Last visit was: 27 Apr 2026, 05:42 It is currently 27 Apr 2026, 05:42
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
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 26 Apr 2026
Posts: 109,928
Own Kudos:
Given Kudos: 105,914
Products:
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 109,928
Kudos: 811,537
 [34]
4
Kudos
Add Kudos
29
Bookmarks
Bookmark this Post
Most Helpful Reply
avatar
bluesquare
Joined: 29 Mar 2015
Last visit: 01 Dec 2016
Posts: 39
Own Kudos:
65
 [7]
Given Kudos: 9
Location: United States
Products:
Posts: 39
Kudos: 65
 [7]
2
Kudos
Add Kudos
5
Bookmarks
Bookmark this Post
General Discussion
User avatar
Harley1980
User avatar
Retired Moderator
Joined: 06 Jul 2014
Last visit: 14 Jun 2024
Posts: 997
Own Kudos:
6,769
 [2]
Given Kudos: 178
Location: Ukraine
Concentration: Entrepreneurship, Technology
GMAT 1: 660 Q48 V33
GMAT 2: 740 Q50 V40
GMAT 2: 740 Q50 V40
Posts: 997
Kudos: 6,769
 [2]
1
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
avatar
ArnavPaliw
Joined: 14 May 2015
Last visit: 14 Sep 2015
Posts: 2
Own Kudos:
2
 [2]
Given Kudos: 2
GMAT Date: 08-01-2015
Posts: 2
Kudos: 2
 [2]
2
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Hello!!

1. Insufficient
2. Sufficient: - By knowing the length of arc we can find the angel AOB. Now can't we apply the sum of all the interior angels of a quadrilateral (in this case AOBC) is equal to 360'.

Here we know angle OAC=OBC=90' and angle AOB is 140' (finding the interior angel by length of the arc). So now equation will be 140+90+90+x=360
320+x=360
Hence x = 40'

Isn't option 2 sufficient to find the angle..Want expert comment.
User avatar
Harley1980
User avatar
Retired Moderator
Joined: 06 Jul 2014
Last visit: 14 Jun 2024
Posts: 997
Own Kudos:
6,769
 [1]
Given Kudos: 178
Location: Ukraine
Concentration: Entrepreneurship, Technology
GMAT 1: 660 Q48 V33
GMAT 2: 740 Q50 V40
GMAT 2: 740 Q50 V40
Posts: 997
Kudos: 6,769
 [1]
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
ArnavPaliw
Hello!!

1. Insufficient
2. Sufficient: - By knowing the length of arc we can find the angel AOB. Now can't we apply the sum of all the interior angels of a quadrilateral (in this case AOBC) is equal to 360'.

Here we know angle OAC=OBC=90' and angle AOB is 140' (finding the interior angel by length of the arc). So now equation will be 140+90+90+x=360
320+x=360
Hence x = 40'

Isn't option 2 sufficient to find the angle..Want expert comment.

Hello ArnavPaliw
You can't find angle only from arc because circle can be different sizes and it will change the angle
If circle really big then this angle can be 1 degree and if circle is very small this arc can be almost all circumference and angle can be for example a 350 degree.
Upd: Not in this case. In this case it can be only less than half of circumference. (thanks to the chetan2u for clarification)

But I like your application of formula (n-2)*180 this is more elegant way to solve last step than I used in my solution
avatar
ArnavPaliw
Joined: 14 May 2015
Last visit: 14 Sep 2015
Posts: 2
Own Kudos:
Given Kudos: 2
GMAT Date: 08-01-2015
Posts: 2
Kudos: 2
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Harley1980 Thanks for correcting !!!! Kudos.. :-)
User avatar
chetan2u
User avatar
GMAT Expert
Joined: 02 Aug 2009
Last visit: 27 Apr 2026
Posts: 11,229
Own Kudos:
Given Kudos: 335
Status:Math and DI Expert
Location: India
Concentration: Human Resources, General Management
GMAT Focus 1: 735 Q90 V89 DI81
Products:
Expert
Expert reply
GMAT Focus 1: 735 Q90 V89 DI81
Posts: 11,229
Kudos: 45,027
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Harley1980
ArnavPaliw
Hello!!

1. Insufficient
2. Sufficient: - By knowing the length of arc we can find the angel AOB. Now can't we apply the sum of all the interior angels of a quadrilateral (in this case AOBC) is equal to 360'.

Here we know angle OAC=OBC=90' and angle AOB is 140' (finding the interior angel by length of the arc). So now equation will be 140+90+90+x=360
320+x=360
Hence x = 40'

Isn't option 2 sufficient to find the angle..Want expert comment.

Hello ArnavPaliw
You can't find angle only from arc because circle can be different sizes and it will change the angle
If circle really big then this angle can be 1 degree and if circle is very small this arc can be almost all circumference and angle can be for example a 350 degree

But I like your application of formula (n-2)*180 this is more elegant way to solve last step than I used in my solution


Hi,
you are absolutely correct when you say that the length of the arc will vary as per the size of circle...
But this arc can never be almost equal to entire circumference..
it can never be greater than half the circumference. say its equal to half the circumference, the lines will never meet being parallel and greater than circumference will meet on the other side
User avatar
chetan2u
User avatar
GMAT Expert
Joined: 02 Aug 2009
Last visit: 27 Apr 2026
Posts: 11,229
Own Kudos:
Given Kudos: 335
Status:Math and DI Expert
Location: India
Concentration: Human Resources, General Management
GMAT Focus 1: 735 Q90 V89 DI81
Products:
Expert
Expert reply
GMAT Focus 1: 735 Q90 V89 DI81
Posts: 11,229
Kudos: 45,027
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Bunuel

In the figure above, two lines are tangent to a circle at points A and B. What is x?

(1) The area of the circle is 81π.
(2) The length of arc ADB is 7π.


Kudos for a correct solution.
Attachment:
2015-06-09_1333.png


1) statement gives us the radius but angle cannot be found by that... insuff
2) statement gives us the length of arc but we know nothing about the radius...in suff
combined suff as we can deduce the centre angle from ratio of arc to entire circumference
ans C
User avatar
Harley1980
User avatar
Retired Moderator
Joined: 06 Jul 2014
Last visit: 14 Jun 2024
Posts: 997
Own Kudos:
Given Kudos: 178
Location: Ukraine
Concentration: Entrepreneurship, Technology
GMAT 1: 660 Q48 V33
GMAT 2: 740 Q50 V40
GMAT 2: 740 Q50 V40
Posts: 997
Kudos: 6,769
Kudos
Add Kudos
Bookmarks
Bookmark this Post
chetan2u
Harley1980
ArnavPaliw
Hello!!

1. Insufficient
2. Sufficient: - By knowing the length of arc we can find the angel AOB. Now can't we apply the sum of all the interior angels of a quadrilateral (in this case AOBC) is equal to 360'.

Here we know angle OAC=OBC=90' and angle AOB is 140' (finding the interior angel by length of the arc). So now equation will be 140+90+90+x=360
320+x=360
Hence x = 40'

Isn't option 2 sufficient to find the angle..Want expert comment.

Hello ArnavPaliw
You can't find angle only from arc because circle can be different sizes and it will change the angle
If circle really big then this angle can be 1 degree and if circle is very small this arc can be almost all circumference and angle can be for example a 350 degree

But I like your application of formula (n-2)*180 this is more elegant way to solve last step than I used in my solution


Hi,
you are absolutely correct when you say that the length of the arc will vary as per the size of circle...
But this arc can never be almost equal to entire circumference..
it can never be greater than half the circumference. say its equal to half the circumference, the lines will never meet being parallel and greater than circumference will meet on the other side


Hello chetan2u, thanks for reprimand.

But I always think that we have two arcs in any case.

In our case we have arc AB with length 7pi and angle 140 degrees
And another arc AB with length 11pi and angle 220 degrees

Maybe I miss something but I never met a rule that arc can't be more than half of circumference.

P.S. I'm not saying that I am right, geometry is my weak place. But if it is not too hard, can you please, give some link on that rule?
User avatar
chetan2u
User avatar
GMAT Expert
Joined: 02 Aug 2009
Last visit: 27 Apr 2026
Posts: 11,229
Own Kudos:
45,027
 [1]
Given Kudos: 335
Status:Math and DI Expert
Location: India
Concentration: Human Resources, General Management
GMAT Focus 1: 735 Q90 V89 DI81
Products:
Expert
Expert reply
GMAT Focus 1: 735 Q90 V89 DI81
Posts: 11,229
Kudos: 45,027
 [1]
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Harley1980
thanks for reprimand.

But I always think that we have two arcs in any case.

In our case we have arc AB with length 7pi and angle 140 degrees
And another arc AB with length 11pi and angle 220 degrees

Maybe I miss something but I never met a rule that arc can't be more than half of circumference.

P.S. I'm not saying that I am right, geometry is my weak place. But if it is not too hard, can you please, give some link on that rule?

Sorry buddy if i have sounded that way..
you are correct that there will be two arcs between two points on circle and the length of one of them could be even 359 degrees.
But this is not possible in this case ...
If you have an angle subtended by the tangents at two points on circle, the arc subtended by those points and enclosed within these two points will be less than 180 and ofcourse the other arc may be 359 also as you have explained..
User avatar
Harley1980
User avatar
Retired Moderator
Joined: 06 Jul 2014
Last visit: 14 Jun 2024
Posts: 997
Own Kudos:
Given Kudos: 178
Location: Ukraine
Concentration: Entrepreneurship, Technology
GMAT 1: 660 Q48 V33
GMAT 2: 740 Q50 V40
GMAT 2: 740 Q50 V40
Posts: 997
Kudos: 6,769
Kudos
Add Kudos
Bookmarks
Bookmark this Post
chetan2u
Harley1980
thanks for reprimand.

But I always think that we have two arcs in any case.

In our case we have arc AB with length 7pi and angle 140 degrees
And another arc AB with length 11pi and angle 220 degrees

Maybe I miss something but I never met a rule that arc can't be more than half of circumference.

P.S. I'm not saying that I am right, geometry is my weak place. But if it is not too hard, can you please, give some link on that rule?

Sorry buddy if i have sounded that way..
you are correct that there will be two arcs between two points on circle and the length of one of them could be even 359 degrees.
But this is not possible in this case ...
If you have an angle subtended by the tangents at two points on circle, the arc subtended by those points and enclosed within these two points will be less than 180 and ofcourse the other arc may be 359 also as you have explained..

Oh, I get it. I missed constraints that impose tangents.
Thanks for clarifying ;)
I've changed my post
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 26 Apr 2026
Posts: 109,928
Own Kudos:
Given Kudos: 105,914
Products:
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 109,928
Kudos: 811,537
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Bunuel

In the figure above, two lines are tangent to a circle at points A and B. What is x?

(1) The area of the circle is 81π.
(2) The length of arc ADB is 7π.


Kudos for a correct solution.
Attachment:
The attachment 2015-06-09_1333.png is no longer available

MANHATTAN GMAT OFFICIAL SOLUTION:

The size of the angle depends on two things:

(1) INSUFFICIENT: From the rubber band picture on the right, we see that for a circle of fixed size, x can still vary with the length of arc ADB (i.e. x varies with the placement of C).

(2) INSUFFICIENT: We can't rely solely on the pictures above, as the arc length varies with both circle size (see left picture) and with placement of C relative to a circle of fixed size (see right picture). We'll draw two cases to prove that x can vary for a given arc ADB:

For a circle with circumference 28π, the arc ADB is 1/4 (= 7π/28π) of the circle, so x is 90°. For a circle with circumference 15π, the arc ADB is nearly half of the circle (= 7π/15π), and the lines tangent to the circle at A and B will be nearly parallel to each other, i.e. x is very small.

(1) AND (2) SUFFICIENT: If the area of the circle is πr^2 = 81π, then r = 9. The circumference of the circle is 2πr = 18. Thus, arc ADB is 7/18 (= 7π/18π) of the circumference of the circle. There is only one way to draw the lines tangent to the circle at A and B, so x can only be one value.


The correct answer is C.

Attachment:
2015-06-15_1649.png
2015-06-15_1649.png [ 59.77 KiB | Viewed 22096 times ]
Attachment:
2015-06-15_1650.png
2015-06-15_1650.png [ 19.39 KiB | Viewed 22004 times ]
Attachment:
2015-06-15_1652.png
2015-06-15_1652.png [ 17.32 KiB | Viewed 21906 times ]
User avatar
mvictor
User avatar
Board of Directors
Joined: 17 Jul 2014
Last visit: 14 Jul 2021
Posts: 2,118
Own Kudos:
1,277
 [1]
Given Kudos: 236
Location: United States (IL)
Concentration: Finance, Economics
GMAT 1: 650 Q49 V30
GPA: 3.92
WE:General Management (Transportation)
Products:
GMAT 1: 650 Q49 V30
Posts: 2,118
Kudos: 1,277
 [1]
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
we can draw 2 radii, from A and B to the origin. and a bisector from C to the origin. we get 2 right triangles with right angles at A and B. angle x is split in 2 similar small angles.
to get X, we need to find one of the two similar angles formed at origin COA or COB. knowing this angle, we can apply the triangle principle that the sum of all angles must equal to 180, and since we already know that 1 angle is 90, finding the second one would crack this bad boy.

1. this one tells that the r=9. or OA and OB=9. doesn't help much. Even if we draw a line from A to B, to form another 2 right triangles, we would still not be able to find the values of the angles formed at the origin. so 1 alone is insufficient. A and D - out.

2. ADC is 7pi. well, clearly this is insufficient. if we knew the circumference of the circle, we would be able to find the value of the two angles formed at the origin. B is out.

1+2 => we know area, we know the length of the chord..we can find the angles formed at the origin..C is sufficient..

just for myself calculations:
7pi/18pi = 7/18. multiply by 360 = 360*7/18 = 140. this would be the sum of the two similar angles formed at the origin. now we need to find the value of only 1.
140/2 = 70.
180 - 90 - 70 = 20 -> value of a small x sub-angle. x will be twice this value -> 40.
User avatar
fskilnik
Joined: 12 Oct 2010
Last visit: 03 Jan 2025
Posts: 883
Own Kudos:
1,889
 [2]
Given Kudos: 57
Status:GMATH founder
Expert
Expert reply
Posts: 883
Kudos: 1,889
 [2]
1
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
Bunuel

In the figure above, two lines are tangent to a circle at points A and B. What is x?

(1) The area of the circle is 81π.
(2) The length of arc ADB is 7π.
Attachment:
The attachment 2015-06-09_1333.png is no longer available

Excellent opportunity to explore our two powerful tools when dealing with geometric-related Data Sufficiency problems:

The GEOMETRIC BIFURCATION and
THE ALGEBRAIC-GEOMETRIC BIFURCATION

Please study the image attached!

This solution follows the notations and rationale taught in the GMATH method.

Regards,
fskilnik.
Attachments

05Set18_9w.gif
05Set18_9w.gif [ 55.61 KiB | Viewed 13579 times ]

User avatar
Karpov92
User avatar
Current Student
Joined: 17 Jun 2020
Last visit: 20 Mar 2023
Posts: 30
Own Kudos:
26
 [1]
Given Kudos: 253
Status:Stuck with 99 points. Call 911!
Location: Bangladesh
Concentration: Finance, General Management
GMAT 1: 630 Q45 V32
WE:General Management (Transportation)
GMAT 1: 630 Q45 V32
Posts: 30
Kudos: 26
 [1]
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Two theories you need to know -

Outside Angle Theorem: The measure of an angle formed by two secants, two tangents, or a secant and a tangent drawn from a point outside the circle is equal to half the difference of the measures of the intercepted arcs.

The size of the Outside angle depends on 2 things -
1. How large the circle is, (the length of its radius)
2. The distance of the outside point from the circle; it is inversely related to the size of the Arc.
Larger the Arc = Smaller the Outside Angle

1 & 2 are insufficient as 1 only indicates the radius of the circle & 2 only indicates the length of arc. We need both the info to find x.

1+2 : Here, r = 9 (from pi*r^2 = 81)
ARC length = 7pi = (Central angle / 360) * 2pi*9
Central Angle = 140

Minor Arc ADB’s Central Angle = 140
Major Arc ADB’s Central Angle = 360-140=220

So, Axxording to Outside Angle Theorem,
x = ½ ( 220-140) = ½ * 80 = 40.
User avatar
Rainman91
Joined: 03 Jul 2020
Last visit: 25 Nov 2025
Posts: 86
Own Kudos:
Given Kudos: 120
Posts: 86
Kudos: 33
Kudos
Add Kudos
Bookmarks
Bookmark this Post
We do assume that the length of each side from C to A and C to B are equal in lengths right? Even though it is not mentioned?
User avatar
bumpbot
User avatar
Non-Human User
Joined: 09 Sep 2013
Last visit: 04 Jan 2021
Posts: 38,987
Own Kudos:
Posts: 38,987
Kudos: 1,118
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Automated notice from GMAT Club BumpBot:

A member just gave Kudos to this thread, showing it’s still useful. I’ve bumped it to the top so more people can benefit. Feel free to add your own questions or solutions.

This post was generated automatically.
Moderators:
Math Expert
109928 posts
498 posts
212 posts