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

It is currently 25 Sep 2018, 14:42

Close

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
Your Progress

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.

Close

Request Expert Reply

Confirm Cancel

Rectangle ABDC is perfectly inscribed inside a circle and has an area

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:

Hide Tags

Manager
Manager
avatar
S
Joined: 19 Dec 2016
Posts: 56
Location: India
Concentration: Technology, Leadership
WE: Consulting (Computer Software)
Rectangle ABDC is perfectly inscribed inside a circle and has an area  [#permalink]

Show Tags

New post 02 Aug 2017, 04:55
1
2
00:00
A
B
C
D
E

Difficulty:

  65% (hard)

Question Stats:

63% (02:42) correct 37% (02:26) wrong based on 79 sessions

HideShow timer Statistics

Attachment:
circle.PNG
circle.PNG [ 8.5 KiB | Viewed 2197 times ]


Rectangle ABDC is perfectly inscribed inside a circle and has an area of 40 square inches. What is the length of minor arc CD?.

(1) The sum of the measures of angle m and angle n is 60 degrees.
(2) The sum of the measures of angle j and angle n is 90 degrees.

_________________

If you want to appreciate, just click on Kudos.

Senior Manager
Senior Manager
User avatar
P
Joined: 02 Jul 2017
Posts: 294
Concentration: Entrepreneurship, Technology
GMAT 1: 730 Q50 V38
GMAT ToolKit User
Re: Rectangle ABDC is perfectly inscribed inside a circle and has an area  [#permalink]

Show Tags

New post 12 Sep 2017, 23:04
Refer :
Circle initial : Initial diagram with extra variables for explanation.
Circle Solution : Diagram with all angels and sides marked

Note: Here m writing full explanation so solution might look lengthy but when we solve without writing all the steps it can be done quickly.

Given: ABCD is a rectangle, inscribed in a circle, area 40
=> By figure : AB||CD and AC|| BD
k+n=90 and m+j=90.
Let length and breadth be l and b . so area =40 = l*b
As cord AD and BC make 90 degree at circle. So AD and CB are diagonals. At O they bisect each other. So OA=OB=OC=OD
length of cord AD and BC = \(\sqrt{l^2 +b^2}\)
OC=OD = \(\frac{\sqrt{l^2 +b^2}}{2}\)

Find: Length of Arc CD
=> Arc length = \(2*pi*r* (\frac{angle at center}{360})\)
=> Arc length = \(2*pi*OD* (\frac{angle COD}{360})\)
=> We just need to find length OD and Angle COD


Option 2: The sum of the measures of angle j and angle n is 90 degrees.
so j+n=90
But As given ABCD is a rectangle. Angle D =90. And in triangle CBD sum of all angles =180
=> j+n+90=180 => j+n =90. We know this equation just by question stem. So this option doesn't give us any additional detail.
So Not sufficient.
And this information is already given in question combining 1 and 2 option will not give us any additional detail.
So our Answer is either A or E.


Option 1: The sum of the measures of angle m and angle n is 60 degrees.
m+n=60.
As AB || CD => angle m=n => m=n=30
=> k=j=60
=> by figure: as OC =OD=> n=x=30 and OD=OB => j=y=60
=> angle COD = 120 and Angle DOB=60.
Now as Triangle DOB becomes equilateral triangle. All sides must be equal. => OB=OD=BD
=> \(\frac{\sqrt{l^2 +b^2}}{2} = b\) => \(l^2 +b^2 =4b^2\) => \(l^2=3b^2\) => \(l=\sqrt{3}b\)

as l*b=40 => \(\sqrt{3}b * b = 40\) => \(b^2= \frac{40}{\sqrt{3}}\) => \(b = \sqrt{\frac{40}{\sqrt[]3}}\)

Also as radius of circle = r = OD = \(\frac{\sqrt{l^2 +b^2}}{2}\) = \(\frac{\sqrt{3b^2 +b^2}}{2}\)
\(r = \frac{2b}{2} = b = \sqrt{\frac{40}{\sqrt[]3}}\)

So we got value of angle COD = 120 and value of radius r = \(\sqrt{\frac{40}{\sqrt[]3}}\)

So we can calculate length of Arc CD =\(2*pi*r* (\frac{angle COD}{360})\)

Sufficient

Answer: A
Attachments

File comment: Circle with marked length and angle values. And brief of solution
Circle Solution.jpg
Circle Solution.jpg [ 2.12 MiB | Viewed 1963 times ]

File comment: Initial Circle diagram with additional variables for proper understanding
circle initial.PNG
circle initial.PNG [ 12.91 KiB | Viewed 1947 times ]

Intern
Intern
avatar
B
Joined: 19 Oct 2015
Posts: 30
Re: Rectangle ABDC is perfectly inscribed inside a circle and has an area  [#permalink]

Show Tags

New post 12 Sep 2017, 23:33
@nikkb Question is about finding length of CD. It seems that both A and B are INSUFFICENT.
For me answer is E

Sent from my A0001 using GMAT Club Forum mobile app
Senior Manager
Senior Manager
User avatar
P
Joined: 02 Jul 2017
Posts: 294
Concentration: Entrepreneurship, Technology
GMAT 1: 730 Q50 V38
GMAT ToolKit User
Re: Rectangle ABDC is perfectly inscribed inside a circle and has an area  [#permalink]

Show Tags

New post 13 Sep 2017, 00:30
Mujeeb wrote:
Nikkb Question is about finding length of CD. It seems that both A and B are INSUFFICENT.
For me answer is E

Sent from my A0001 using GMAT Club Forum mobile app



Yes i know we have to find length of Arc CD.
Please see the comments i have written on top on the solution. Copy pasting it here again:

Find: Length of Arc CD
=> Arc length = \(2*pi*r* (\frac{angle at center}{360})\)
=> Arc length = \(2*pi*OD* (\frac{angle COD}{360})\)
=> We just need to find length OD and Angle COD

As per formula to calculate length of Arc we should know radius of circle and the angle subtended by the arc at the center.
So here if we calculate radius and Angle COD suptended by arc CD at center O, using formula we can calculate the length of arc.

So as per our calculation Angle COD = 120 and radius = \(\sqrt{\frac{40}{\sqrt[]3}}\)


So we can put this in above formula to calculate length of arc.

Arc CD length = \(2*pi*OD* (\frac{angle COD}{360})\) = \(2*pi* \sqrt{\frac{40}{\sqrt[]3}} *(\frac{120}{360})\)

Now we don't have to calculate exact value here.

So Answer is A as option 1 alone is sufficient.

Hope m clear.
Intern
Intern
avatar
B
Joined: 28 Apr 2016
Posts: 11
Location: India
GMAT 1: 700 Q48 V37
GPA: 3.85
Re: Rectangle ABDC is perfectly inscribed inside a circle and has an area  [#permalink]

Show Tags

New post 13 Sep 2017, 00:33
In this question we should know that the diagonal is diameter hence the angle BDC is 90.

S1- when m+n =60 we get n =30 as both the angle need to be same. So by using tan 30 we can get both BD and CD.

Jyotipravat
Manager
Manager
User avatar
S
Joined: 01 Dec 2016
Posts: 113
Concentration: Finance, Entrepreneurship
GMAT 1: 650 Q47 V34
WE: Investment Banking (Investment Banking)
GMAT ToolKit User
Re: Rectangle ABDC is perfectly inscribed inside a circle and has an area  [#permalink]

Show Tags

New post 19 Oct 2017, 05:41
Another quick way to solve the question:

Lets r be the radius of the cercle, and O be the center.
To get the length of "minor arc CD" we need the measure of "Angle COD" and r.

From stmt(1), m+n=60. n and m has the same measure, hence n=30
we now know that CBD is a (30:60:90) right triangle, which hypthenus measure 2*r .
Therefore, CBD has dimensions (r : sqrt(3)*r : 2*r) --- BD=r and CD=sqrt(3)*r
Area of rectangle ABCD is CD* DB= r * sqrt(3)*r = 40. Hence, we find r.
We also know that triangle COD is isocèles in O and OCD is 30. Hence, "Angle COD" is 180.

Stmt(1) is suff.
Stmt(2) is insuff since it repeats information from the question stem. Nothing new.

Hope it helps...
_________________

What was previously considered impossible is now obvious reality.
In the past, people used to open doors with their hands. Today, doors open "by magic" when people approach them

GMAT Club Bot
Re: Rectangle ABDC is perfectly inscribed inside a circle and has an area &nbs [#permalink] 19 Oct 2017, 05:41
Display posts from previous: Sort by

Rectangle ABDC is perfectly inscribed inside a circle and has an area

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  

Events & Promotions

PREV
NEXT


Copyright

GMAT Club MBA Forum Home| About| Terms and Conditions and Privacy Policy| GMAT Club Rules| Contact| Sitemap

Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne

Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.