GMAT Question of the Day: Daily via email | Daily via Instagram New to GMAT Club? Watch this Video

 It is currently 31 Mar 2020, 02:30

### 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

# A Circle inscribed in a equilateral triangle ABC so that poi

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

### Hide Tags

Intern
Joined: 19 Oct 2009
Posts: 36
A Circle inscribed in a equilateral triangle ABC so that poi  [#permalink]

### Show Tags

01 Dec 2009, 11:01
2
10
00:00

Difficulty:

95% (hard)

Question Stats:

58% (03:02) correct 42% (03:18) wrong based on 120 sessions

### HideShow timer Statistics

A Circle inscribed in a equilateral triangle ABC so that point D lies on the circle and on line segment AC and point E lies on circle and on line segment AB. If line segment AB=6, what is the area of the figure created by line segments AD, AE and minot Arc DE.

A. 3√3 - (9/4) π
B. 3√3 - π
C. 6√3 - 3 π
D. 9√3 - 3 π
5. cannot be determined
Manager
Joined: 29 Oct 2009
Posts: 170
GMAT 1: 750 Q50 V42
Re: Tough Geometry  [#permalink]

### Show Tags

01 Dec 2009, 12:08
7
Since it is an equilateral triangle, the required area can be expressed as : [Area of Triangle - Area of Circle]/3

Now we know that the side of the triangle $$a$$ = 6

Therefore, area of triangle = $$\frac{\sqrt{3}}{4}a^2$$ = $$9\sqrt{3}$$

Radius of circle inscribed in an equilateral triangle (r) = $$a\frac{\sqrt{3}}{6}$$ = $$\sqrt{3}$$

Therefore, area of circle = $$\pi*r^2$$ = $$3\pi$$

Thus, required area = $$\frac{9\sqrt{3}-3\pi}{3}$$ = $$3\sqrt{3} - \pi$$

_________________
Click below to check out some great tips and tricks to help you deal with problems on Remainders!
http://gmatclub.com/forum/compilation-of-tips-and-tricks-to-deal-with-remainders-86714.html#p651942

Word Problems Made Easy!
1) Translating the English to Math : http://gmatclub.com/forum/word-problems-made-easy-87346.html
2) 'Work' Problems Made Easy : http://gmatclub.com/forum/work-word-problems-made-easy-87357.html
3) 'Distance/Speed/Time' Word Problems Made Easy : http://gmatclub.com/forum/distance-speed-time-word-problems-made-easy-87481.html
##### General Discussion
Math Expert
Joined: 02 Sep 2009
Posts: 62378
Re: Tough Geometry  [#permalink]

### Show Tags

10 Jan 2010, 09:38
3
gottabwise wrote:
:?
sriharimurthy wrote:
gmat620 wrote:
Ok thanks to you ppl !!

is the formula below correct ?

radius of circle in equilateral triangle = 1 /3 height

thanks once again ):

For an equilateral triangle with altitude 'h' :

Radius of inscribed circle = h/3

Radius of circumscribed circle = 2h/3

Thank you for the formulas. This is the first time I've come across them and I don't know if that's good or bad.

The property given by sriharimurthy is absolutely correct. But I'd like to add couple of things to this:

1. In any triangle the three medians intersect at a single point, called centroid.
2. In any triangle two-thirds of the length of each median is between the vertex and the centroid, while one-third is between the centroid and the midpoint of the opposite side.
3. In any triangle the bisectors intersect at a single point, called incenter. The incenter is also the center of the triangle's incircle - the largest circle that will fit inside the triangle.
4. In equilateral triangle altitude(height)=bisector=median.

From above:
With an equilateral triangle, the radius of the incircle is exactly half the radius of the circumcircle --> R=2r.

Radius of inscribed circle = height/3=bisector/3=median/3

Radius of circumscribed circle = height*2/3=bisector*2/3=median*2/3

For more see the Triangle chapter of Math Book in my signature.

Edited.
_________________
Manager
Joined: 30 Aug 2009
Posts: 213
Location: India
Concentration: General Management
Re: Tough Geometry  [#permalink]

### Show Tags

01 Dec 2009, 11:58
1
gmat620 wrote:
A Circle inscribed in a equilateral triangle ABC so that point D lies on the circle and on line segment AC and point E lies on circle and on line segment AB. If line segment AB=6, what is the area of the figure created by line segments AD, AE and minot Arc DE.

A. 3√3 - (9/4) π
B. 3√3 - π
C. 6√3 - 3 π
D. 9√3 - 3 π
5. cannot be determined

B - 3√3 - π
Area of equilateral triangle - √3/4 (a^2) = √3/4 (6^2) = 9√3
to find the inradius we have r = A/s where s is semiperimeter = 9√3/9 = √3

Area of the incircle = π r^2 = π (√3)^2 = 3π

Area of Triangle - Area of Incircle = 9√3 -3π

This area will be equal from all three sides hence (9√3 -3π)/3 = 3√3 - π
Manager
Joined: 17 Aug 2009
Posts: 128
Re: Tough Geometry  [#permalink]

### Show Tags

12 Jan 2010, 06:44
1
Bunuel, Correct me if i am wrong

Any Triangle in a circumscribed circle can be broken down into the median property (2/3rd, 1/3rd length). It does not have to restrict itself to equilateral triangles alone.
The medians intersect at a point called the centroid, which is also the center of the circumscribed circle.

Any Triangle can have an inscribed circle. Here, rather than the medians, we have the bisectors drawn from each vertex, with all bisectors meeting at the center of the circle, also called incircle.
However, if equilateral triangles have inscribed circles, the bisectors = medians and hence we can calculate the radius of the incircle

But how do we calculate the radius of a circle inscribed within a scalene triangle?
Intern
Joined: 19 Oct 2009
Posts: 36
Re: Tough Geometry  [#permalink]

### Show Tags

01 Dec 2009, 12:47
Ok thanks to you ppl !!

is the formula below correct ?

radius of circle in equilateral triangle = 1 /3 height

thanks once again ):
Manager
Joined: 29 Oct 2009
Posts: 170
GMAT 1: 750 Q50 V42
Re: Tough Geometry  [#permalink]

### Show Tags

01 Dec 2009, 13:21
gmat620 wrote:
Ok thanks to you ppl !!

is the formula below correct ?

radius of circle in equilateral triangle = 1 /3 height

thanks once again ):

For an equilateral triangle with altitude 'h' :

Radius of inscribed circle = h/3

Radius of circumscribed circle = 2h/3
_________________
Click below to check out some great tips and tricks to help you deal with problems on Remainders!
http://gmatclub.com/forum/compilation-of-tips-and-tricks-to-deal-with-remainders-86714.html#p651942

Word Problems Made Easy!
1) Translating the English to Math : http://gmatclub.com/forum/word-problems-made-easy-87346.html
2) 'Work' Problems Made Easy : http://gmatclub.com/forum/work-word-problems-made-easy-87357.html
3) 'Distance/Speed/Time' Word Problems Made Easy : http://gmatclub.com/forum/distance-speed-time-word-problems-made-easy-87481.html
Manager
Joined: 24 Jul 2009
Posts: 163
Location: Anchorage, AK
Schools: Mellon, USC, MIT, UCLA, NSCU
Re: Tough Geometry  [#permalink]

### Show Tags

09 Jan 2010, 23:34
sriharimurthy wrote:
gmat620 wrote:
Ok thanks to you ppl !!

is the formula below correct ?

radius of circle in equilateral triangle = 1 /3 height

thanks once again ):

For an equilateral triangle with altitude 'h' :

Radius of inscribed circle = h/3

Radius of circumscribed circle = 2h/3

Thank you for the formulas. This is the first time I've come across them and I don't know if that's good or bad.
Manager
Joined: 24 Jul 2009
Posts: 163
Location: Anchorage, AK
Schools: Mellon, USC, MIT, UCLA, NSCU
Re: Tough Geometry  [#permalink]

### Show Tags

11 Jan 2010, 20:06
1
The property given by sriharimurthy is absolutely correct. But I'd like to add couple of things to this:

1. In any triangle the three medians intersect at a single point, called centroid.
2. In any triangle two-thirds of the length of each median is between the vertex and the centroid, while one-third is between the centroid and the midpoint of the opposite side. So R=Median*2/3.
3. In any triangle the bisectors intersect at a single point, called incenter. The incenter is also the center of the triangle's incircle - the largest circle that will fit inside the triangle.
4. In equilateral triangle altitude(height)=bisector=median.

From above:
With an equilateral triangle, the radius of the incircle is exactly half the radius of the circumcircle --> R=2r.

Radius of inscribed circle = height/3=bisector/3=median/3

Radius of circumscribed circle = height*2/3=bisector*2/3=median*2/3

For more see the Triangle chapter of Math Book in my signature.[/quote]

Thanks again. Let's say someone forgets the formula for the radius of a circle inscribed in an equilateral triangle (that's a mouthful in itself). How else can the radius be found? Asking because the area you're looking for is pretty simple (A of triangle - A of circle)/3. I want to be able to do the work if necessary.

And yeah, I know I'm better off memorizing the formula since the clock's my enemy.
Manager
Joined: 27 Apr 2008
Posts: 155
Re: Tough Geometry  [#permalink]

### Show Tags

11 Jan 2010, 20:21
gottabwise wrote:
The property given by sriharimurthy is absolutely correct. But I'd like to add couple of things to this:

1. In any triangle the three medians intersect at a single point, called centroid.
2. In any triangle two-thirds of the length of each median is between the vertex and the centroid, while one-third is between the centroid and the midpoint of the opposite side. So R=Median*2/3.
3. In any triangle the bisectors intersect at a single point, called incenter. The incenter is also the center of the triangle's incircle - the largest circle that will fit inside the triangle.
4. In equilateral triangle altitude(height)=bisector=median.

From above:
With an equilateral triangle, the radius of the incircle is exactly half the radius of the circumcircle --> R=2r.

Radius of inscribed circle = height/3=bisector/3=median/3

Radius of circumscribed circle = height*2/3=bisector*2/3=median*2/3

For more see the Triangle chapter of Math Book in my signature.

Thanks again. Let's say someone forgets the formula for the radius of a circle inscribed in an equilateral triangle (that's a mouthful in itself). How else can the radius be found? Asking because the area you're looking for is pretty simple (A of triangle - A of circle)/3. I want to be able to do the work if necessary.

And yeah, I know I'm better off memorizing the formula since the clock's my enemy.[/quote]

Probably won't have enough time in less than 2 minutes after reading the question though.
Math Expert
Joined: 02 Sep 2009
Posts: 62378
Re: Tough Geometry  [#permalink]

### Show Tags

12 Jan 2010, 18:22
zaarathelab wrote:
Bunuel, Correct me if i am wrong

Any Triangle in a circumscribed circle can be broken down into the median property (2/3rd, 1/3rd length). It does not have to restrict itself to equilateral triangles alone.
The medians intersect at a point called the centroid, which is also the center of the circumscribed circle.

Any Triangle can have an inscribed circle. Here, rather than the medians, we have the bisectors drawn from each vertex, with all bisectors meeting at the center of the circle, also called incircle.
However, if equilateral triangles have inscribed circles, the bisectors = medians and hence we can calculate the radius of the incircle

But how do we calculate the radius of a circle inscribed within a scalene triangle?

It will depend on what is given. Generally radius of a circle inscribed in a triangle is $$r=\frac{\sqrt{s(s-a)(s-b)(s-c)}}{s}$$, OR $$r=\frac{2A}{a+b+c}$$, where $$s=\frac{1}{2}(a+b+c)$$, half of a perimeter and $$A$$ is an area of the triangle.
Attachment:

CircleInscribeTriangleGen0.gif [ 1.88 KiB | Viewed 9013 times ]

I should mention here that I've never seen the GMAT question requiring this formula. I wouldn't worry about this case at all.

NOTE: The intersection of medians is not the center of circumscribed circle.
_________________
Manager
Joined: 17 Aug 2009
Posts: 128
Re: Tough Geometry  [#permalink]

### Show Tags

14 Jan 2010, 05:40
Bunuel, Thanks for the info

You epitomize the name of the person who derived the above formula (Hero's formula using triangle's semi-perimeter)
Math Expert
Joined: 02 Sep 2009
Posts: 62378
Re: Tough Geometry  [#permalink]

### Show Tags

14 Jan 2010, 17:00
zaarathelab wrote:
Bunuel, Correct me if i am wrong

Any Triangle in a circumscribed circle can be broken down into the median property (2/3rd, 1/3rd length). It does not have to restrict itself to equilateral triangles alone.
The medians intersect at a point called the centroid, which is also the center of the circumscribed circle.

Any Triangle can have an inscribed circle. Here, rather than the medians, we have the bisectors drawn from each vertex, with all bisectors meeting at the center of the circle, also called incircle.
However, if equilateral triangles have inscribed circles, the bisectors = medians and hence we can calculate the radius of the incircle

But how do we calculate the radius of a circle inscribed within a scalene triangle?

Red part is not correct. The intersection of medians is not the center of circumscribed circle. There was a typo in the text.
_________________
Manager
Joined: 17 Aug 2009
Posts: 128
Re: Tough Geometry  [#permalink]

### Show Tags

16 Jan 2010, 12:46
Bunuel,

Got it! I was confusing circumcenter with centroid. Centroid is related to the medians of a triangle alone while circumcenter is related to the center of a circle that has a triangle inscribed within it, which i believe is constructing perpendicular bisectors (all meeting at circumcenter).

But is there a formulaic approach to finding the radius or the circumcenter of a circle that has a triangle inscribed within it?
Math Expert
Joined: 02 Sep 2009
Posts: 62378
Re: Tough Geometry  [#permalink]

### Show Tags

17 Jan 2010, 16:25
zaarathelab wrote:
Bunuel,

Got it! I was confusing circumcenter with centroid. Centroid is related to the medians of a triangle alone while circumcenter is related to the center of a circle that has a triangle inscribed within it, which i believe is constructing perpendicular bisectors (all meeting at circumcenter).

But is there a formulaic approach to finding the radius or the circumcenter of a circle that has a triangle inscribed within it?

Usually in GMAT questions when triangle is inscribed in circle it's either right triangle or equilateral. I've never seen the GMAT question involving the radius of the circle circumscribing scalene triangle.

For equilateral triangle inscribed in circle $$R=a\frac{\sqrt{3}}{3}$$, where $$a$$ is the side of the triangle.

For right triangle inscribed in circle $$R=\frac{hypotenuse}{2}$$.

For scalene triangle inscribed in circle $$R=\frac{abc}{4A}=\frac{abc}{4\sqrt{s(s-a)(s-b)(s-c)}}$$, where A is the area of the triangle, $$a,b,c$$ are the three sides of the triangle and $$s = \frac{a+b+c}{2}$$ which is the semi perimeter of the triangle.
_________________
Senior Manager
Joined: 13 May 2013
Posts: 395
Re: A Circle inscribed in a equilateral triangle ABC so that poi  [#permalink]

### Show Tags

16 Dec 2013, 10:33
A Circle inscribed in a equilateral triangle ABC so that point D lies on the circle and on line segment AC and point E lies on circle and on line segment AB. If line segment AB=6, what is the area of the figure created by line segments AD, AE and minot Arc DE.

A. 3√3 - (9/4) π
B. 3√3 - π
C. 6√3 - 3 π
D. 9√3 - 3 π
5. cannot be determined

I am a bit confused with this problem...

This is an equilateral triangle so we know that AE=AC=CB=6. We also know that we can draw a altitude from angle C to nine AB and create a 30:60:90 triangle. With this, we know that the height of the large triangle is 6/√3 and the base (AE) = 3.

Here is where I may be making a mistake. I believe that if a circle is inscribed in an equilateral triangle, the points at which the circle touches the line also signify the midpoint of the line (i.e. D = midpoint AC and E = midpoint AB) If that is the case then triangle ADE is also an equilateral triangle with a base of 3 and a height of 3/√3. Thus, we can find the area of the triangle. Also, because the circle is inscribed in an equilateral triangle, the midpoints of the triangle touch at intervals of 1/3rd of the circles circumference which means the arc DE =1/3 of the circle or 120 degrees. Despite doing what I think is right, I am getting the wrong answer. Can someone explain?
Math Expert
Joined: 02 Sep 2009
Posts: 62378
Re: A Circle inscribed in a equilateral triangle ABC so that poi  [#permalink]

### Show Tags

17 Dec 2013, 00:19
WholeLottaLove wrote:
A Circle inscribed in a equilateral triangle ABC so that point D lies on the circle and on line segment AC and point E lies on circle and on line segment AB. If line segment AB=6, what is the area of the figure created by line segments AD, AE and minot Arc DE.

A. 3√3 - (9/4) π
B. 3√3 - π
C. 6√3 - 3 π
D. 9√3 - 3 π
5. cannot be determined

I am a bit confused with this problem...

This is an equilateral triangle so we know that AE=AC=CB=6. We also know that we can draw a altitude from angle C to nine AB and create a 30:60:90 triangle. With this, we know that the height of the large triangle is 6/√3 and the base (AE) = 3.

Here is where I may be making a mistake. I believe that if a circle is inscribed in an equilateral triangle, the points at which the circle touches the line also signify the midpoint of the line (i.e. D = midpoint AC and E = midpoint AB) If that is the case then triangle ADE is also an equilateral triangle with a base of 3 and a height of 3/√3. Thus, we can find the area of the triangle. Also, because the circle is inscribed in an equilateral triangle, the midpoints of the triangle touch at intervals of 1/3rd of the circles circumference which means the arc DE =1/3 of the circle or 120 degrees. Despite doing what I think is right, I am getting the wrong answer. Can someone explain?

All is correct but the red parts.

The height of equilateral triangle is $$side*\frac{\sqrt{3}}{2}$$.

Thus the height of ABC is $$side*\frac{\sqrt{3}}{2}=3\sqrt{3}$$ and the height of ADE is $$side*\frac{\sqrt{3}}{2}=\frac{3\sqrt{3}}{2}$$.

BTW, easier solution is here: a-circle-inscribed-in-a-equilateral-triangle-abc-so-that-poi-87483.html#p657675
_________________
Senior Manager
Joined: 13 May 2013
Posts: 395
Re: A Circle inscribed in a equilateral triangle ABC so that poi  [#permalink]

### Show Tags

17 Dec 2013, 09:02
Ahh...Simple mistake! Thanks!

Bunuel wrote:
WholeLottaLove wrote:
A Circle inscribed in a equilateral triangle ABC so that point D lies on the circle and on line segment AC and point E lies on circle and on line segment AB. If line segment AB=6, what is the area of the figure created by line segments AD, AE and minot Arc DE.

A. 3√3 - (9/4) π
B. 3√3 - π
C. 6√3 - 3 π
D. 9√3 - 3 π
5. cannot be determined

I am a bit confused with this problem...

This is an equilateral triangle so we know that AE=AC=CB=6. We also know that we can draw a altitude from angle C to nine AB and create a 30:60:90 triangle. With this, we know that the height of the large triangle is 6/√3 and the base (AE) = 3.

Here is where I may be making a mistake. I believe that if a circle is inscribed in an equilateral triangle, the points at which the circle touches the line also signify the midpoint of the line (i.e. D = midpoint AC and E = midpoint AB) If that is the case then triangle ADE is also an equilateral triangle with a base of 3 and a height of 3/√3. Thus, we can find the area of the triangle. Also, because the circle is inscribed in an equilateral triangle, the midpoints of the triangle touch at intervals of 1/3rd of the circles circumference which means the arc DE =1/3 of the circle or 120 degrees. Despite doing what I think is right, I am getting the wrong answer. Can someone explain?

All is correct but the red parts.

The height of equilateral triangle is $$side*\frac{\sqrt{3}}{2}$$.

Thus the height of ABC is $$side*\frac{\sqrt{3}}{2}=3\sqrt{3}$$ and the height of ADE is $$side*\frac{\sqrt{3}}{2}=\frac{3\sqrt{3}}{2}$$.

BTW, easier solution is here: a-circle-inscribed-in-a-equilateral-triangle-abc-so-that-poi-87483.html#p657675
SVP
Joined: 06 Sep 2013
Posts: 1506
Concentration: Finance
Re: A Circle inscribed in a equilateral triangle ABC so that poi  [#permalink]

### Show Tags

02 Jan 2014, 05:48
gmat620 wrote:
A Circle inscribed in a equilateral triangle ABC so that point D lies on the circle and on line segment AC and point E lies on circle and on line segment AB. If line segment AB=6, what is the area of the figure created by line segments AD, AE and minot Arc DE.

A. 3√3 - (9/4) π
B. 3√3 - π
C. 6√3 - 3 π
D. 9√3 - 3 π
5. cannot be determined

To find the area of that region we need to learn three things

First the area of an equilateral triangle is s^2 sqrt (3) / 4 therefore area will be 9sqrt (3)

Next, radius of an inscribed circle will be '6'* sqrt (3) / 6 , where '6' is the side of the equilateral triangle

Therefore, radius will be sqrt (3) and thus the area of the circle will be 3 (pi)

Finally, since the circle divides the equilateral triangle evenly then Area of region ADE = (Area of equilateral triangle - Area of circle)/3 = 3 sqrt (3) - pi

Answer is hence B

Hope it helps
Let me know if you have any doubts OK?

Cheers!
J
Manager
Joined: 29 May 2017
Posts: 192
Location: Pakistan
Concentration: Social Entrepreneurship, Sustainability
Re: A Circle inscribed in a equilateral triangle ABC so that poi  [#permalink]

### Show Tags

12 Oct 2018, 23:59
I did something different and it would be helpful if my mistake is pointed out

1. the shape ADOEA is a kite. thus area between AD, AE and the minor arc DE will be: area of kite - area of the minor arc DE
2. OD EQUALS OE equals r (radius of the inscribed circle)
3. since side-->2 x r x sqrt(3) this implies 3--> 2 x r x sqrt(3), thus r--> 3/sqrt(3)
4. area of the minor arc DE is (2/3) x pi x r^2--> (2/3) x pi x (9/3)-->2pi
5. area of kite: ab x Sin C--> 3 x 3/sqrt(3) x Sin 90--> 9/sqrt(3)
6. answer: 9/sqrt(3) - 2pi

will be grateful for feedback
Re: A Circle inscribed in a equilateral triangle ABC so that poi   [#permalink] 12 Oct 2018, 23:59

Go to page    1   2    Next  [ 21 posts ]

Display posts from previous: Sort by

# A Circle inscribed in a equilateral triangle ABC so that poi

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

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