Author 
Message 
TAGS:

Hide Tags

Director
Joined: 07 Jun 2004
Posts: 609
Location: PA

The measures of the interior angles in a polygon are [#permalink]
Show Tags
03 Feb 2011, 14:45
7
This post received KUDOS
17
This post was BOOKMARKED
Question Stats:
68% (01:52) correct 32% (04:42) wrong based on 315 sessions
HideShow timer Statistics
The measures of the interior angles in a polygon are consecutive integers. The smallest angle measures 136 degrees. How many sides does this polygon have? A) 8 B) 9 C) 10 D) 11 E) 13
Official Answer and Stats are available only to registered users. Register/ Login.
_________________
If the Q jogged your mind do Kudos me : )



Math Expert
Joined: 02 Sep 2009
Posts: 43894

Re: Geometry [#permalink]
Show Tags
03 Feb 2011, 15:02
4
This post received KUDOS
Expert's post
6
This post was BOOKMARKED
rxs0005 wrote: The measures of the interior angles in a polygon are consecutive integers. The smallest angle measures 136 degrees. How many sides does this polygon have?
A) 8 B) 9 C) 10 D) 11 E) 13 Sum of Interior Angles of a polygon is \(180(n2)\) where \(n\) is the number of sides (so is the number of angles); We are told that the smallest angle is 136 degrees, the next will be 136+1 degrees, ..., and the largest one, \(n_{th}\) angle, will be \(136+(n1)\) degrees. The sum of the \(n\) consecutive integers (the sum of \(n\) angles) is given by \(\frac{first+last}{2}*# \ of \ terms=\frac{136+(136+n1)}{2}*n=\frac{271+n}{2}*n\); So we have that \(180(n2)=\frac{271+n}{2}*n\) > \(360(n2)=(271+n)*n\), now try the answer choices: in order RHS to end with zero (as LHS is because of 360) then \(n\), out of the options listed, could be either 10 or 9, \(n=9\) fits. Answer: B.
_________________
New to the Math Forum? Please read this: Ultimate GMAT Quantitative Megathread  All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 7955
Location: Pune, India

Re: Geometry [#permalink]
Show Tags
03 Feb 2011, 20:05
8
This post received KUDOS
Expert's post
1
This post was BOOKMARKED
rxs0005 wrote: The measures of the interior angles in a polygon are consecutive integers. The smallest angle measures 136 degrees. How many sides does this polygon have?
A) 8 B) 9 C) 10 D) 11 E) 13 When I look at this question, I say, "I know how to find the Interior angle of a regular polygon. But this is not a regular polygon since it has angles 136, 137, 138, 139, 140, 141, 142 ....etc." Also, Interior angle of 8 sided regular polygon = 180*6/8 = 135 Interior angle of 9 sided regular polygon = 180*7/9 = 140 Interior angle of 10 sided regular polygon = 180*8/10 = 144 etc The average of the given angles can only match 140 (such that effectively, all the angles are 140) Hence, the polygon must have 9 sides. Remember, capitalize on what you know.
_________________
Karishma Veritas Prep  GMAT Instructor My Blog
Get started with Veritas Prep GMAT On Demand for $199
Veritas Prep Reviews



Senior Manager
Joined: 28 Aug 2010
Posts: 257

Re: Geometry [#permalink]
Show Tags
04 Feb 2011, 09:47
Karishma, Can you explain in more details when you say "The average of the given angles can only match 140 (such that effectively, all the angles are 140)". I did not get your point. Thanks
_________________
Verbal:newtotheverbalforumpleasereadthisfirst77546.html Math: newtothemathforumpleasereadthisfirst77764.html Gmat: everythingyouneedtoprepareforthegmatrevised77983.html  Ajit



Math Forum Moderator
Joined: 20 Dec 2010
Posts: 1945

Re: Geometry [#permalink]
Show Tags
04 Feb 2011, 10:20
4
This post received KUDOS
1
This post was BOOKMARKED
Sum of interior angles for n sides: (n2)*180 And each interior angle are increasing by 1. This can be written as : 136+137+138+139+... = 136+(136+1)+(136+2)+(136+3)+... For n sides 136n+(1+2+3+4....+n1) 136n+(n1)n/2  Sum of n1 natural number Equate: 136n+(n1)n/2 = (n2)*180 n^289n+720=0 n^280n9n+720=0 n(n80)9(n80)=0 (n9)(n80)=0 n=9 n=80 Since 9 is one of the options. Ans: "B"
_________________
~fluke
GMAT Club Premium Membership  big benefits and savings



Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 7955
Location: Pune, India

Re: Geometry [#permalink]
Show Tags
04 Feb 2011, 11:56
ajit257 wrote: Karishma, Can you explain in more details when you say "The average of the given angles can only match 140 (such that effectively, all the angles are 140)". I did not get your point. Thanks Sum of interior angles of a polygon = (n2)*180 (not necessarily regular polygon) Why? See the figure below: Attachment:
Ques2.jpg [ 5.14 KiB  Viewed 22655 times ]
A 6 sided polygon can be split into 4 triangles each of which has a sum of interior angles 180 degrees. An n sided polygon can be split into n  2 triangles. When the polygon is regular, each angle is same so the sum is divided by the number of sides to get the measure of each angle e.g. in a regular hexagon, each interior angle = 4*180/6 = 120 degrees. Now if I have a hexagon whose angles are 115, 117, 119, 121, 123 and x, what will be the angle x? We can see it in two ways  1. The sum of all angles should be 4*180 = 720 So 115 + 117 + 119 + 121 + 123 + x = 720 or x = 125 2. The average of the angles should be 120. (Since the sum of the angles is 720 and there are 6 sides) 119 and 121 average out as 120. (119 is 1 less than 120 and 121 is 1 more than 120) 117 and 123 average out as 120. So 115 and x should average out as 120 too. Therefore, x should be 125. In the question, the average of the given angles of the polygon can only be 140. So it must have 9 sides. To confirm, 136, 137, 138, 139, 140, 141, 142, 143, 144  9 angles with average 140. So the polygon must have 9 sides. (It cannot be 144 or anything else because 10 angles (136, 137, 138, 139, 140, 141, 142, 143, 144, 145) will not give an average of 144)
_________________
Karishma Veritas Prep  GMAT Instructor My Blog
Get started with Veritas Prep GMAT On Demand for $199
Veritas Prep Reviews



Senior Manager
Joined: 08 Nov 2010
Posts: 385
WE 1: Business Development

Re: Geometry [#permalink]
Show Tags
05 Feb 2011, 08:01
tough one. simple solution, but its not an ez one. am i right?
_________________
GMAT Club Premium Membership  big benefits and savings



Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 7955
Location: Pune, India

Re: Geometry [#permalink]
Show Tags
06 Feb 2011, 06:25
144144 wrote: tough one. simple solution, but its not an ez one. am i right? Actually, I wouldn't say it is very tough if we know that the sum of interior angles of a polygon is 180(n 2) (I explained above why it is so). We can also use a very straight forward but long approach. Interior angles of the given polygon: 136, 137, 138, 139, 140, 141, 142, 143.... Using options: If the polygon had 8 sides, it would have had 8 interior angles. Their sum: (82)180 = 1080 Sum of 8 angles: 136 +137 + 138 + 139 + 140 + 141 + 142 + 143 is more than 1080 hence this polygon does not have 8 sides. If the polygon had 9 sides, it would have had 9 interior angles. Their sum: (92)180 = 1260 Sum of 9 angles: 136 +137 + 138 + 139 + 140 + 141 + 142 + 143 + 144 is 1260. Hence this polygon does have 9 sides.
_________________
Karishma Veritas Prep  GMAT Instructor My Blog
Get started with Veritas Prep GMAT On Demand for $199
Veritas Prep Reviews



Math Expert
Joined: 02 Sep 2009
Posts: 43894

Re: Geometry [#permalink]
Show Tags
06 Feb 2011, 06:45
144144 wrote: tough one. simple solution, but its not an ez one. am i right? I wouldn't say it's very hard either, though I wouldn't recommend trial and error in this cases: ... Calculating the sum of the interior angles of a polygon with 7 sides then checking whether 7 consecutive integers starting from 136 add up to that value; Calculating the sum of the interior angles of a polygon with 8 sides then checking whether 8 consecutive integers starting from 136 add up to that value; Calculating the sum of the interior angles of a polygon with 9 sides then checking whether 9 consecutive integers starting from 136 add up to that value; Whereas equating two formulas and working on answer choices should give an answer in less time: \(180(n2)=\frac{271+n}{2}*n\) > \(360(n2)=(271+n)*n\) > \(n=9\).
_________________
New to the Math Forum? Please read this: Ultimate GMAT Quantitative Megathread  All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



Senior Manager
Joined: 08 Nov 2010
Posts: 385
WE 1: Business Development

Re: Geometry [#permalink]
Show Tags
06 Feb 2011, 13:03
1
This post received KUDOS
ye... well i guess i need to work on my geometry more so the way i process these things will be better... thanks both as usual.
_________________
GMAT Club Premium Membership  big benefits and savings



VP
Status: Been a long time guys...
Joined: 03 Feb 2011
Posts: 1372
Location: United States (NY)
Concentration: Finance, Marketing
GPA: 3.75

Re: Geometry [#permalink]
Show Tags
09 Aug 2012, 09:08
1
This post received KUDOS
Hey guys...please let me know whether this method is correct or not. the stem says that the angles are consecutive integers, 136 being the smallest one. Also going by the formula of sum of interior angles, we know atleast this much that sum of angles cant be number like 729, 847, 653, 542 but it can be a number of the form xx0. So going by this knowledge, if the smallest angle is 136 then the sum of angles can only be of the form xx0 when the largest angle is 144. That is 9 sides. Yeah the largest angle could be 154, 164 but in that case the number of sides must be 19 and 29 respectively. Since the largest option is 13, hence the answer is 9. Please correct me if I am wrong.
_________________
Prepositional Phrases ClarifiedElimination of BEING Absolute Phrases Clarified Rules For Posting www.UnivScholarships.com



NonHuman User
Joined: 09 Sep 2013
Posts: 13776

Re: The measures of the interior angles in a polygon are [#permalink]
Show Tags
11 Feb 2014, 08:52
Hello from the GMAT Club BumpBot! Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up  doing my job. I think you may find it valuable (esp those replies with Kudos). Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________
GMAT Books  GMAT Club Tests  Best Prices on GMAT Courses  GMAT Mobile App  Math Resources  Verbal Resources



Current Student
Joined: 06 Sep 2013
Posts: 1954
Concentration: Finance

Re: The measures of the interior angles in a polygon are [#permalink]
Show Tags
05 Apr 2014, 15:32
I dunno if this method is valid, please advice But a polygon has sum of sides equal to (n2)*180, thus sum will always have 0 as units digit. Therefore if we begin with 136+137 etc... only 9 fits the bill since 4 is not a valid answer choice Thanks! Cheers J



Manager
Joined: 17 Mar 2014
Posts: 160

Re: The measures of the interior angles in a polygon are [#permalink]
Show Tags
03 Nov 2014, 07:36
Bunuel wrote: rxs0005 wrote: The measures of the interior angles in a polygon are consecutive integers. The smallest angle measures 136 degrees. How many sides does this polygon have?
A) 8 B) 9 C) 10 D) 11 E) 13 Sum of Interior Angles of a polygon is \(180(n2)\) where \(n\) is the number of sides (so is the number of angles); We are told that the smallest angle is 136 degrees, the next will be 136+1 degrees, ..., and the largest one, \(n_{th}\) angle, will be \(136+(n1)\) degrees. The sum of the \(n\) consecutive integers (the sum of \(n\) angles) is given by \(\frac{first+last}{2}*# \ of \ terms=\frac{136+(136+n1)}{2}*n=\frac{271+n}{2}*n\); So we have that \(180(n2)=\frac{271+n}{2}*n\) > \(360(n2)=(271+n)*n\), now try the answer choices: in order RHS to end with zero (as LHS is because of 360) then \(n\), out of the options listed, could be either 10 or 9, \(n=9\) fits.
Answer: B. Hi Bunnel, In above post, i could not get " in order RHS to end with zero (as LHS is because of 360) then \(n\), out of the options listed, could be either 10 or 9, \(n=9\) fits'. Why are you saying that LHS is 0. Regards, Ammu



Math Expert
Joined: 02 Sep 2009
Posts: 43894

Re: The measures of the interior angles in a polygon are [#permalink]
Show Tags
03 Nov 2014, 07:39
ammuseeru wrote: Bunuel wrote: rxs0005 wrote: The measures of the interior angles in a polygon are consecutive integers. The smallest angle measures 136 degrees. How many sides does this polygon have?
A) 8 B) 9 C) 10 D) 11 E) 13 Sum of Interior Angles of a polygon is \(180(n2)\) where \(n\) is the number of sides (so is the number of angles); We are told that the smallest angle is 136 degrees, the next will be 136+1 degrees, ..., and the largest one, \(n_{th}\) angle, will be \(136+(n1)\) degrees. The sum of the \(n\) consecutive integers (the sum of \(n\) angles) is given by \(\frac{first+last}{2}*# \ of \ terms=\frac{136+(136+n1)}{2}*n=\frac{271+n}{2}*n\); So we have that \(180(n2)=\frac{271+n}{2}*n\) > \(360(n2)=(271+n)*n\), now try the answer choices: in order RHS to end with zero (as LHS is because of 360) then \(n\), out of the options listed, could be either 10 or 9, \(n=9\) fits.
Answer: B. Hi Bunnel, In above post, i could not get " in order RHS to end with zero (as LHS is because of 360) then \(n\), out of the options listed, could be either 10 or 9, \(n=9\) fits'. Why are you saying that LHS is 0. Regards, Ammu Because the left hand side (LHS) is 360(n2) = 10*(36(n2)) = 10*integer = something with the units digit of 0. Does this make sense?
_________________
New to the Math Forum? Please read this: Ultimate GMAT Quantitative Megathread  All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



Manager
Joined: 17 Mar 2014
Posts: 160

Re: The measures of the interior angles in a polygon are [#permalink]
Show Tags
03 Nov 2014, 08:22
Hi Bunnel,
In above post, i could not get " in order RHS to end with zero (as LHS is because of 360) then \(n\), out of the options listed, could be either 10 or 9, \(n=9\) fits'. Why are you saying that LHS is 0.
Regards, Ammu[/quote]
Because the left hand side (LHS) is 360(n2) = 10*(36(n2)) = 10*integer = something with the units digit of 0.
Does this make sense?[/quote]
I understood it now. Thank you Actually I had misinterpreted it . I thought you were saying "RHS to end with zero >(271+n)*n= 0" . Its my Bad. Thanks for clarification.
Regards, Ammu



NonHuman User
Joined: 09 Sep 2013
Posts: 13776

Re: The measures of the interior angles in a polygon are [#permalink]
Show Tags
26 Nov 2015, 04:06
Hello from the GMAT Club BumpBot! Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up  doing my job. I think you may find it valuable (esp those replies with Kudos). Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________
GMAT Books  GMAT Club Tests  Best Prices on GMAT Courses  GMAT Mobile App  Math Resources  Verbal Resources



Board of Directors
Joined: 17 Jul 2014
Posts: 2736
Location: United States (IL)
Concentration: Finance, Economics
GPA: 3.92
WE: General Management (Transportation)

The measures of the interior angles in a polygon are [#permalink]
Show Tags
07 Oct 2016, 06:10
rxs0005 wrote: The measures of the interior angles in a polygon are consecutive integers. The smallest angle measures 136 degrees. How many sides does this polygon have?
A) 8 B) 9 C) 10 D) 11 E) 13 oh..nice question.. i haven't reviewed the geometry and i took some additional time to solve it... total sum of the interior angle of a polygon is 180(n2) i started by trial... suppose we have 10 sides, then 180(102) = 180*8 = 1440 sum of the angles in this case... i solved this way  10*136 = 1360 then, since we have consecutive numbers, let's find the sum of the numbers from 1 to 9. best way, to use the formula n(n+1)/2 > 9*10/2 = 45. 1360+45 = 1405. the sum doesn't match with what we need. in this case, we know for sure that C, D, and E are not correct. we are left with 2 answers, and we need to test one more time... 9 sides > 180(92) = 1260 now..don't need to go all the way through with what we did earlier... sum of the consecutive numbers is 1405, and the greatest number is 136+9 or 145. 1405  145 = 1260. we have a match! answer is B. woah, so many more ways to solve it here! awesome!



Senior SC Moderator
Joined: 14 Nov 2016
Posts: 1277
Location: Malaysia

The measures of the interior angles in a polygon are [#permalink]
Show Tags
30 Mar 2017, 17:37
rxs0005 wrote: The measures of the interior angles in a polygon are consecutive integers. The smallest angle measures 136 degrees. How many sides does this polygon have?
A) 8 B) 9 C) 10 D) 11 E) 13 OFFICIAL SOLUTION We are told that the smallest angle measures 136 degreesthis is the first term in the consecutive set. If the polygon has S sides, then the largest anglethe last term in the consecutive setwill be (S  1) more than 136 degrees. The sum of consecutive integers = (Average Term) * (# of Term)= \(\frac{First + Last}{2}\) * (# of Terms). Given that there are S terms in the set, we can plug in for the first and last term as follows: \(\frac{136 + 136 + (S1)}{2} * S\) = sum of the angles in the polygon. We also know that the sum of the angles in a polygon = 180 (S2) where S represents the number of sides. Therefore: \(180(S2) = \frac{136+136+(S1)}{2} * S\). We can solve for S by crossmultiplying and simplifying as follows: \(2(180)(S2) = [272 + (S1)] S\) \(360S  720 = (271 + S)S\) \(360S  720 = 271S + S^2\) \(S^2  89S + 720 = 0\) A look at the answer choices tells you to try (S  8), (S  9), or (S  10) in factoring. As it turns out (S9)(S80)=0, which means S can be 9 or 80. However S cannot be 80 because this creates a polygon with angles greater than 180. Therefore S equals 9; there are 9 sides in the polygon. The correct answer is B.
_________________
"Be challenged at EVERY MOMENT."
“Strength doesn’t come from what you can do. It comes from overcoming the things you once thought you couldn’t.”
"Each stage of the journey is crucial to attaining new heights of knowledge."
Rules for posting in verbal forum  Please DO NOT post short answer in your post!



Intern
Joined: 15 Jun 2013
Posts: 47
GPA: 3.82

Re: The measures of the interior angles in a polygon are [#permalink]
Show Tags
31 Mar 2017, 06:50
Bunuel wrote: rxs0005 wrote: The measures of the interior angles in a polygon are consecutive integers. The smallest angle measures 136 degrees. How many sides does this polygon have?
A) 8 B) 9 C) 10 D) 11 E) 13 Sum of Interior Angles of a polygon is \(180(n2)\) where \(n\) is the number of sides (so is the number of angles); We are told that the smallest angle is 136 degrees, the next will be 136+1 degrees, ..., and the largest one, \(n_{th}\) angle, will be \(136+(n1)\) degrees. The sum of the \(n\) consecutive integers (the sum of \(n\) angles) is given by \(\frac{first+last}{2}*# \ of \ terms=\frac{136+(136+n1)}{2}*n=\frac{271+n}{2}*n\); So we have that \(180(n2)=\frac{271+n}{2}*n\) > \(360(n2)=(271+n)*n\), now try the answer choices: in order RHS to end with zero (as LHS is because of 360) then \(n\), out of the options listed, could be either 10 or 9, \(n=9\) fits. Answer: B. This is the best and the most proper way of doing this task but may I suggest another way: the sum of all angles must be divisible by 10. Let's sum last figures of consecutive angles until we'll get 0 (but the number of angles should be more then 7): we have angles #1  136, #2  137, #3  138, 139, 140, #6  141, #7  142. So 6+7=13, 8 (next angle) + 3 (last figure in 13)=11, 1>0. 9+1=0 great! but this is only angle #4, and we don't have this in our answers so we move on. 0+0=0, but see above. 0+1=1, 1+2(angle #7)=3. Now we are in the area of answer therefore we should be more accurate. So 3 (angle #8  143) +3=6. 4+6=0! great. This correspondes to the angle #9. Just to check if we don't have other answers for angles 1013: 0+5=5>0 5+6=11,1>0 1+7=8>0, 8+8=16, 6>0. Here we go, the only possible answer is B. This is pretty long description but this is definitely is not time consuming way.




Re: The measures of the interior angles in a polygon are
[#permalink]
31 Mar 2017, 06:50



Go to page
1 2
Next
[ 22 posts ]



