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

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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.

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Re: A pentagon with 5 sides of equal length and 5 interior angles of equal  [#permalink]

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1
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.
Attachments Screen Shot 2019-04-28 at 3.02.23 PM.png [ 31.01 KiB | Viewed 10674 times ]

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Re: A pentagon with 5 sides of equal length and 5 interior angles of equal  [#permalink]

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I agree with how Ian is looking at this – definitely prefer this approach over analyzing arc lengths for statement (1) and solving the quadratic equation $$d^{2} – ds – s^{2} = 0$$ with the quadratic formula for statement (2), as the official solution suggests.

Here are a few more thoughts on how to get through this beast!

Statement (1) pins down the radius of the circle, so there’s only one way to draw the pentagon, and the perimeter will be definitively greater than 26cm or not. Thankfully, we don’t have to determine whether it is or isn’t – statement (1) is sufficient.

Statement (2) demands that we determine the relationship between the side s and diagonal d in a regular pentagon. This will help us figure out what d < 8cm tells us about s. That’s driving the analysis. Of course, to figure out whether the perimeter is > 26cm, we need to figure out whether s > 5.2.

First, this definitely requires some good sketching! In the figure below, we label what we can to piece together a relationship between d and s. Here’s how you can determine the angles in the diagram.

1. There are (5 – 2) x 180˚ = 540˚ in any pentagon. In this case, since all interior angles are equal, each interior angle = 108˚.
2. Any central angle formed by connecting adjacent vertices of the pentagon to the center of the circle must be 1/5 of 360˚ = 72˚.
3. Each of the angles labeled 36˚ is an inscribed angle, and therefore equal to half of the corresponding 72˚ central angle.
4. Of course, we also need to use the fact that triangles have a total of 180˚, and opposite angles are equal.

Next, because the angles in triangle 1 match the angles in triangle 2, these two triangles are similar, and because the ratio of their sides is 1:1, these two triangles are congruent. Because the angles in triangle 3 match the angles in triangles 1 and 2, triangle 3 is similar to triangles 1 and 2. As a result, corresponding sides are in a constant ratio, and we get

$$\frac{d}{s} = \frac{s}{d – s}$$

This yields

$$s^{2} = d(d – s) = d^{2} – ds$$

Since we’re trying to figure out what d < 8 tells us about s, we can set that up as:

$$s^{2} = d^{2} – ds < 8^{2} – 8s$$

So the following must be true

$$s^{2} < 8^{2} – 8s$$

Re-arranging, the following must be true

$$s^{2} + 8s < 64$$

Since we’re trying to see whether s > 5.2cm, we can test a case near the borderline (s = 5cm) to get some insight.

If we assume s = 5, we get 25 + 40 < 64, which is a contradiction. Therefore, the assumption that s = 5 must be false.

Since s must be positive, $$s^{2} + 8s$$ strictly increases as s increases. Therefore, any s > 5 will also yield a contradiction. Therefore s is definitively NOT greater than 5.2, and statement (2) is sufficient.

Attachments OG2020 DS369.jpg [ 44.3 KiB | Viewed 9579 times ]

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*** The following solution has been revised thanks to Ian's comment below. Thanks, Ian.

The Alternative approach to this question - since we know very little about a regular pentagon - is to try using specific numbers and more familiar shapes to assist us.
Statement (1): we can use the formula of circular area and find that the radius is 4. Now, if it were a regular HEXAGON inscribed in the circle, its side would have been equal to the radius (since it can be divided into 6 equilateral triangles), and thus its perimeter would have been 6x4=24. But since the more sides there are to a regular polygon inscribed in a circle, the larger its perimeter, then the perimeter of the pentagon must be less than 24. That's enough. Answer choices (B), (C) and (E) are eliminated.
Statement (2): As the attached figure shows, the green and the red triangles created by the diagonals are identical (that's because the pentagon's 108 angle is divided into three equal angles - opposite the same arc - of 36 degrees, so all of their angles are equal; and their largest angle is opposite the same side - the diagonal). Thus, since if the pentagon's perimeter was 26 each side would have been just a bit above 5, we'll mark all sides as 5 to see whether this is possible. Also, since the statement relates to the diagonal as less than 8, we'll try 8. If that's true, the sides of the blue triangle are 8 - 5 = 3. Now, how do we know if that's possible? The green and the blue triangles are similar triangles (having 2 identical angles of 36 degrees), so
3/5 = 5/8 We'll multiply by 5 & 8:
24 = 25
This is wrong, of course, and since the diagonal is smaller than 8 and the side is larger than 5, the left side of the equation is even smaller than 24. Thus, in order to create a correct equation, the side of the pentagon must be smaller than 5, and so its perimeter is smaller than 5x5=25 >>> smaller than 26.
So statement (2) also gives us enough information, and the correct answer is (D).
Attachments GC_circumscribed_2.png [ 46.52 KiB | Viewed 10577 times ]

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Originally posted by DavidTutorexamPAL on 26 Apr 2019, 14:42.
Last edited by DavidTutorexamPAL on 06 May 2019, 00:05, edited 3 times in total.
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Re: A pentagon with 5 sides of equal length and 5 interior angles of equal  [#permalink]

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DavidTutorexamPAL wrote:
Statement (2): If it were a SQUARE inscribed in the circle, the diagonal would have been (root2) times the side, so with a diagonal of 8 the side would have been 8/(root2), which is 4(root2) >>> 4x1.3 >>> 5.2. But since the more sides there are to a regular polygon inscribed in a circle, the smaller its side, then the side of the pentagon must be less than 5.2, and thus its perimeter must be less than 5x5.2,which is less than 26. That's enough. Answer choice (A) is also eliminated.

I mentioned in my post above that this solution isn't correct (or if it is, I have not correctly understood what you're trying to say), and in case it's confusing for test takers, I think I should explain. For one thing, I think there's a logical issue with your inequalities. If s is the side of the square, and p the side of the pentagon, you seem to have concluded from these two inequalities:

s > 5.2
s > p

that 5.2 > p is true. There's no way from those two inequalities to compare p and 5.2.

Nor is it even the case that we can directly compare the square and pentagon you're describing. It's true that if you inscribe a square and a pentagon in the same circle, the square will have longer sides. But we can't inscribe them in the same circle here, if the square and pentagon both have a diagonal of 8. The pentagon is too big to fit in the circle you'd use for the square.

There are other ways to prove quickly in this question that p < 4√2 (and maybe you have and I haven't followed) but that isn't enough to answer the question if we genuinely need to prove that the perimeter is less than 26, which is why I had to resort to similar triangles in my post above.
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Re: A pentagon with 5 sides of equal length and 5 interior angles of equal  [#permalink]

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dng992 wrote:
your sentences dont even make sense, why would i subscribe to your exam service

"But since the more sides there are to a regular polygon inscribed in a circle,
the smaller its perimeter, [ok so youre saying more sides smaller perimeter]
then the perimeter of the pentagon must be less than 24. That's enough" [pentagon has fewer sides then hexagon, less sides, larger perimeter]

triggers students looking for solutions and seeing this

This is a typo, it should say 'larger' and not smaller.
Thanks for pointing this out, fixed!

Note also that a potentially shorter way to address stmt (2) is to notice that drawing a height in the 'green triangle' in my answer above (or the triangle labeled '1' in Ian's answer) divides this triangle into two 36-54-90 triangles. This is very nearly a 30-60-90 triangle and so we have a decent approximation for the length of the diagonal. In particular, if the perimeter were exactly 26, each side of the pentagon would be 5.2, meaning that the hypotenuse of our 36-54-90 triangle is also 5.2. Then half of the diagonal, which is the side in front of the 54-degree angle, is a bit shorter than sqrt(3)*5.2/2, and doing the math gives that the diagonal must be a bit shorter than 9 (more precisely, a bit shorter than 8.84). Since (2) tells us the diagonal is shorter than 8, and since 8 is quite a ways off from our approximation of 9, then the perimeter must be shorter than 26, which is sufficient.
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Re: A pentagon with 5 sides of equal length and 5 interior angles of equal  [#permalink]

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First of all, I want to thank all of you guys for the detailed answer but I feel that the method used for the second statement would not be suitable during the exam.

To answer the question we need the length of the Pentagon's sides, therefore we do not need the exact solution but just to know that given the information provided we would be able to determine that length.
I would state that the second statement is sufficient just by applying the following rule: you can determine each of the measures of triangle 1 (referring to Ian sketch) if you know the length of one side (given by statement two) and the width of two angles (which can be derived given the fact that it is a regular pentagon).
Now we know that is possible to determine the sides of the pentagon and therefore answering the question, so statement 2 is SUFFICIENT
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GMAT 1: 710 Q50 V36 GMAT 2: 710 Q48 V40 GMAT 3: 720 Q50 V37 Re: A pentagon with 5 sides of equal length and 5 interior angles of equal  [#permalink]

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Is there a good solution as to why statement 2 is sufficient? Most of the solutions for St 2 is difficult to implement in test-environment.
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A pentagon with 5 sides of equal length and 5 interior angles of equal  [#permalink]

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If you are good at trigonometry, you can use this approach

Let side of the pentagon is x and diagonal 8cm
then x=4/cos(36)
Now, if we know value of cos 36, we can find x very easily.

sin(108) = sin(72+36)
Sin(pi-72)=sin72 co36 + cos72 sin36 {sin(A+B)= sinA cosB + cosA sinB)
sin72= 2sin36 cos36 cos36 + (2cos^2(36) -1) sin36 [sin2A= 2sinAcosA and cos2A= 2cos^2 (A) - 1]
2sin36cos36 = sin36 {2cos^2(36)+2cos^2(36) -1}
2cos36= 4cos^2(36) -1
4cos^2(36) - 2cos36 -1 =0
Let cos36=y
4y^2 - 2y -1 = 0
we know cos36 is positive, hence we will consider only positive root of this equation.
By solving above equation, we will get
cos36= 0.8 nearly
Hence x= 4/0.8 =5
Now we know that diagonal is less than 8, hence length of any side of pentagon is less than 5.
Perimeter will be less than 25

OR

There is one more approach, though it's not accurate one, to find approximate value of cos36
We know cos 30 is 3^{1/2}/3 = 0.866
and cos 45= 1/(2)^1/2 = 0.7
Use interpolation to find the cos 36, and you will get value equals to 0.8

I know cosine curve in not linear, hence interpolation technique is not a good approach, but it gives nearly exact value in this case.
If you're running out of time in exam, you can use it. Otherwise i won't recommend it.

souvikgmat1990 wrote:
Is there a good solution as to why statement 2 is sufficient? Most of the solutions for St 2 is difficult to implement in test-environment.

Attachments fig 5.png [ 5.54 KiB | Viewed 10026 times ]

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Re: A pentagon with 5 sides of equal length and 5 interior angles of equal  [#permalink]

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sebastianm wrote:
First of all, I want to thank all of you guys for the detailed answer but I feel that the method used for the second statement would not be suitable during the exam.

To answer the question we need the length of the Pentagon's sides, therefore we do not need the exact solution but just to know that given the information provided we would be able to determine that length.
I would state that the second statement is sufficient just by applying the following rule: you can determine each of the measures of triangle 1 (referring to Ian sketch) if you know the length of one side (given by statement two) and the width of two angles (which can be derived given the fact that it is a regular pentagon).
Now we know that is possible to determine the sides of the pentagon and therefore answering the question, so statement 2 is SUFFICIENT

Hey @sebastinam,
Unfortunately this doesn't work, as (2) gives you a range of potential diagonal lengths and not a specific length. Then you have a range of potential pentagon side lengths, and without knowing its lower bound cannot know if the perimeter is larger than 26 or not. So you need to find the lower bound by using one of the methods of calculation offered above.

Best,
David
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Re: A pentagon with 5 sides of equal length and 5 interior angles of equal  [#permalink]

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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.

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Re: A pentagon with 5 sides of equal length and 5 interior angles of equal  [#permalink]

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Can someone please explain part 1!
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Re: A pentagon with 5 sides of equal length and 5 interior angles of equal  [#permalink]

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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.

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OG2020 NEW QUESTION

All the solutions provided are difficult to understand. Is there any other way to solve statement 2?
After looking all these, I thought I will and guess and move on if I encounter such question in exam.
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Re: A pentagon with 5 sides of equal length and 5 interior angles of equal  [#permalink]

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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

The real question is whether the side length of this regular pentagon is greater than 5.2.

(1) This tells us all we could ever hope to know about the circle, and therefore the regular pentagon therein inscribed. Sufficient.

(2) The ratio of a regular pentagon's diagonal to its side length is equal to the Golden ratio, which is quite close to 1.6 = 8/5. If the diagonal is less than 8, then the side length is less than 5. Sufficient.

Easy, next.
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Re: A pentagon with 5 sides of equal length and 5 interior angles of equal  [#permalink]

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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
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Re: A pentagon with 5 sides of equal length and 5 interior angles of equal  [#permalink]

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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)
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Re: A pentagon with 5 sides of equal length and 5 interior angles of equal  [#permalink]

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DavidTutorexamPAL wrote:
*** The following solution has been revised thanks to Ian's comment below. Thanks, Ian.

The Alternative approach to this question - since we know very little about a regular pentagon - is to try using specific numbers and more familiar shapes to assist us.
Statement (1): we can use the formula of circular area and find that the radius is 4. Now, if it were a regular HEXAGON inscribed in the circle, its side would have been equal to the radius (since it can be divided into 6 equilateral triangles), and thus its perimeter would have been 6x4=24. But since the more sides there are to a regular polygon inscribed in a circle, the larger its perimeter, then the perimeter of the pentagon must be less than 24. That's enough. Answer choices (B), (C) and (E) are eliminated.
Statement (2): As the attached figure shows, the green and the red triangles created by the diagonals are identical (that's because the pentagon's 108 angle is divided into three equal angles - opposite the same arc - of 36 degrees, so all of their angles are equal; and their largest angle is opposite the same side - the diagonal). Thus, since if the pentagon's perimeter was 26 each side would have been just a bit above 5, we'll mark all sides as 5 to see whether this is possible. Also, since the statement relates to the diagonal as less than 8, we'll try 8. If that's true, the sides of the blue triangle are 8 - 5 = 3. Now, how do we know if that's possible? The green and the blue triangles are similar triangles (having 2 identical angles of 36 degrees), so
3/5 = 5/8 We'll multiply by 5 & 8:
24 = 25
This is wrong, of course, and since the diagonal is smaller than 8 and the side is larger than 5, the left side of the equation is even smaller than 24. Thus, in order to create a correct equation, the side of the pentagon must be smaller than 5, and so its perimeter is smaller than 5x5=25 >>> smaller than 26.
So statement (2) also gives us enough information, and the correct answer is (D).

Hi, thanks so much for such a detailed explanation. Unfortunately, I am still struggling with statement 1, could you please explain how st-1 is sufficient? How does it help us to estimate the length of the pentagon's side by knowing the radius of the circle it is inscribed in? Will appreciate if you could delve a little deeper in this. Thanks!
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Re: A pentagon with 5 sides of equal length and 5 interior angles of equal  [#permalink]

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guys there is a much quicker solution
We know that the diameter is given by the formula d=(a^2+b^2)^(1/2). Also we know from the question that the pentagon has all of their sides equal, so d=(2*a^2)^(1/2) and then the problem is solved with the following statements. Very simple.
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Re: A pentagon with 5 sides of equal length and 5 interior angles of equal  [#permalink]

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arban77 wrote:
guys there is a much quicker solution
We know that the diameter is given by the formula d=(a^2+b^2)^(1/2). Also we know from the question that the pentagon has all of their sides equal, so d=(2*a^2)^(1/2) and then the problem is solved with the following statements. Very simple.

It's not right (both the triangle and this shortcut).
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Re: A pentagon with 5 sides of equal length and 5 interior angles of equal  [#permalink]

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1
1
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.

For that very reason, I wondered if the question contained a typo, and meant to ask about a perimeter of 28 (instead of 26). Then you could do the problem easily, just using estimates with 30-60-90 and 45-45-90 triangles. I haven't seen OG2020 yet, but I gather from a post above that the official solution uses the quadratic formula, so they don't seem to have a fast solution for this either. The fastest way, as someone pointed out above, is to use the 'golden ratio', but I didn't post that solution because there's no way test takers could be expected to know anything about 'the golden ratio' (and since this is the only GMAT question you'll ever see where it would be useful, don't bother learning about it).

It is important to understand why Statement 1 is sufficient alone - a few people have asked about it since the first solutions were posted. From Statement 1, we know we have a circle with radius 4. There is only one size of regular pentagon you could fit in that circle, with the corners of the pentagon precisely on the circumference of the circle. Since only one size of pentagon can fit in the circle, and since we know exactly how big the circle is, there must be some way to work out how big that pentagon is too. We don't care what that method is (even if that method were just "take a ruler and measure it", that would be fine), because it's a DS question, and we don't actually need to answer the question; we only need to know that the question can be answered.
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