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

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01 Dec 2009, 11:01
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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
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Re: Tough Geometry [#permalink]

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01 Dec 2009, 11:58
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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
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Re: Tough Geometry [#permalink]

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01 Dec 2009, 12:08
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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 = $$\frac{9\sqrt{3}-3\pi}{3}$$ = $$3\sqrt{3} - \pi$$

Answer : B
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Re: Tough Geometry [#permalink]

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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 ):
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Re: Tough Geometry [#permalink]

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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
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Re: Tough Geometry [#permalink]

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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.
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Re: Tough Geometry [#permalink]

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10 Jan 2010, 09:38
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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.
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Re: Tough Geometry [#permalink]

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11 Jan 2010, 20:06
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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.
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Re: Tough Geometry [#permalink]

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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.
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Re: Tough Geometry [#permalink]

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12 Jan 2010, 06:44
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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?
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Re: Tough Geometry [#permalink]

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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 5887 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.
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Re: Tough Geometry [#permalink]

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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)
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Re: Tough Geometry [#permalink]

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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.
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Re: Tough Geometry [#permalink]

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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?
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Re: Tough Geometry [#permalink]

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17 Jan 2010, 16:25
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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.
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Re: A Circle inscribed in a equilateral triangle ABC so that poi [#permalink]

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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?
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Re: A Circle inscribed in a equilateral triangle ABC so that poi [#permalink]

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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
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Re: A Circle inscribed in a equilateral triangle ABC so that poi [#permalink]

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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
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Re: A Circle inscribed in a equilateral triangle ABC so that poi [#permalink]

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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
Re: A Circle inscribed in a equilateral triangle ABC so that poi   [#permalink] 02 Jan 2014, 05:48

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