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

Is there a formula for the relationship between the diagonals/sides of the pentagon and the radius of the circle?

Edit: I found it:

diagonal d of pentagon given side s, d = s * (1 + SQR5) / 2

height h of pentagon given side s, h = s * SQR(5 + 2SQR5) / 2

area of pentagon given side s = s^2 * SQR(25 + 10SQR5) / 4

circumcircle radius R of inscribed pentagon with side s, R = s/(2sin36) = 0,851s

incircle radius R, circle inside of pentagon with side s, R = s/(2tan36) = 0,688s

Not of very much use I guess, but who knows...

https://www.omnicalculator.com/math/pentagon

We are given a regular pentagon inscribed in a circle. We need to find whether the perimeter of pentagon is less than 26 cms.
What makes this problem hard is they have used “pentagon”, instead of for example, a square or a hexagon. At least, we don’t know much about pentagons if we stay within the scope of GMAT.

But we can work with what we know within the scope of GMAT. Let’s see.

Let’s recall some of the facts that we know.

We are given a regular pentagon – all the 5 sides and 5 angles are equal.

The regular pentagon is inscribed in a circle.

So, certainly the perimeter of pentagon will be less than perimeter (circumference) of the circle.

Now, think of this. If we have a regular polygon with let’s say, 100 or 1000 sides, it will nearly look like a circle and the perimeter of the polygon will be very close to that of a circle’s perimeter.

If we get the above point, we will also understand that as we keep on increasing the number of sides of a regular polygon (inscribed in a circle) the perimeter of that polygon also increases and approaches the perimeter (circumference) of circle.

So, we can say that for a given circle, the perimeter of an inscribed regular pentagon should lie between perimeter of a regular quadrilateral (square) and perimeter of an inscribed regular hexagon.

Bingo! Now we have something within the scope of GMAT to work with.

Perimeter of square inscribed in a circle of radius r < Perimeter of regular pentagon inscribed in a circle of radius r < Perimeter of regular hexagon inscribed in a circle of radius r.

Let’s analyse statements now.

Statement 1: The area of the circle is 16π square centimetres.

This gives us the radius of the circle as 4cm.
Per our conceptual understanding, we know that a regular hexagon is composed of 6 equilateral triangles. If we draw it, we will instantly realize that the side-length of a regular hexagon is also 4cm and hence perimeter of regular hexagon is 24cm.

So, with certainty, we can say that perimeter of regular pentagon is less than 24 cm. We are getting a definite NO to the question. Hence statement 1 is sufficient.

Statement 2: The length of each diagonal of the pentagon is less than 8 centimetres.

Couple of things:

This statement has more to do with the origin of \(/pi\). What is \(/pi\)? It’s the ratio of a circle’s circumference to its diameter. The ratio is always a fixed value, \(/pi\).

We know that the longest diagonal of a regular polygon inscribed in a circle can never exceed the diameter.

For square, hexagon etc. – basically any polygon with even number of sides inscribed in a circle the longest diagonal will be equal to the diameter of the circle (owing to its symmetry).

For polygon with odd number of sides inscribed in a circle, longest diagonal will be less than the diameter of the circle.

So, whatever polygon you inscribe within a circle,

As we keep on increasing the number of sides of the polygon inscribed in a circle (think of a circle with 100 or 1000 sides) – we know that as the number of side increases, the figure almost approaches a circle, and the ratio of perimeter to the longest diagonal would also approach π.
In other words,
for a regular pentagon inscribed in a circle, the ratio of its perimeter to its longest diagonal will be lesser than that of a regular hexagon.
for a regular pentagon inscribed in a circle, the ratio of its perimeter to its longest diagonal will be greater than that of a regular quadrilateral (square).

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edit: I realized that the strikeout portion above is not precise as pointed out by IanStewart in the post below. As shown in the example below for quadrilateral and hexagon, the ratios are 2√2 and 3 respectively. For pentagon (5 sides) and heptagon (7 sides), it can be calculated and found out to be ~3.09 & ~3.11 respectively. (see that the ratio for pentagon is slightly greater than that of hexagon)
But the general idea that we need to retain is: the ratio (perimeter to longest diagonal) approaches pi in both the cases. For polygon with even number of sides the ratio of perimeter to longest diagonal approaches pi as number of sides increases. Similarly, for polygon with odd number of sides the ratio of perimeter to longest diagonal approaches pi as number of side increases. It's just that polygon with even number of sides and odd number of sides needs a different treatment.
-------

For example:
For a square inscribed in a circle, longest diagonal = diameter = 2r = a√2
\(\frac{(Perimeter-of-inscribed-square)}{(Longest-diagonal)}\) = \(\frac{4a}{a√2}\) = 2√2 < π
For a hexagon inscribed in a circle of radius ‘r’, longest diagonal = diameter = 2r = 2a
\(\frac{(Perimeter-of-inscribed-hexagon)}{(Longest diagonal)}\) = \(\frac{6a}{2a} \)= 3 < π

Once again, as the number of sides increases, the ratio approaches π.

Since for a regular pentagon inscribed in a circle, the ratio of its perimeter to its longest diagonal will be lesser than that of a regular hexagon, we can state
(Perimeter-of-inscribed-regular-pentagon)/(Its-longest-diagonal)< 3

If we assume longest diagonal to be 8 (it is given to us that it less than 8), then we get

5a/8< 3

5a < 24

Since for a regular pentagon inscribed in a circle, the ratio of its perimeter to its longest diagonal will be lesser than pi, we can state
\(\frac{(Perimeter-of-inscribed-regular-pentagon)}{(Its-longest-diagonal)}< 3.14 \)

If we assume longest diagonal to be 8 (it is given to us that it less than 8), then we get

\(\frac{5a}{8} < 3.14 \)

5a < 25.12

There you go! Perimeter must be less than 26.

Statement 2 alone is also sufficient.

Each statement alone is sufficient.

Correct Answer: D

Hope this helps.

The logic you're using here would be correct if you were comparing polygons with an even number of sides, where the longest diagonal is the diameter of the circle, or if you were comparing polygons with lots of sides, where the longest diagonal will always closely approximate the diameter, even with an odd number of sides. But in a regular pentagon, the longest diagonal is not even close to being a diameter. If you actually calculate the ratios you're describing, you'll find that your perimeter-diagonal ratio for a regular pentagon is actually significantly larger (by about 40%) than the perimeter-diagonal ratio for a regular hexagon (because in the case of the pentagon, we're dividing by a much smaller diagonal length). It's for that reason that your numerical conclusion at the end isn't correct; the perimeter here can be larger than 24 (it can be close to 24.8).

Thank you IanStewart for pointing out the error. In the process of simplifying the explanation as much as possible, I ended up overlooking that part. (good learning, btw!!) Example of hexagon and quadrilateral was taken to drive the point that ratio (perimeter to diagonal) is less than 3.14 for all polygons inscribed in a circle. I should have just stated the ratio must be less than 3.14 in the case of a pentagon too. So, assuming the longest diagonal of pentagon to be 8, the perimeter would be less than 25.12.

Thanks once again!

Where I've highlighted part of your post, I think you might mean "perimeter to diameter"? In a regular pentagon, the ratio of the perimeter to the length of the longest diagonal is roughly 4.1, so is larger than 3.14 -- it's big because the diagonal is quite a bit shorter than the diameter.

edit - my numbers here aren't right; I just took them from a website that had the wrong diagonal length.

Hi IanStewart, I am not sure if this is correct.


diagonal is shorter, so is the perimeter.

Per my understanding, all the diagonals of a regular pentagon are equal. So, longest diagonal = diagonal.

Though out of scope of GMAT to use any formula here to find the diagonal, just for the sake of putting this discussion to an end (This question is something!!), I just double checked this with the formula. For a regular pentagon of side 1cm, the diagonal ~1.62. Ratio of perimeter to diagonal would be ~3.09. For a regular pentagon of side 15 cm, the diagonal ~ 24.27. Ratio of perimeter to diagonal would be ~3.09, a fixed value less than 3.14.

I was trying to figure out if there is a general rule that states ratio of perimeter to longest diagonal of a regular polygon approaches pi as number of sides increases. Interestingly, I found this: https://vixra.org/pdf/1709.0144v3.pdf

The ratio of perimeter to longest diagonal of lot many polygons are found out here. For all cases, value is less than pi. Not seen it published anywhere else, so can't comment much.

I'm so sorry - you're right with the numbers here. Instead of just taking the minute to do the trigonometry, I trusted the internet to tell me the diagonal length, and the site I looked at (here) had it wrong, unless I misunderstand what that page says. I should have just calculated it by hand (the diagonal length is just 2*sine(54 degrees), which is easy enough). Lesson learned!

Video solution from Quant Reasoning:
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Statement 1 is sufficient
It's sufficient because the sides of this pentagon can have only one length.

What is the length of each pentagon side? Don't know
Is the perimeter of the pentagon greater or less than 26? Don't know

This is Data Sufficiency, and we don't need to know the answers to the above questions.
We just need to figure out whether the questions can be answered. And the questions can certainly be answered by someone (though not by me)

Statement 2 is sufficient
More work is needed to show that Statement 2 is sufficient.

We know: The length of each diagonal of the pentagon is less than 8 centimeters.
We want to find out: Is the perimeter greater than 26? Or, is each side of the pentagon greater than 5.2?

Logic:
Each diagonal is less than 8.
When the diagonals become smaller, the sides and perimeter will become smaller too.
Take a pentagon with diagonals of length 8.
Drop a perpendicular line from a vertex to the nearest diagonal.
We get a right triangle with angles 36-90-54. The long side is 4. We want to find out or estimate the length of the hypotenuse.
A 36-90-54 triangle is close enough to a 30-90-60 triangle. The hypotenuse in a 30-60-90 triangle with side 4 would be less than 5 (see image)
The hypotenuse of the 36-90-54 triangle, which is the side of the pentagon, would also be less than 5, and the perimeter would be less than 26.

PS: Is this logic acceptable?

nick1816 posted the same ideas on the first page of this thread. When we have one leg of length 4 in a right triangle with a 45 degree angle, the hypotenuse is 4√2 ~ 5.6. When we have one leg of length 4 opposite a 60 degree angle in a right triangle, the hypotenuse is 8/√3 ~ 4.6. So when we have a leg of length 4 opposite a 54 degree angle in a right triangle, we can be sure the hypotenuse is somewhere between 4.6 and 5.6. The problem is, 54 isn't all that close to 45 or to 60 (52.5 is the midpoint of those two numbers, so 54 is close to halfway between them), and we're dealing with a trigonometric function (here we're really dealing with the cosecant (csc) function, the reciprocal of the sine function -- the hypotenuse is equal to 4*csc(x) where x is the opposite angle). Trigonometric functions don't change linearly, so it's hard to guess exactly what csc(54) will be from the values of csc(45) and csc(60). So unless you know trigonometry so well that you can confidently interpolate the values of csc(x) between csc(45) and csc(60), this solution gives you a pretty good basis for a guess at this question, but isn't a conclusive proof of the answer.

Thank you for the feedback.
This estimate was the best I could do without trigonometry or calculations. But I just scrolled through some of the earlier solutions and saw some good simple ones.

Posted from my mobile device

Hi IanStewart, can you please elaborate how special triangles, such as 30-60-90 and 45-45-90, would help here?

I actually just had a conversation about that last month, with vv65, so if you scroll up you can see my reply there. The idea is if you look at the 36-54-90 triangle in vv65's diagram, where the horizontal base is of length 4, its hypotenuse needs to be shorter than the hypotenuse in a 45-45-90 triangle with legs of length 4. In such a 45-45-90 triangle, the hypotenuse would be 4√2, which is roughly 5.6, so in the 36-54-90 triangle, the hypotenuse is less than 5.6. But that's the side of the pentagon, so the perimeter is less than 5*5.6 = 28 (well, I rounded off slightly, but it actually needs to be a fair bit less than that because 54 degrees is not all that close to 45 degrees). That's why I said the question would be a lot easier to answer if it asked about a perimeter of 28 instead of a perimeter of 26, because just estimating with a 45-45-90 triangle tells you the answer must be less than 28.

You can also use 30-60-90 triangles to work out that the pentagon's perimeter needs to be greater than roughly 23 when you make the pentagon's diagonal exactly 8, so estimating with those special triangles, we learn the maximum perimeter is somewhere between 23 and 28, but because we're dealing with trigonometry, and trig functions don't change linearly, it's hard to be completely confident the maximum perimeter is less than 26 this way.

So if we let x = the length of each side of the pentagon,
d = the length of diagonal
As many solutions discussed above, we have:
x.x + d.x = d.d
=> x.x + d.x + d.d/4 = 5.d.d/4
=> (x+d/2).(x+d/2) = 5.d.d/4
=> x = d.(V5 -1)/2 < 8 x (2.3 - 1)/2
=> x < 5.2
=> 5x < 26
Hope that helps!

Hi Ian,

Can we logically approach this question like this: for any regular polygon, the limit between the perimeter and the longest straight line possible is pi (which is when the number of sides approach infinity- i.e. a circle). The diagonal is the longest line possible in a regular pentagon. So the ratio of this diagonal to the perimeter of the regular pentagon has to be less than pie. Therefore, if the diagonal of this pentagon is less than 8, its perimeter has to be less than 26.

ps. Just trying to connect the dots between the GMAT question maker’s mindset, and the enigmatic pi.

That's a great way to think about what's happening here, and I think that reasoning might let you confidently guess the right answer to this question, but using only GMAT-level math, I don't see a way to be completely sure just using this logic. Say a regular polygon has a longest diagonal of d, a perimeter of P, and we inscribe it in a circle of diameter D. If the polygon has an even number of sides, then its longest diagonal is a diameter, so d = D. The perimeter is clearly less than the circumference of the circle, so with an even number of sides, it's always true that P < πd = πD. So if this question asked about a hexagon or decagon instead of a pentagon, we could answer it easily this way.

But with an odd number of sides, things are a lot more complicated. Then the longest diagonal is not a diameter, so it's just a chord of the circle, and d < D. Certainly the perimeter is still less than the circumference of the inscribing circle, so P < πD. But we don't immediately know if P < πd is also true; we could have πd < P < πD or P < πd < πD.

It actually turns out that you're right, and P/d < π is true even when you have an odd number of sides, but that's something I doubt anyone taking the GMAT would know, and I don't think it's possible to prove without using trigonometry (I haven't tried to prove it, but I glanced at another proof, and it looked complicated). If you did work out the ratio P/d for each regular polygon, of course you'll find that gets very close to π when the number of sides of the regular polygon gets large, because then the polygon approximates the circle. But the ratio doesn't constantly increase as you increase the number of sides -- for example, with a square, the ratio is roughly 2.83, for a regular pentagon it's roughly 3.09, but for a regular hexagon it drops to 3. After that the ratio goes up and down again and again as the side number switches from odd to even. It's not going to be relevant for the GMAT, but I took those numbers from this paper (which also proves the result that P/d < π) : https://vixra.org/pdf/1709.0144v3.pdf

Thanks, Ian. I just logically thought about it like this - on why the diagonal would be the longest line in a regular pentagon - If I eyeball the isoscles triangle formed by two of the diagonals and one of the sides of the pentagon, I can see that the sides of that isoscles triangle will always be greater than any other line drawn from the vertex of that triangle to the base of the triangle. Therefore, I can infer that the diagonal would indeed be the longest line possible in a regular pentagon.

Kind regards,
BT