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A pentagon with 5 sides of equal length and 5 interior angles of equal
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26 Apr 2019, 03:21
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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. DS75271.01 OG2020 NEW QUESTION
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A pentagon with 5 sides of equal length and 5 interior angles of equal
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Updated on: 06 May 2019, 00:05
*** 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).
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Re: A pentagon with 5 sides of equal length and 5 interior angles of equal
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28 Apr 2019, 12:07
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 3636108 triangles. Triangles 1 and 2 are identical ('congruent' in mathspeak) 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 ds 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 ds, ds, 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/(ds) 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
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28 Apr 2019, 12:47
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
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03 May 2019, 18:51
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 smaller 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). 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



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Re: A pentagon with 5 sides of equal length and 5 interior angles of equal
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03 May 2019, 18:54
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 do you have a solution for this brunel?



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Re: A pentagon with 5 sides of equal length and 5 interior angles of equal
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06 May 2019, 00:18
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 365490 triangles. This is very nearly a 306090 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 365490 triangle is also 5.2. Then half of the diagonal, which is the side in front of the 54degree 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
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11 May 2019, 12:06
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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Re: A pentagon with 5 sides of equal length and 5 interior angles of equal
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11 May 2019, 12:36
Is there a good solution as to why statement 2 is sufficient? Most of the solutions for St 2 is difficult to implement in testenvironment.



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A pentagon with 5 sides of equal length and 5 interior angles of equal
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11 May 2019, 18:00
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(pi72)=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 testenvironment.
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Re: A pentagon with 5 sides of equal length and 5 interior angles of equal
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12 May 2019, 00:10
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
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22 May 2019, 09:29
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 3636108 triangles. Triangles 1 and 2 are identical ('congruent' in mathspeak) 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 ds 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 ds, ds, 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/(ds)
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. Please explain how you got the answer to part 1!




Re: A pentagon with 5 sides of equal length and 5 interior angles of equal
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