GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 13 Dec 2019, 02:55 ### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

#### Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.  # In the figure above, inscribed polygon PQRST is equilateral. If the

Author Message
TAGS:

### Hide Tags

Math Expert V
Joined: 02 Sep 2009
Posts: 59722
In the figure above, inscribed polygon PQRST is equilateral. If the  [#permalink]

### Show Tags 00:00

Difficulty:   55% (hard)

Question Stats: 72% (02:05) correct 28% (01:55) wrong based on 57 sessions

### HideShow timer Statistics In the figure above, inscribed polygon PQRST is equilateral. If the diameter of the circle is 10, then the length of arc PQR is

(A) 10π
(B) 4π
(C) 2π
(D) 4π/5
(E) 2π/5

Attachment: 2017-11-15_1035.png [ 10.25 KiB | Viewed 2700 times ]

_________________
Manager  S
Joined: 05 Dec 2016
Posts: 232
Concentration: Strategy, Finance
GMAT 1: 620 Q46 V29 Re: In the figure above, inscribed polygon PQRST is equilateral. If the  [#permalink]

### Show Tags

2
Circumference of PQRST= P*D=10P
Given that polygon is equilateral we understand that circumference of PQR would be 2/5 of PQRST, hence 2/5*10P=4P
Manager  B
Joined: 11 Feb 2017
Posts: 181
Re: In the figure above, inscribed polygon PQRST is equilateral. If the  [#permalink]

### Show Tags

Alexey1989x wrote:
Circumference of PQRST= P*D=10P
Given that polygon is equilateral we understand that circumference of PQR would be 2/5 of PQRST, hence 2/5*10P=4P

Can u plz explain this?
Manager  S
Joined: 05 Dec 2016
Posts: 232
Concentration: Strategy, Finance
GMAT 1: 620 Q46 V29 Re: In the figure above, inscribed polygon PQRST is equilateral. If the  [#permalink]

### Show Tags

1
rocko911 wrote:
Alexey1989x wrote:
Circumference of PQRST= P*D=10P
Given that polygon is equilateral we understand that circumference of PQR would be 2/5 of PQRST, hence 2/5*10P=4P

Can u plz explain this?

I'll try. The answer to the question boils down to finding the circumference of the circle, which can be found by using the formula: 2*P (3,14)*r and finding the proportion of arc PQR to the circle PQRST.

1) Let's calculate the circumference
2*P (3,14)*r =10P since we are given the value of the diameter.
2) Let's find the relationship between arc PQR to circumference of the PQRST
Let's assume that circle has point O as its centre. Then we cut polygon into 5 equal triangles with one shared vertice O: PQO; QOR;ROS;SOT;TPO. Each trangle is 1/5 of the polygon and its length equals to 1/5 of the circumference of the circle.
So, we see that arc PQR covers two triangles and hence we can deduce that PQR would constitute 2/5 of the circumference of the circle.

2/5*10P=4P

Hope it's clear.
If not, you should go to the basics of geometry.
Senior SC Moderator V
Joined: 22 May 2016
Posts: 3747
In the figure above, inscribed polygon PQRST is equilateral. If the  [#permalink]

### Show Tags

1
1
Bunuel wrote: In the figure above, inscribed polygon PQRST is equilateral. If the diameter of the circle is 10, then the length of arc PQR is

(A) 10π
(B) 4π
(C) 2π
(D) 4π/5
(E) 2π/5

Attachment:
2017-11-15_1035.png

rocko911 - in fairness, if I recall correctly (and I may not), many properties of pentagons inscribed in circles involve trig.

Not true here, but you do have to know (or derive): An equilateral $$n$$-gon inscribed in a circle divides the circle's circumference into $$n$$ pieces.

The distance between the vertices is the same. The vertices lie on the circle. The chords that intercept the arcs are equal. If two chords are equal in measure, then their corresponding minor arcs (PQ and QR) are equal in measure.

Here, an inscribed equilateral pentagon's vertices divide the circumference into 5 arcs.

We are asked for the arc length of arc PQR. That length is a fraction of the circumference's length.

Count arc segments: from P to Q, then Q to R, there are 2 arc segments out of 5 total. Arc PQR hence is $$\frac{2}{5}$$ of the circumference of the circle.

Circumference of circle?
Diameter = 10 = 2r. So r = 5
Circumference = $$2\pi*r=10\pi$$

Length of arc PQR?
$$(\frac{2}{5} * 10\pi) = 4\pi$$

Hope that helps.
_________________
SC Butler has resumed! Get two SC questions to practice, whose links you can find by date, here.

Never doubt that a small group of thoughtful, committed citizens can change the world; indeed, it's the only thing that ever has -- Margaret Mead
Target Test Prep Representative G
Affiliations: Target Test Prep
Joined: 04 Mar 2011
Posts: 2809
Re: In the figure above, inscribed polygon PQRST is equilateral. If the  [#permalink]

### Show Tags

1
Bunuel wrote: In the figure above, inscribed polygon PQRST is equilateral. If the diameter of the circle is 10, then the length of arc PQR is

(A) 10π
(B) 4π
(C) 2π
(D) 4π/5
(E) 2π/5

Attachment:
2017-11-15_1035.png

Since polygon PQRST is equilateral, it breaks the circle into 5 arcs of equal length. Since PQR represents 2 arcs, that is equivalent to 2/5 of the circumference. Since the diameter of the circle is 10, the circumference is 10π and arc PQR is:

10π x 2/5 = 4π

_________________

# Jeffrey Miller

Jeff@TargetTestPrep.com

See why Target Test Prep is the top rated GMAT quant course on GMAT Club. Read Our Reviews

If you find one of my posts helpful, please take a moment to click on the "Kudos" button.

Manager  B
Joined: 04 May 2014
Posts: 150
Location: India
WE: Sales (Mutual Funds and Brokerage)
In the figure above, inscribed polygon PQRST is equilateral. If the  [#permalink]

### Show Tags

1
Made a mistake edited the same

The sum of all the angles of a regular(equal sides) polygon=(n-2)*180
n=total number of sides
here n=5
The sum of angles=(5-2)*180=540
Each angle=540/5=108
Let us create a triangle POQ with OP and OQ being its sides and they split the angle P and Q to half each ie 54.
Now the sum of the triangle is 180 hence the angle O will be 72.
Arc PQR has 2 such angles=72+72=144
Length of arc=144/360*(2pi5)-Here 5=radius of the circle which is given-half of diameter(10/2)

=4Pi

Originally posted by gps5441 on 27 Nov 2017, 22:18.
Last edited by gps5441 on 28 Nov 2017, 02:36, edited 1 time in total.
Intern  B
Joined: 20 Jun 2016
Posts: 4
Location: United States (MA)
Concentration: Finance, Technology
Schools: Carroll '20
GMAT 1: 650 Q40 V38 GPA: 3.25
In the figure above, inscribed polygon PQRST is equilateral. If the  [#permalink]

### Show Tags

1
gps5441 wrote:
The sum of all the angles of a regular(equal sides) polygon=(n-2)*180
n=total number of sides
here n=5
The sum of angles=(5-2)*180=360
Each angle=360/5=72
Arc PQR has 2 such angles=72+72=144(if o is the center of the circle than angle opq +angle orq)

Length of arc=144/360*(2pi5)-Here 5=radius of the circle which is given-half of diameter(10/2)
=4Pi

If n is 5, then (n-2) * 180 = (5-2) * 180 = 3 * 180 = 540. Each Angle of the polygon would have to be 540/5 = 108.

Still, going by the idea we create an isosceles triangle POQ, we split the 108's in two = leaving two angles of 54 and a remainder of 72 stemming from the origin of the circle.
Double the 72 to acquire the central angle that encompasses the arc PQR, 144.
144 / 360 = (12 * 12) / (12 * 30) = (12 / 30) = (6 / 15) = (2 / 5)

Multiply that by the circumference 10pi to get 4pi

genxer123's tidbit about n-sided, equilateral polygons inscribed in a circle though would've saved a lot more time; thanks for sharing that.
Manager  B
Joined: 04 May 2014
Posts: 150
Location: India
WE: Sales (Mutual Funds and Brokerage)
Re: In the figure above, inscribed polygon PQRST is equilateral. If the  [#permalink]

### Show Tags

Silver65x wrote:
gps5441 wrote:
The sum of all the angles of a regular(equal sides) polygon=(n-2)*180
n=total number of sides
here n=5
The sum of angles=(5-2)*180=360
Each angle=360/5=72
Arc PQR has 2 such angles=72+72=144(if o is the center of the circle than angle opq +angle orq)

Length of arc=144/360*(2pi5)-Here 5=radius of the circle which is given-half of diameter(10/2)
=4Pi

If n is 5, then (n-2) * 180 = (5-2) * 180 = 3 * 180 = 540. Each Angle of the polygon would have to be 540/5 = 108.

Still, going by the idea we create an isosceles triangle POQ, we split the 108's in two = leaving two angles of 54 and a remainder of 72 stemming from the origin of the circle.
Double the 72 to acquire the central angle that encompasses the arc PQR, 144.
144 / 360 = (12 * 12) / (12 * 30) = (12 / 30) = (6 / 15) = (2 / 5)

Multiply that by the circumference 10pi to get 4pi

genxer123's tidbit about n-sided, equilateral polygons inscribed in a circle though would've saved a lot more time; thanks for sharing that.

Thank you for pointing out the mistake in my mistake
Non-Human User Joined: 09 Sep 2013
Posts: 13742
Re: In the figure above, inscribed polygon PQRST is equilateral. If the  [#permalink]

### Show Tags

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________ Re: In the figure above, inscribed polygon PQRST is equilateral. If the   [#permalink] 25 Nov 2019, 03:14
Display posts from previous: Sort by

# In the figure above, inscribed polygon PQRST is equilateral. If the  