I am SO stumped on 23! I tried to work it out by setting the formula for the measures of the interior angels of an n-gon equal to 1080 and then multiplying both sides by (n/1) and then multiplying both side by 1/(n-2) but it hasn't gotten me anywhere.. Help??
Is <2 on #27 referring to the angle between the square and the side of the smaller octagon? Because if it is, I have no idea how to solve for its measure!
For #33, shouldn't we find the number of sides before finding the measure of an interior angle? Because don't you need to know the number of sides to find the measure of an interior angle?
#2 is a trick question, iff (if and only if) Theorem 6-2 is a trick theorem.
#23 (or #22-26 for that matter) For this problem, you will have to use a little known sub-topic of mathematics known as ALGEBRA... 180(n-2)=1080 ... can you solve for n... please gosh-gee-willakers I hope so...
For #27, yes, you identified the angle correctly. (You must assume - shame on the text book - that the small quadrilateral is a square). You simply need to be able to find the measure of an interior angle of a regular octagon (you can do that correctly) and then perhaps use another magical mathematical tool, some call it subtraction?
For #33, answer me these questions five (I mean three!): 1) What is a name for the relationship between the interior angle and exterior angle of a polygon? (hint: it begins with 's') 2) Given the measure of the exterior angle, how can you calculate the measure of the adjacent interior angle? 3) What's your favorite color? Perhaps this video tutor can help... http://www.youtube.com/watch?v=IMxWLuOFyZM
In #33, they are giving you the measure of the exterior angle, 36, of a regular polygon. The interior angle is the supplement, so it must be 180-36 or 144.
The number of sides of an n-gon is n, yes? (duh!). One formula for the sum of the interior angles of THIS PARTICULAR REGULAR polygon is 144n, yes? Another formula (already on our tool-belt) is 180(n-2), yes?
So, we just need to know when: 144n = 180(n-2)
Once again, ALGEBRA to the rescue, yes??
(hmmm, 'n' on both sides... sounds like October of 2010 to me!)
Part of the challenge of this chapter will be recognizing how to put these geometric situations into algebraic balance (i.e. equations). Not to worry... re-learning Algebra is just like riding a bicycle!
I'm really confused on what to do on #10.. The only thing I can possibly think of a way to answer is C., using the Polygon Exterior Angle Theorem, so it would be 360!
We will go over this in class... I haven't made any videos because, incredibly, my voice has not fully recovered although I think I will be ready for Tuesday (ARGH!). I finally went to the see a doc yesterday. I am a dope-a-potamus. That said...
Here is why 180(n-2) works for the PolygonAngleSum theorem. Get a paper and pencil. Draw a quadrilateral, any quadrilateral. Now draw one diagonal. What is the quadrtrl composed of now? Two triangles, right? We have faith in our bible that the sum of the angles of a triangle is 180, yes. So, the two triangles contain angles that sum to 360, yes? So the four-sided figure contains (4-2=2) triangles, so the sum of the interior angles is (4-2)(180) or 2(180). Draw a pentagon. Draw non-intersecting diagonals that create triangles. How many non-overlapping triangles do you see? Three, right? So a five-sided polygon creats (5-2=3) triangles, so the sum is (5-2)(180) or 3(180)=540. Keep going... for an n-sided polygon you will be able to create (n-2) non-overlapping triangles (by drawing (n-3) diagonals. The sum of the interior angles of an n-gon is equal to the sum of the angles of all of the non-overlapping triangles (180) that you can draw (you can draw n-2 of them). So 180(n-2) is the formula, yes?
Can you tell me why can you only draw n-3 diagonals to create n-2 non-overlapping triangles in any n-gon? (Hint: all of the diagonals must "radiate" from a single vertex)
I understand how 180(n-2) works, but I can't figure out how 180-360/n relates. As well, I can't figure out how the given information in #39 would allow you to draw a conclusion about the angle measures.
Fair enough... I can't figure out where you gotsk 180-360/n...
Here's how I see #39... see if you can agree with me point by point:
a) OK, we agree that there are n exterior angles to an n-gon. And if we take the int/ext angle combos, there are n of those as well, yes? Since the int/ext combos are supplementary, we now have n straight angles aka 180n worth of angle measure, yes? All that agree, say "HRMPH-HRMPH!" and bang the table hard once!
b) OK, so we understand the (n-2)180 is the sum of the measures of the interior angles of a polygon, cuz we can make (n-2) non-overlapping triangles inside the polygon, AND CUZ WE TRUST Theorem 6-1 since it's in our bible, yes? All that agree, say "HRMPH-HRMPH!" and bang the table hard twice!
c) So, if we took the sum of all of the int/ext combo/straight angles "180n" and subtracted the sum of all of the interior angles "180(n-2)" THEN we would have the sum of all of the exterior angles as a difference, yes?
Re-stated, the sum of the exterior angles must be equal to the sum of the int/ext angles less the sum of the int angles, yes??? Let's call it the Angle Addition on Steroids Postulate.
Algebra to the rescue!! Let E equal the sum of the ext angles:
E = 180n - 180(n-2) E = 180n - 180n - (-360) E = 360
Therefore, logically and algebraically, the sum of the measures of the exterior angles of any polygon is 360 degrees.
All that agree, say "HRMPH-HRMPH!" and bang the table hard 3 times!
P.S. I'm not buying anyone a new table OR paying for any hand surgeries... Mr. C.
It is imperative that you grab a pencil and paper and do the exercise I suggest in my Jan 1st 5:35 comment above. You should even be able to understand what I mean by n-3 diagonals...
For a given REGULAR n-gon, 360/n is the measure of an exterior angle. This makes sense since we know the sum of the ext. angles is 360, and there are n ext. angles.
Given this info, to find the measure of the (or any) interior angle of this same REGULAR n-gon, we need only subtract the exterior angle measure (360/n) from 180. Remember, interior angles and exterior angles at the same vertex are supplementary by definition.
Hence, the measure on an int angle of a REGULAR n-gon is 180-(360/n).
This is a question for #6-2: What do they mean on #5? "What are____"? What do they mean "What are", am I supposed to name their relationship or state that they are transversals/segments?
Sorry if I'm missing something here if it sounds stupid: But what could I put for c. on #4? I know it doesn't state anywhere in the definition of a parallelogram that it has opposite congruent sides...
fyi... I will be out of the MathChamber office until Thursday afternoon... hope you are enjoying your time off... don't get too used to it ;)
ReplyDeleteOoooo yay! My logo is up! :)
ReplyDeleteIs #2 a trick question? Isn't it always 360?
ReplyDeleteI am SO stumped on 23! I tried to work it out by setting the formula for the measures of the interior angels of an n-gon equal to 1080 and then multiplying both sides by (n/1) and then multiplying both side by 1/(n-2) but it hasn't gotten me anywhere.. Help??
ReplyDeleteIs <2 on #27 referring to the angle between the square and the side of the smaller octagon? Because if it is, I have no idea how to solve for its measure!
ReplyDeleteFor #33, shouldn't we find the number of sides before finding the measure of an interior angle? Because don't you need to know the number of sides to find the measure of an interior angle?
ReplyDelete#2 is a trick question, iff (if and only if) Theorem 6-2 is a trick theorem.
ReplyDelete#23 (or #22-26 for that matter)
For this problem, you will have to use a little known sub-topic of mathematics known as ALGEBRA...
180(n-2)=1080
... can you solve for n... please gosh-gee-willakers I hope so...
For #27, yes, you identified the angle correctly. (You must assume - shame on the text book - that the small quadrilateral is a square). You simply need to be able to find the measure of an interior angle of a regular octagon (you can do that correctly) and then perhaps use another magical mathematical tool, some call it subtraction?
For #33, answer me these questions five (I mean three!):
1) What is a name for the relationship between the interior angle and exterior angle of a polygon? (hint: it begins with 's')
2) Given the measure of the exterior angle, how can you calculate the measure of the adjacent interior angle?
3) What's your favorite color?
Perhaps this video tutor can help...
http://www.youtube.com/watch?v=IMxWLuOFyZM
Mr. C.
Thanks!
ReplyDeleteAnd for some reason I can't figure out now I was using the [(n-2)180]/n formula instead of just the (n-2)180.
And for your answer to #33, I still don't understand.. ): Nor do I understand how any of those questions help!
ReplyDeleteFormulas are shortcuts... shortcuts can be problematic unless we understand...
ReplyDeleteDo you NOW know why to use (n-2)180 vs [(n-2)180]/n?
In #33, they are giving you the measure of the exterior angle, 36, of a regular polygon. The interior angle is the supplement, so it must be 180-36 or 144.
ReplyDeleteThe number of sides of an n-gon is n, yes? (duh!).
One formula for the sum of the interior angles of THIS PARTICULAR REGULAR polygon is 144n, yes?
Another formula (already on our tool-belt) is 180(n-2), yes?
So, we just need to know when:
144n = 180(n-2)
Once again, ALGEBRA to the rescue, yes??
(hmmm, 'n' on both sides... sounds like October of 2010 to me!)
Part of the challenge of this chapter will be recognizing how to put these geometric situations into algebraic balance (i.e. equations). Not to worry... re-learning Algebra is just like riding a bicycle!
ReplyDeleteOhhh.. I understand!!
ReplyDeleteI'm really confused on what to do on #10.. The only thing I can possibly think of a way to answer is C., using the Polygon Exterior Angle Theorem, so it would be 360!
ReplyDeleteSorry, not #10, #39!
ReplyDeleteIs the only method to figure out #41 guess and check?
ReplyDeleteI just know the formulas but don't know why they work! Can't figure out #43? :o
ReplyDeleteWe will go over this in class... I haven't made any videos because, incredibly, my voice has not fully recovered although I think I will be ready for Tuesday (ARGH!). I finally went to the see a doc yesterday. I am a dope-a-potamus. That said...
ReplyDeleteHere is why 180(n-2) works for the PolygonAngleSum theorem. Get a paper and pencil. Draw a quadrilateral, any quadrilateral. Now draw one diagonal. What is the quadrtrl composed of now? Two triangles, right? We have faith in our bible that the sum of the angles of a triangle is 180, yes. So, the two triangles contain angles that sum to 360, yes? So the four-sided figure contains (4-2=2) triangles, so the sum of the interior angles is (4-2)(180) or 2(180). Draw a pentagon. Draw non-intersecting diagonals that create triangles. How many non-overlapping triangles do you see? Three, right? So a five-sided polygon creats (5-2=3) triangles, so the sum is (5-2)(180) or 3(180)=540. Keep going... for an n-sided polygon you will be able to create (n-2) non-overlapping triangles (by drawing (n-3) diagonals. The sum of the interior angles of an n-gon is equal to the sum of the angles of all of the non-overlapping triangles (180) that you can draw (you can draw n-2 of them). So 180(n-2) is the formula, yes?
Can you tell me why can you only draw n-3 diagonals to create n-2 non-overlapping triangles in any n-gon? (Hint: all of the diagonals must "radiate" from a single vertex)
Do you UNDERSTAND now??
I understand how 180(n-2) works, but I can't figure out how 180-360/n relates. As well, I can't figure out how the given information in #39 would allow you to draw a conclusion about the angle measures.
ReplyDeleteFair enough... I can't figure out where you gotsk 180-360/n...
ReplyDeleteHere's how I see #39... see if you can agree with me point by point:
a) OK, we agree that there are n exterior angles to an n-gon. And if we take the int/ext angle combos, there are n of those as well, yes? Since the int/ext combos are supplementary, we now have n straight angles aka 180n worth of angle measure, yes? All that agree, say "HRMPH-HRMPH!" and bang the table hard once!
b) OK, so we understand the (n-2)180 is the sum of the measures of the interior angles of a polygon, cuz we can make (n-2) non-overlapping triangles inside the polygon, AND CUZ WE TRUST Theorem 6-1 since it's in our bible, yes? All that agree, say "HRMPH-HRMPH!" and bang the table hard twice!
c) So, if we took the sum of all of the int/ext combo/straight angles "180n" and subtracted the sum of all of the interior angles "180(n-2)" THEN we would have the sum of all of the exterior angles as a difference, yes?
Re-stated, the sum of the exterior angles must be equal to the sum of the int/ext angles less the sum of the int angles, yes??? Let's call it the Angle Addition on Steroids Postulate.
Algebra to the rescue!! Let E equal the sum of the ext angles:
E = 180n - 180(n-2)
E = 180n - 180n - (-360)
E = 360
Therefore, logically and algebraically, the sum of the measures of the exterior angles of any polygon is 360 degrees.
All that agree, say "HRMPH-HRMPH!" and bang the table hard 3 times!
P.S. I'm not buying anyone a new table OR paying for any hand surgeries... Mr. C.
^ For b), I don't know what you mean by (n-2) overlapping triangles..
ReplyDeleteAnd it's still kind of fuzzy on both problems.
(n-2) NON-overlapping triangles...
ReplyDeleteIt is imperative that you grab a pencil and paper and do the exercise I suggest in my Jan 1st 5:35 comment above. You should even be able to understand what I mean by n-3 diagonals...
I sense an ah-hah moment coming on...
Seeya in class!
OHH! I understand why (n-2)180 works. But not the other formula, like Paul said.
ReplyDeleteWhich other formula? Keep drawing and asking questions and I think you'll get it... me thinks you've got to be very close to the next OHHHH!!
ReplyDeleteGo back to #39 a,b,&c... at what point are you not banging the table??
180-360/n
ReplyDeleteand for #39, I think I'm starting to get it now, I was just confused on b) because of the (n-2) which I didn't understand before.
I still don't understand why you are referencing 180-360/n or where it came from or to what you are trying to relate it. ????
ReplyDeleteAre you saying it has something to do with #39??? cuz I don't think it does...
You ask which other formula it was that I could not figure out why it worked, and 180-(360/n) is the one that confuses me.
ReplyDeleteAnd no, I'm not saying it has something to do with #39!
ReplyDeleteOHHHH!!!! That's what you're asking!
ReplyDeleteFor a given REGULAR n-gon, 360/n is the measure of an exterior angle. This makes sense since we know the sum of the ext. angles is 360, and there are n ext. angles.
Given this info, to find the measure of the (or any) interior angle of this same REGULAR n-gon, we need only subtract the exterior angle measure (360/n) from 180. Remember, interior angles and exterior angles at the same vertex are supplementary by definition.
Hence, the measure on an int angle of a REGULAR n-gon is 180-(360/n).
Capeesh??
YES!! and Capeesh! :)
ReplyDeletePhew!!
ReplyDeleteThis is a question for #6-2:
ReplyDeleteWhat do they mean on #5? "What are____"? What do they mean "What are", am I supposed to name their relationship or state that they are transversals/segments?
Sorry if I'm missing something here if it sounds stupid: But what could I put for c. on #4? I know it doesn't state anywhere in the definition of a parallelogram that it has opposite congruent sides...
ReplyDelete