$
+1 I want THIS heart for my valentine gravatar - (background blue?) ;-) Than it be made so that all the lengths of the sides are congruent? (hint: challenge!)
$endgroup$
– Namaste
Feb 2 '13 at 14:55
$begingroup$
@amWhy: I started out making all the sides congruent, but as the angles are fixed (and therefore all the sides must be parallel to those as drawn), the bottom two sides need to be a different length.
$endgroup$
– robjohn♦
Feb 2 '13 at 15:01
1
$begingroup$
Would the downvoter care to comment? I explained why I was posting a non-answer, although I realize that that doesn't make it an answer.
$endgroup$
– robjohn♦
Feb 2 '13 at 18:14
$begingroup$
Thanks for the display picture!
$endgroup$
– Parth Kohli
Feb 2 '13 at 14:46
$begingroup$
Thanks for the display picture!
$endgroup$
– Parth Kohli
Feb 2 '13 at 14:46
$begingroup$
Prediction #0129384: This will get 100+ upvotes.
$endgroup$
– Parth Kohli
Feb 2 '13 at 14:50
$begingroup$
Prediction #0129384: This will get 100+ upvotes.
$endgroup$
– Parth Kohli
Feb 2 '13 at 14:50
1
1
$begingroup$
+1 I want THIS heart for my valentine gravatar - (background blue?) ;-) Than it be made so that all the lengths of the sides are congruent? (hint: challenge!)
$endgroup$
– Namaste
Feb 2 '13 at 14:55
$begingroup$
+1 I want THIS heart for my valentine gravatar - (background blue?) ;-) Than it be made so that all the lengths of the sides are congruent? (hint: challenge!)
$endgroup$
– Namaste
Feb 2 '13 at 14:55
$begingroup$
@amWhy: I started out making all the sides congruent, but as the angles are fixed (and therefore all the sides must be parallel to those as drawn), the bottom two sides need to be a different length.
$endgroup$
– robjohn♦
Feb 2 '13 at 15:01
$begingroup$
@amWhy: I started out making all the sides congruent, but as the angles are fixed (and therefore all the sides must be parallel to those as drawn), the bottom two sides need to be a different length.
$endgroup$
– robjohn♦
Feb 2 '13 at 15:01
1
1
$begingroup$
Would the downvoter care to comment? I explained why I was posting a non-answer, although I realize that that doesn't make it an answer.
$endgroup$
– robjohn♦
Feb 2 '13 at 18:14
$begingroup$
Would the downvoter care to comment? I explained why I was posting a non-answer, although I realize that that doesn't make it an answer.
$endgroup$
– robjohn♦
Feb 2 '13 at 18:14
|
show 1 more comment
$begingroup$
Hint: An octagon has $1080^circ$, and you have the equation $$2theta+theta+cdots+theta=9theta=1080^circ.$$
answered Feb 2 '13 at 14:08
ClaytonClayton
19.6k33288
$endgroup$
add a comment |
$begingroup$
Hint: An octagon has $1080^circ$, and you have the equation $$2theta+theta+cdots+theta=9theta=1080^circ.$$
answered Feb 2 '13 at 14:08
ClaytonClayton
19.6k33288
$endgroup$
add a comment |
$begingroup$
Hint: An octagon has $1080^circ$, and you have the equation $$2theta+theta+cdots+theta=9theta=1080^circ.$$
answered Feb 2 '13 at 14:08
ClaytonClayton
19.6k33288
$endgroup$
Hint: An octagon has $1080^circ$, and you have the equation $$2theta+theta+cdots+theta=9theta=1080^circ.$$
answered Feb 2 '13 at 14:08
ClaytonClayton
19.6k33288
answered Feb 2 '13 at 14:08
ClaytonClayton
19.6k33288
answered Feb 2 '13 at 14:08
ClaytonClayton
19.6k33288
answered Feb 2 '13 at 14:08
ClaytonClayton
19.6k33288
19.6k33288
add a comment |
add a comment |
$begingroup$
The sum of the interior angles of an octagon is $1080^circ$.
There are eight interior angles, one twice the measures of all others:
$7 theta + (2theta) = 1080^circ$
Now solve for $theta$ (which is the measure of the 7 equal angles),
and then compute $2theta$, the measure of the angle that is twice the size of the other 7 interior angles.
answered Feb 2 '13 at 16:04
NamasteNamaste
1
$endgroup$
add a comment |
$begingroup$
The sum of the interior angles of an octagon is $1080^circ$.
There are eight interior angles, one twice the measures of all others:
$7 theta + (2theta) = 1080^circ$
Now solve for $theta$ (which is the measure of the 7 equal angles),
and then compute $2theta$, the measure of the angle that is twice the size of the other 7 interior angles.
answered Feb 2 '13 at 16:04
NamasteNamaste
1
$endgroup$
add a comment |
$begingroup$
The sum of the interior angles of an octagon is $1080^circ$.
There are eight interior angles, one twice the measures of all others:
$7 theta + (2theta) = 1080^circ$
Now solve for $theta$ (which is the measure of the 7 equal angles),
and then compute $2theta$, the measure of the angle that is twice the size of the other 7 interior angles.
answered Feb 2 '13 at 16:04
NamasteNamaste
1
$endgroup$
The sum of the interior angles of an octagon is $1080^circ$.
There are eight interior angles, one twice the measures of all others:
$7 theta + (2theta) = 1080^circ$
Now solve for $theta$ (which is the measure of the 7 equal angles),
and then compute $2theta$, the measure of the angle that is twice the size of the other 7 interior angles.
answered Feb 2 '13 at 16:04
NamasteNamaste
1
answered Feb 2 '13 at 16:04
NamasteNamaste
1
answered Feb 2 '13 at 16:04
NamasteNamaste
1
answered Feb 2 '13 at 16:04
NamasteNamaste
1
1
add a comment |
add a comment |
1
$begingroup$
It seems like you have posted the same question twice.
$endgroup$
– Parth Kohli
Feb 2 '13 at 14:48
2
$begingroup$
It is up to you what you accept, however, my post is not really an answer. It is a comment that can't be posted as a comment. Clayton and amWhy have given actual answers.
$endgroup$
– robjohn♦
Feb 2 '13 at 19:50