GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 22 Sep 2019, 03:18 ### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

#### Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here. ### Request Expert Reply # A pentagon with 5 sides of equal length and 5 interior angles of equal

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics
Author Message
TAGS:

### Hide Tags

Intern  B
Joined: 26 Mar 2019
Posts: 3
Re: A pentagon with 5 sides of equal length and 5 interior angles of equal  [#permalink]

### Show Tags

Solution for Statement 1:

Area (Circle) = 16П
--> r=4
--> Circumference =8П=~8*3.14, which is less than 26

Area of the pentagon have to be less than circumference of the circle --> it is less than 26.
Intern  B
Joined: 18 Sep 2016
Posts: 1
Location: India
WE: Operations (Energy and Utilities)
Re: A pentagon with 5 sides of equal length and 5 interior angles of equal  [#permalink]

### Show Tags

Essentially the question is to test if given statements are sufficient enough to find the side of the pentagon.

Do you have sufficient data to determine the side of the pentagon? The perimeter can be >26cm(YES) and for all other cases, NO. In each case you have a definite answer. There is no ambiguity.

Statement1: area of circle is 16pi sq.cm. So radius of the circle is 4cm. Since the pentagon is cyclic, the distance between vortex of pentagon and center of the circle is 4cm. As all sides of pentagon are of equal length, each side forms an isosceles triangle with center as the 3rd vortex (subtends 72 degrees, i.e., =360/5). Hence, length (unique not multiple) of the side of the pentagon can be determined. Doesn't matter if it is >26cm or <26cm or =26cm.
Therefore Statement1 is sufficient.

Statement2: All diagonals are of equal length, because pentagon is cyclic and each side is of equal length. Since two diagonals (8cm each) subtend 36 degrees (angle subtended by a chord in the major arc is half the angle subtended by the chord at the center), one can determine the side (unique not multiple) of the pentagon. Hence this statement too is sufficient. Doesn't matter if it is >26cm or <26cm or =26cm.

IMO no complex math or trigonometry is required to solve this question.
Intern  B
Joined: 26 Mar 2019
Posts: 3
Re: A pentagon with 5 sides of equal length and 5 interior angles of equal  [#permalink]

### Show Tags

sassyisnasty wrote:
Essentially the question is to test if given statements are sufficient enough to find the side of the pentagon.

Do you have sufficient data to determine the side of the pentagon? The perimeter can be >26cm(YES) and for all other cases, NO. In each case you have a definite answer. There is no ambiguity.

Statement1: area of circle is 16pi sq.cm. So radius of the circle is 4cm. Since the pentagon is cyclic, the distance between vortex of pentagon and center of the circle is 4cm. As all sides of pentagon are of equal length, each side forms an isosceles triangle with center as the 3rd vortex (subtends 72 degrees, i.e., =360/5). Hence, length (unique not multiple) of the side of the pentagon can be determined. Doesn't matter if it is >26cm or <26cm or =26cm.
Therefore Statement1 is sufficient.

Statement2: All diagonals are of equal length, because pentagon is cyclic and each side is of equal length. Since two diagonals (8cm each) subtend 36 degrees (angle subtended by a chord in the major arc is half the angle subtended by the chord at the center), one can determine the side (unique not multiple) of the pentagon. Hence this statement too is sufficient. Doesn't matter if it is >26cm or <26cm or =26cm.

IMO no complex math or trigonometry is required to solve this question.

Statement 2 says the diagonal is less than 8, not equal. Hence the “unique” side cannot be determined. Actual computation is required to solve this.

Posted from my mobile device
Manager  B
Joined: 24 Jun 2019
Posts: 108
Re: A pentagon with 5 sides of equal length and 5 interior angles of equal  [#permalink]

### Show Tags

All solutions for (2) (Including OG) seem to be too cumbersome for the actual test.

I prefer Ians solution amongst all that have been posted.

However, here is my take on how I would guess the solution on the actual test:

See attached diagram.

Essentially I would try to eyeball it.

I would assume that diagonal is 8 and get the maximum possible perimeter. If MAX is less than 26... then it is definitely less than 26. If MAX is greater than 26... then it would be insufficient.

So, If diagonal = 8.... Base of triangle would be 4.

height is less than 4 (by eyeballing).... so lets assume it is 3ish

then it becomes a 3-4-5 right triangle, and side of pentagon will be 5ish

so MAX perimeter would be 25ish. hence sufficientISH?.

Attachment: ApplicationFrameHost_2019-07-11_06-30-51.png [ 104.82 KiB | Viewed 387 times ]

I Know that this is very far from fool-proof. The margin of error between 25 and 26 is WAY TOO CLOSE. I would be much more comfortable if the question had something like 30 instead of 26.

I have practically no knowledge of advanced stuff like trigonometry / golden ration etc mentioned above. So using tools of mere mortals, this is the best I could come up with that is feasible in the actual test. :p

PS: On Googling i found that the 2 angles of 3-4-5 right triangle are approximately 36.87° and 53.13° .... which is pretty close.

Please let me know if I have made any mistakes!
Manager  G
Joined: 12 Jul 2017
Posts: 206
GMAT 1: 570 Q43 V26 GMAT 2: 660 Q48 V34 A pentagon with 5 sides of equal length and 5 interior angles of equal  [#permalink]

### Show Tags

shobhitkh wrote:
Bunuel wrote:
A pentagon with 5 sides of equal length and 5 interior angles of equal measure is inscribed in a circle. Is the perimeter of the pentagon greater than 26 centimeters?

(1) The area of the circle is 16π square centimeters.
(2) The length of each diagonal of the pentagon is less than 8 centimeters.

DS75271.01
OG2020 NEW QUESTION

Hi Bunuel and VeritasKarishma, Can anyone of you please provide a good explanation for this question? Thanks

Here is the thing about this question and why I took time to get back - It is an OG problem so I expected a good, quick solution. I did invest a bit of time in this but unfortunately couldn't come up with anything satisfactory. Then I checked with another expert but still no luck.
The fastest way was trigonometry though I had to approximate the value of cos 36 about which I wasn't happy at all.
The better solution is the one Ian suggested above (using similar triangles). It just seems a bit too elaborate as the only solution to an official problem.
Also, I certainly cannot rely on knowing some obscure fact about pentagon to solve an official question.
All in all, I hope this was an experimental question that was later discarded, but found its way into the OG. In case we find some more such questions, we might need to reassess our understanding of the official problems (but I will wait for @Bunuel's input before making my mind about it)

I solved this question with this approach.

Statement 1 - since it's a regular polygon and inscribed in a circle, so max it can be that it can take equidistance from the center of circle i.e radius of circle. If we visualise it, then we divide the pentagon into 5 triangles with angles - 59,59,72 ( since it's an isosceles triangle)

Now this is POSSIBLY close to an equilateral triangle ( since the angles are close length has to be close) so the sides of polygon could not go beyond 5.2 (SUFFICIENT)

Statement 2 - When we say that the POLYGON's diagonal is NOT bigger than 8 then we mean that DEFINETELY the radius of circle is < 4. Based on our findings from statement 1, we know that the side will never exceed 5.2 ( SUFFICIENT)

So answer is D.

Is there any FLAW with this line of reasoning? I thought the question to be fairly simple Regards,
Rishav
Manager  S
Joined: 29 Dec 2018
Posts: 80
Location: India
WE: Marketing (Real Estate)
Re: A pentagon with 5 sides of equal length and 5 interior angles of equal  [#permalink]

### Show Tags

IanStewart wrote:
Statement 1 is immediately sufficient, because if we know exactly how big the circle is, we know exactly how big the pentagon is, so we can compute its exact perimeter and answer the question.

If there's an extremely fast way to prove Statement 2 is sufficient as written, I don't see it (the solution someone else posted above isn't correct), though it would be much easier with slightly different numbers. If you draw the three diagonals I draw in the picture below, and label all of the relevant angles, you'll see that the triangles I've labeled '1', '2' and '3' all are 36-36-108 triangles. Triangles 1 and 2 are identical ('congruent' in math-speak) because they have the same angles, and they share a side (opposite the 108 degree angle in both). So the sides I've labeled with blue tickmarks are all equal. Triangle 3 is similar to, but smaller than, triangles 1 and 2.

If we call the side of the pentagon 's' and the diagonal 'd', then the sides of triangle 1 are just s, s, and d. Notice that in triangle 3, the two equal sides, the sloping ones, are both d-s in length (because each is just part of a sloping diagonal of the pentagon, but not counting the 's' part that belongs to triangle 2, so we just need to subtract that). So the sides of triangle 3 are d-s, d-s, and s. Since triangle 1 and triangle 3 are similar, their sides are in the same ratio. So if we divide the longer side by the shorter side in triangle 1, we get the same ratio as we get when we do the same for triangle 3:

d/s = s/(d-s)

And rewriting this:

d^2 - sd = s^2
d^2 = s^2 + sd

Now say d = 8, the largest the diagonal is allowed to be according to Statement 2. Then plugging in,

64 = s^2 + 8s

and now you can see if you just plug in s = 5, the right side of this equation is very slightly too big. So s cannot be 5 -- the equation will only work if s is slightly less than 5. And if the diagonal is shorter than 8, then s will need to be even smaller of course. In any case, the perimeter is certainly less than 25. So Statement 2 is also sufficient, and the answer is D.

For those who love exact calculations and are satisfied only when they get to some approximation value (NOT RECOMMENDED ON A TEST DAY)

If we assume diagonal d=7
then $$d^2 = s^2 + sd$$
$$7^2 = s^2 + 7s$$
s = 4.326

If we take the diagonal d=8
then $$d^2 = s^2 + sd$$
$$8^2 = s^2 + 8s$$
s = 4.944

As IanStewart explained earlier in no case can the perimeter be more than 26 hence each statement alone is sufficient - answer choice D

For more curious math lovers kindly refer https://sites.math.washington.edu/~king ... swers.html
_________________
Keep your eyes on the prize: 750
Manager  S
Joined: 29 Oct 2015
Posts: 199
A pentagon with 5 sides of equal length and 5 interior angles of equal  [#permalink]

### Show Tags

Bunuel wrote:
A pentagon with 5 sides of equal length and 5 interior angles of equal measure is inscribed in a circle. Is the perimeter of the pentagon greater than 26 centimeters?

(1) The area of the circle is 16π square centimeters.
(2) The length of each diagonal of the pentagon is less than 8 centimeters.

DS75271.01
OG2020 NEW QUESTION

Let me explain why the statement 1 is sufficient.

Pi * r ^2 = 16 * pi
r^2 = 16
r = 4 ... (where r = radius)

For a regular hexagon , the radius is equal to the side of each side of the hexagon.
We can prove it too...
Please refer to the attachment of the image...

In triangle ABC and DBC ,
BC = common..
AC = CD = radius ..
AB = BD = side of a regular pentagon
Triangle ABC is congruent to triangle DBC.

So angle ABC = Angle DBC

Since angle ABD = 120 degree ... ( interior angle of a hexagon )

Angle ABC = Angle DBC = 60 degree
Now Angle ABC = Angle CAB = Angle BCA = 60 degree
So Triangle ABC is an equilateral triangle and the radius is equal to the side of each side of the hexagon.

So if the figure was a hexagon its perimeter would have been = 6 * 4 = 24 cms... ( each side is 4 cms = radius = 4 cms)
Now 24 cm s is already lesser than 26 cms.
Since our figure is a pentagon its perimeter will be even smaller than that of a hexagon and hence will be smaller than 24 cms and hence will definitely be lesser than 26 cms...

So the answer is NO .
Option A is sufficient.

In my next post , I will write why option B is also sufficient.
Please give me KUDO s if you liked my explanation.
Please find the diagram attached herewith.
Please let me know if there is any doubt/discrepancy anywhere.
Attachments 20190811_071106.jpg [ 2.75 MiB | Viewed 199 times ] 20190811_071106.jpg [ 2.75 MiB | Viewed 241 times ]

Originally posted by sayan640 on 10 Aug 2019, 18:49.
Last edited by sayan640 on 11 Aug 2019, 02:28, edited 3 times in total.
Intern  B
Joined: 09 Jun 2018
Posts: 1
Re: A pentagon with 5 sides of equal length and 5 interior angles of equal  [#permalink]

### Show Tags

My approach for Solution for Statement 2:

Given: Diagonal of pentagon is less than 8 centimetres. We can safely say that the diagonal of the circle will be greater than the diagonal of pentagon. Let's take the smallest such possible value for diagonal of circle, ie. 8 cm. Hence the perimeter of circle is 8x3.14 = 25 ish. Since pentagon is inscribed inside the circle, its perimeter will be lesser than that of the circle. Hence the perimeter of the pentagon will be lesser than 26.
Manager  S
Joined: 29 Oct 2015
Posts: 199
A pentagon with 5 sides of equal length and 5 interior angles of equal  [#permalink]

### Show Tags

Bunuel wrote:
A pentagon with 5 sides of equal length and 5 interior angles of equal measure is inscribed in a circle. Is the perimeter of the pentagon greater than 26 centimeters?

(1) The area of the circle is 16π square centimeters.
(2) The length of each diagonal of the pentagon is less than 8 centimeters.

DS75271.01
OG2020 NEW QUESTION

Please find the diagram attached herewith.
Lets see why option B is also sufficient.
Triangle ABC and Triangle BCF and Trinagle FDE are 36 , 36 , 108 degree triangle.
( I explained the reason in the diagram attached )
So I am keeping it brief here.
Angle BAC = 108 degree ( interior angle of a pentagon)
Since AB = AC (sides of a regular pentagon are equal ) , Angle ABC = angle ACB = (180 - 108) / 2 = 72 / 2 = 36 degree
ANgle BDE = 108 degree ( interior angle of a pentagon)
Angle DBE = Angle DFB = (180 degree - angle BDE ) / 2 = (180 - 108 )/ 2 = 72 / 2 = 36 degree
Angle CBE = 108 - ( angle ABC + angle DBE) = 108 - 72 = 36 degree
Similarly , Angle BCD = 36 degree
ANgle BFC = 180 - ( CBE + BCD) = 180 - ( 36 + 36 ) = 108 degree

Say , The length of each diagonal = c
So BE = c

Now Triangle ABC and traingle BCF are congruent .( because they have two angles equal and one side common So ASA condition satisifes)
ANgle ABC = ANgle CBE = 36 degree
Angle ACB = angle BCF = 36 degree
SIde BC common
So Triangle ABC and traingle BCF are congruent.
SO the corresponding sides AB = BF
and AC = CF

Say AB = a
So , BF = a
Now , FE = BE - BF = c - a

SImilarly FD = c - a
DE = a

Now Triangles ABC and triangle FDE are similar. ( All the angles are same for them...36 , 36 , 108 )
So corresponding sides are of same proportion...
So , c/a = a / c-a
so c^2 - a*c = a^2
For the maximum length of the diagonal i.e 8 cm,

64 - 8*a = a^2
Now Say , side of the pentagon = 5 cm
it does not satisfy the equation above as (64 - 8 * 5 ) is smaller than 5^2 = 25 cm

So 'a' has to a bit smaller than 5 cm.
Now if the diagonal is less than 8 cm , then it will be even less than 64 . And side i.e the value of 'a' will be even smaller.
Even if the side was 5 cm , the perimeter of the pentagon would have been 5 *5 = 25 cm , which is smaller than 26 cm.
If the sides are even smaller , the perimeter will be even smaller and hence less than 26 cm.

Hence , Option B is also sufficient.

Please give me KUDO s if you liked my explanation.
Attachments 20190811_154626.jpg [ 3.11 MiB | Viewed 198 times ]

Intern  B
Joined: 24 Jul 2019
Posts: 36
A pentagon with 5 sides of equal length and 5 interior angles of equal  [#permalink]

### Show Tags

Is it really realistic to solve such kind of question in a Test situation given that one has no Math / Physics background and is not a ultra High IQ person?

I got that question right now as "medium-quant" and it blew my mind. Pentagons weren't even covered in my Study guides. For people with a business BsC background this is a completely overkill.

To formulate it better: Does it make sense to focus on such kind of questions when in an early study phase? Usually the medium-quant I practice everyday flow much easier. Given that there were other geometry question where you basically just had to solve a pythagoran theorem I don't really get how they rate the questions.
_________________
Don't wait for the opportunity to knock.
Go kick in the door. A pentagon with 5 sides of equal length and 5 interior angles of equal   [#permalink] 19 Aug 2019, 01:21

Go to page   Previous    1   2   [ 30 posts ]

Display posts from previous: Sort by

# A pentagon with 5 sides of equal length and 5 interior angles of equal

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics

 Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne  