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).-----
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 < 24Since 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.
01 Jun 2021, 00:30
