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

This is one of a perfect example of the testing of HIGHER ORDER REASONING SKILL what exactly the GMAT exam measures on. So use your skill of mathematical reasoning rather wasting time in determining the exact value. Let us see how to reasonably approach this question.

GIVEN DATA: Regular Pentagon inscribed in circle.

FIND:
Is the perimeter (P) of pentagon greater than 26 centimeters?
ie., Is P > 26 ?
This can be considered in another way as to find, Is the side (S) of a Pentagon greater than \(\frac{26}{5}\) centimeters?
ie., Is S > 5.2 ?

STATEMENT 1: The area of the circle is 16\(\pi\) square centimeters.

So, Radius of circle (R) = 4 sq.cm

Since a regular pentagon of side (S) is inscribed in a circle of radius (R = 4), the side of a pentagon (S) should be less than \(\frac{1}{5}\)th of the circumference of circle.

Circumference of circle = 2\(\pi\)R = 2\(\pi\)(4) = 8\(\pi\)
Therefore, S < \(\frac{1}{5}\)(8\(\pi\))
ie., S < \(\frac{176}{35}\)
S < 5.03 (appx)

Therefore, S is NOT GREATER THAN 5.2, which is a definite NO to the question.

Hence STATEMENT 1 - SUFFICIENT

STATEMENT 2: The length of each diagonal of the pentagon is less than 8 centimeters.

Let us consider a regular hexagon of maximum diagonal 8 centimeters.

For GMAT, We should know that a regular hexagon is made of equilateral triangles on joining the opposite vertices. therefore the length of the longest diagonal for a regular hexagon will be 2 times the side of a regular hexagon.

So, 2*(Side of the regular hexagon) = 8
So, Side of the regular hexagon = 4
So, Perimeter of the regular hexagon will be 4*6 = 24

If you notice that we have considered a regular hexagon (six sided polygon) with a maximum diagonal of 8 centimeters itself gives perimeter just 24. So it is obvious that a Regular pentagon (five sided polygon) with diagonal less than 8 centimeters will definitely have perimeter less than 24, is a clear indication that the perimeter of a regular pentagon is NOT GREATER THAN 26 centimeter, which is a definite NO to the question.

Still not convinced ?

Let us consider a regular quadrilateral (square) of maximum diagonal 8 centimeters.
So, Side of the square will be \(\frac{8}{\sqrt{2\)
So, Perimeter of the square will be 4\(\frac{8}{\sqrt{2\), P = 18.5 (appx)

It is evident, as regular quadrilateral with maximum diagonal 8 centimeter posses perimeter of 18.5 (appx) where a regular hexagon of maximum diagonal posses perimeter of 24, then a regular pentagon of diagonal less than 8 centimeter will have perimeter between 18.5 and 24 is a clear indication that the perimeter of a regular pentagon is NOT GREATER THAN 26 centimeter, which is a definite NO to the question.

Hence STATEMENT 2 - SUFFICIENT

Answer Option: D

If the circle was a constant size, this would be correct, but you are not comparing circles of equal size. If you inscribe a square in a circle, a diagonal of the square is a diameter of the circle. If you inscribe a regular hexagon in a circle, any of its longest diagonals is a diameter of the circle. But if you inscribe a regular pentagon in a circle, none of its diagonals are diameters. They are just chords, so they are shorter than the diameter. So you are comparing the perimeters of shapes inscribed in circles of diameter 8 with the perimeter of a pentagon inscribed in a circle with diameter larger than 8. That's why the inequality you've produced is not correct, which you can see if you calculate the perimeter of the pentagon here numerically. If a regular pentagon has diagonals of length 8, its perimeter is, to one decimal place, 24.7, which is not between 18.5 and 24.

Solution:

We are given a regular pentagon inscribed in a circle. We need to determine whether the perimeter is greater than 26 cm. If it is, then each side has to be greater than 26/5 = 5.2 cm. We need to know the following fact:

If a regular pentagon with side length s is inscribed in a circle of radius r, then:

s ≈ 1.18r

Statement One Only:

The area of the circle is 16π square centimeters.

We see that the radius of the circle is √(16π/π) = 4 cm. So a side of the inscribed regular pentagon is approx. 4(1.18) = 4.72 cm, which is less than 5.2 cm. Therefore, the perimeter of the pentagon is not greater than 26 cm. Statement one alone is sufficient.

Statement Two Only:

The length of each diagonal of the pentagon is less than 8 centimeters.

Each diagonal of a regular pentagon has the same length, and we need to use the following fact:

If a regular pentagon has a side length of s and a diagonal of length d, then:

d ≈ 1.62s

Since each diagonal of the pentagon is less than 8 cm, each side of the pentagon is less than 8/1.62 ≈ 4.94 cm, which is less than 5.2 cm. Therefore, the perimeter of the pentagon is not greater than 26 cm. Statement two alone is sufficient.

Answer: D

Here is a much quicker solution;

We know that on the GMAT DS, statements 1 and statement 2 do not contradict each other. A quick solution for this problem will leverage what we know from statement 1.

Things we know from the stem: A regular pentagon is inscribed inside a circle. We know that a regular pentagon can be divided into 5 equal triangles. Given that the pentagon is inscribed inside the circle, one of its triangles will share a side with the diameter of the circle. The length of the diagonal of a "regular" pentagon inscribe inside a circle can never exceed the diameter of the circle

Statement 1: Tells us the diameter of the circle is 8. The diameter crosses the center of pentagon. Since the pentagon is inscribed inside of the circle, the length of the side of one of the pentagon's triangle inside the circle has to be less than or equal to 4. If we assume 4, then the perimeter of the pentagon is 4*5 = 20, which is less than 26. Sufficient.

Statement 2: Using statement 1, we already know that the perimeter is less than or equal to 20 when the diameter of the circle is 8. Since the pentagon is inscribed inside a circle, the diameter of the circle is greater than or equal to the length of the diagonal of a pentagon. This follows statement 1 that the length of each side of the pentagon has to be less than or equal to 4, which means that the perimeter is less than or equal to 20. Sufficient

IanStewart
ishanliberhan

You seem to be assuming that you can divide the pentagon into five equilateral triangles. That is not the case. If you draw lines connecting the center of the circle to each corner of the pentagon, then you do indeed divide the pentagon into five triangles of equal size. In each triangle, two of the sides are radii of the circle. But we've divided the 360 degree angle at the center of the circle into five equal parts, so each of these triangles has a 72 degree angle at the center of the circle. We're also dividing each 108 degree angle of the pentagon in half. So the five equal triangles we're making have angles of 72, 54 and 54 degrees. Because the 72 degree angle is the largest angle in each triangle, it is opposite the longest side, and since the 72 degree angle is opposite an edge of the pentagon, while each 54 degree angle is opposite a radius, we know by looking at these triangles that the edges of the pentagon are longer than the radii of the circle.

So when you know the radius is 4, you know from the above that the perimeter of the pentagon must be greater than 5*4 = 20. Without doing more work, we can't be sure if it's also greater than 26, which is the question we need to answer.

As I explained in a much earlier post, using Statement 1 alone, we actually don't need to calculate anything - if we know the exact size of the circle, we know the exact size of the pentagon, and can thus answer the question without doing any work at all. But Statement 2 here does require a lot of work, and there's a reason there are no simple solutions earlier in the thread (or at least none relying only on GMAT-level math). The perimeter of the pentagon in this question can be as large as 24.7, rounding off to one decimal place, so if a solution concludes that the perimeter must lie in some range of values that does not have ~24.7 as its maximum, that solution cannot be correct.

Please find the official explanation attached in the pictures below.






Let us know if you can come up with a easier solution that can be applied to save time. Tx

Does this solution work?

#1) Radius = 4, Circumference = 8pi (~24.8)
For each side the arc is longer than the side of the pentagon.
Therefore,
sum of pentagon side (perimeter) < circumference (24.8)
Therefore Perimeter < 26

#2) Each side of pentagon = x.
Diagonal & 2 sides make a triangle.
Using the triangle side rule,
x + x > Diagonal
Therefore, Diagonal < 2x
Statement: Diagonal < 8
Therefore x could be 4.
5 sides = 4*5 = 20
Therefore perimeter <26.

I agree x could be 4, but that only proves that the perimeter can be 20. The perimeter can be larger or smaller than 20 as well. For Statement 2 to be sufficient, you need to prove the perimeter can never be greater than 26, and unfortunately the triangle inequalities alone can't be used to prove that.

My gmat mind says that the answer should be E.. let me explain why..

Each angle of the mentioned pentagon is a 108 degrees..

Statement one — this is basically saying that the length of each diagonal is 8 units..now if I consider all the diagonals, I have isosceles triangle(108-36-36 degrees), one side of each triangle is 8 units..so in order to find the perimeter, we must find the two equal sides ie the sides of the pentagon.. now 8 units is the longest side of the triangle, side opposite to 108 degree angle, each of the other two equal sides , sides opposite to 36 degree angles, shall be less than 8.. this is it.. all I can infer is that each side of the pentagon is less than 8 units, which will give me a perimeter both greater and smaller than 26 units hence this option is not sufficient.

Statement 2– this basically is the same as statement 1 ie length of each diagonal is 8 unites ie the radius is 4 .... again we have the same isosceles triangles (108 -36 -36 degrees) .. not sufficient again..

Even if I consider both statement1 and statement 2, I have the same info, no change...hence not sufficient.

The only way, in my opinion, to make both the statements sufficient is—-we will have to assume that the length of each of the equal sides of the isosceles triangles is pretty smaller than 8, considering that 108 degree angle is quite large as compared to a 36 degree angle, and that side opposite to 108 degree angle will be much larger than the side opposite to 36 degree angle.. and obviously making this assumption is something that shouldn’t be done.

Is this analysis fine??

Posted from my mobile device

There are a couple of issues in your post. If you draw a line of length 8, and then from either end draw two lines at exactly 36 degrees to make a triangle, there's only one triangle you could possibly draw. Once you decide on one length and all of your angles in a triangle, the lengths of the other two sides are completely determined - they can't vary. With these specific angles (36-36-108), it's not easy to find the precise lengths of the sides (which is why this question is so hard), but it is possible to solve for them.

For the GMAT, test takers typically learn that when you have 30-60-90 triangles, or 45-45-90 triangles, then when you know one side, you can find all three sides. There's actually nothing special about those angles. One thing you learn in trigonometry is that if you know all three angles in a triangle, and you know one side, it is always possible to find all three sides, no matter what the angles are. So if you know trigonometry, if you had a triangle with angles of 87, 52 and 41 degrees, say, and you knew the longest side had length 10, you'd be able to find the other two side lengths.

Fortunately you never need to do that on the GMAT, except when you have very simple angles like 30-60-90, because you never need to use sines and cosines on the GMAT. But your assumption, that the equal sides in the 36-36-108 triangle in this question can have various lengths, is not correct, once you choose a length for the long side. And that's why the answer does not need to be E for this question (in fact it's not E -- the answer is D).



I think you're assuming the diagonals of the pentagon are also diameters of the circle, and that's not the case. The diagonals of the pentagon are shorter than the diameter of the circle.

This is an inequality involving problem we should look for some upper bound we can compare our perimeter to. This problem involves circles and typically if we have info about radius we are sufficient

Now, if you cannot find the upper bound quantity look to the information in the statements to see if they point you in the direction of it

the perimeter of our pentagon will be less than the circumference of the circle circumscribing it, so if we can show the circumference of the circle <26 cm we are good

(1) area of circle is 16pi--->r=4 and c=2(3.14)(4) = about 25.1 <26 sufficient

(2) all diagonals of our pentagon <8. all sides and angles are equal, so this is a regular polygon, BUT each diagonal is NOT a diameter of our circle so we cannot know the radius of our circle. We are going to have to use some other approach. Start by drawing all of the diagonals as shown in the attached drawing. Also, it helps to draw partial figures of some triangles formed of interest that might be helpful in determining sufficient .

Now, each angle of the pentagon will be 540/5=108 degrees, and with the diagonals drawn will each be split into 3 36 degree angles. Lets call the length of the side of our pentagon S. Lets also assume our diagonal = 8 (All diagonals of a regular polygon are congruent. we will use 8 for calculation purposes, but remembering that our final answer will use a less than sign). Now, you should notice that triangles EAB and EGB are congruent by ASA. Therefore Sides EG and BG are congruent to corresponding sides EA and AB. So we can label sides EG and BG with variable S as well. So, if EC = 8 and EG = S, then GC = 8-s. The same holds for diagonal BD and segment GD, so GD=8-S as well

Now, <GDC = <GCD = 36. This is enough to conclude that triangle GDC is similar to triangle AEB. So, a proportional relationship will hold. So, GD/AE = DC/EB, so 8-S/S = S/8, so we get 64-8s=S^2 -->S^2+8S=64---> S^2+8s+16=80-->(S+4)^2=80 --> S+4=root(80) ----> S=root(80)-4 approx 5. So, the a side of our pentagon <5, so our perimeter <25<26 Sufficient.

It's unfortunately not that easy. If you read the last sentence of my post just above yours, you'll see why the logic you're using here is not correct. The radius can be greater than 4.

I can't believe I missed that haha. So the similar triangles route it is (2). I will edit my post

From statement 1, we will definitely get an answer since it tells us the actual circle dimension.

From statement 2, the reasoning is as follows:

Posted from my mobile device

Answer: Option D

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Answer - D,

Why (1) is correct is explained beautifully by the above posts.
An explanation why (2) is correct - While studying for engineering competitive examinations, we are constantly asked to remember the angles of the most basic right-angle triangle (3-4-5) (as it was used in physics too and to tackle the pentagon questions as the values are ~36 degrees).



Hence, using these values, if we take the maximum value of the diagonal of the pentagon as 8 unit, thus the maximum length of the base of below mentioned triangle can be 4 (8/2) --->




Using the above logic of 3-4-5 triangle, The maximum value of the side of pentagon(hypotenuse) is 5 units, which will give the perimeter as 25, and is smaller than 26. Therefore, The perimeter of this pentagon can NEVER be greater than 26. Hence, answer is D.

However, as this solution requires a concept which is not tested usually, it makes the question too difficult to attempt in exam pressure. Still, learning the angular values of 3-4-5 triangle has been useful to me in eliminating some options in other questions as well

Bunuel Is the parameter of an inscribed polygon always lesser than the circumference of the inscribing circle? If that is so, Statement 1 gets very easy