A circle is inscribed in equilateral triangle ABC so that point D lies on the circle and on line segment AC and point E lies on the circle and line segment AB. If AB=6, what is the area of the figure created by line segments AD, AE and minor arc DE?
Request one of the experts to please let us know how to solve this optimally?
Thanks!
5-Day Free Trial
5-day free, full-access trial TTP
Available with Beat the GMAT members only code
MORE DETAILS
Inscribed Circle - Equilateral Triangle
This topic has expert replies
-
PGPEX_aspirant
- Newbie | Next Rank: 10 Posts
- Posts: 2
- Joined: Tue Mar 30, 2010 4:48 am
- Thanked: 1 times
-
thephoenix
- Legendary Member
- Posts: 1560
- Joined: Tue Nov 17, 2009 2:38 am
- Thanked: 137 times
- Followed by:5 members
GMAT/MBA Expert
-
Rahul@gurome
- GMAT Instructor
- Posts: 1179
- Joined: Sun Apr 11, 2010 9:07 pm
- Location: Milpitas, CA
- Thanked: 447 times
- Followed by:88 members
The triangle ABC is an equilateral. So all angles are 60 and all sides are equal.
Let O be the centre of the circle and join A to O.
Let the point where BC touches the circle be F.
So A is the point of tangency and AD and AE are tangents.
Hence AD = AE.
Also B and C are points of tangency and BE = BF and CD = CF.
Now AB = AE + EB = AC = AD + DC = BC = BF+ FC.
So AE + EB = AD + DC = AE + DC
So EB = DC
Or BF = CF.
Since BF + CF = BC = 6 we have BF = CF = 3 = BE = CD
Since AC = AD + DC, we have that AD = AC-DC = 6 - 3 = 3 = AE.
Now consider triangle AOE and AOD.
Both are congruent.
Hence they have equal areas.
Also angle EAO = angle OAD = 60/2 = 30.
Angle ODA is 90.
So triangle AOD is 30 - 60 - 90.
So length of sides are in the ratio 1:sqrt(3):2.
So if AD = 3, then OD = sqrt(3).
So area of triangle AOD is ½*3*sqrt(3).
So area of triangle AEO is also ½*3*sqrt(3).
So area of quadrilateral AEOD is 2*{½*3*sqrt(3)} = 3*sqrt(3).
Now angle AOD = AOE = 60.
So angle EOD is 120.
So area of the sector EOD is 120/360*pi*{sqrt(3)}^2 = pi
Hence area formed by AE, AD and minor arc ED is 3*sqrt(3) - pi = 3*sqrt(3) - 22/7.
Let O be the centre of the circle and join A to O.
Let the point where BC touches the circle be F.
So A is the point of tangency and AD and AE are tangents.
Hence AD = AE.
Also B and C are points of tangency and BE = BF and CD = CF.
Now AB = AE + EB = AC = AD + DC = BC = BF+ FC.
So AE + EB = AD + DC = AE + DC
So EB = DC
Or BF = CF.
Since BF + CF = BC = 6 we have BF = CF = 3 = BE = CD
Since AC = AD + DC, we have that AD = AC-DC = 6 - 3 = 3 = AE.
Now consider triangle AOE and AOD.
Both are congruent.
Hence they have equal areas.
Also angle EAO = angle OAD = 60/2 = 30.
Angle ODA is 90.
So triangle AOD is 30 - 60 - 90.
So length of sides are in the ratio 1:sqrt(3):2.
So if AD = 3, then OD = sqrt(3).
So area of triangle AOD is ½*3*sqrt(3).
So area of triangle AEO is also ½*3*sqrt(3).
So area of quadrilateral AEOD is 2*{½*3*sqrt(3)} = 3*sqrt(3).
Now angle AOD = AOE = 60.
So angle EOD is 120.
So area of the sector EOD is 120/360*pi*{sqrt(3)}^2 = pi
Hence area formed by AE, AD and minor arc ED is 3*sqrt(3) - pi = 3*sqrt(3) - 22/7.
Rahul Lakhani
Quant Expert
Gurome, Inc.
https://www.GuroMe.com
On MBA sabbatical (at ISB) for 2011-12 - will stay active as time permits
1-800-566-4043 (USA)
+91-99201 32411 (India)
Quant Expert
Gurome, Inc.
https://www.GuroMe.com
On MBA sabbatical (at ISB) for 2011-12 - will stay active as time permits
1-800-566-4043 (USA)
+91-99201 32411 (India)
Did get the same result, though used a different method..can someone verify?
I) Calculate height of the triangle - dropping an altitude to the base, the right triangle would be in the ratio of 1:sqrt(3):2. Since we have an equilateral triangle, the median from A would create a right triangle with one side (hyp) as 6, base as 3 (6/2) and therefore, height as 3*sqrt(3)
II)For an inscribed circle in an equilateral triangle, the center of the circle would coincide with the intersection of the medians of the triangle, the medians dividing themselves in a ratio of 1:2. So, the shorter median would be 1/3rd of the total median length. This would also be the radius of the circle = [3*sqrt(3)]/3 (from I) = sqrt(3)
III) Area of circle = pi*[sqrt(3)]^2 = pi*3
IV) Area of Triangle = 1/2*b*h = [1/2]*[6]*[3*sqrt(3)] = 9*sqrt(3)
V) Subtracting the two, Area of Tri - Area of Cir = 9*sqrt(3) - pi*3 = 3 [3sqrt(3) - pi]
VI) Since all the figures are symmetric (Equilateral traingle and the circle inscribed, the three areas created by the vertices of the triangle and the minor arc of the circle would be the same. Hence, for one area, we divide this figure by three
= [3{3sqrt(3)} - pi] / 3 = 3sqrt(3) - pi
Makes sense? Or did I go wrong somewhere?
I) Calculate height of the triangle - dropping an altitude to the base, the right triangle would be in the ratio of 1:sqrt(3):2. Since we have an equilateral triangle, the median from A would create a right triangle with one side (hyp) as 6, base as 3 (6/2) and therefore, height as 3*sqrt(3)
II)For an inscribed circle in an equilateral triangle, the center of the circle would coincide with the intersection of the medians of the triangle, the medians dividing themselves in a ratio of 1:2. So, the shorter median would be 1/3rd of the total median length. This would also be the radius of the circle = [3*sqrt(3)]/3 (from I) = sqrt(3)
III) Area of circle = pi*[sqrt(3)]^2 = pi*3
IV) Area of Triangle = 1/2*b*h = [1/2]*[6]*[3*sqrt(3)] = 9*sqrt(3)
V) Subtracting the two, Area of Tri - Area of Cir = 9*sqrt(3) - pi*3 = 3 [3sqrt(3) - pi]
VI) Since all the figures are symmetric (Equilateral traingle and the circle inscribed, the three areas created by the vertices of the triangle and the minor arc of the circle would be the same. Hence, for one area, we divide this figure by three
= [3{3sqrt(3)} - pi] / 3 = 3sqrt(3) - pi
Makes sense? Or did I go wrong somewhere?
