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# An equilateral triangle ABC is inscribed in square ADEF,

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An equilateral triangle ABC is inscribed in square ADEF, [#permalink]

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17 Jul 2008, 07:11
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An equilateral triangle ABC is inscribed in square ADEF, forming three right triangles: ADB, ACF and BEC. What is the ratio of the area of triangle BEC to that of triangle ADB?

A. 4/3
B. $$sqrt(3)$$
C. 2
D. 5/2
E. $$sqrt(5)$$
[Reveal] Spoiler: OA

Last edited by Bunuel on 25 Jul 2013, 03:38, edited 2 times in total.
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Joined: 27 May 2008
Posts: 543
Re: PS: Geometry [#permalink]

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17 Jul 2008, 08:45
side of triangle BCE = a
side of square = (a+b)
two sides of triangle ACF are b and (a+b)

side of equilatrel triangle = $$a*sqrt(2)$$

sum of area of all triangles = area of square

sqrt(3)/4 * 2a^2 + 1/2 a^2 + 2 * 1/2 * b(a+b) = (a+b)^2

simplify this

b = (sqrt(3) - 1)/2 * a
a+b = a * ( sqrt(3) + 1) / 2

area of BCE = 1/2 a^2
area of AFC = 1/2 * b * (a+b)
= 1/2 * a^2 (3-1) / 4

ratio = 2: 1

time to solve : 7 minutes
SVP
Joined: 30 Apr 2008
Posts: 1874
Location: Oklahoma City
Schools: Hard Knocks
Re: PS: Geometry [#permalink]

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17 Jul 2008, 10:22
Here is how I approached the problem.

AD (side of square) = 1
BE (side of 45:45:90 triangle) = y
CB (side of equilateral triangle) = z

We're looking for the ratio of the area of BEC to ADB, or $$\frac{BEC}{ADB}$$

If area of Tri. BEC = $$\frac{y^2}{2}$$ and Tri.ADB = $$\frac{1(1-y)}{2}$$ then we have
$$\frac{y^2}{2} * \frac{2}{1-y} = \frac{y^2}{1-y}$$

if y = $$\frac{1}{2}$$, then y^2 = 1/4 and 1-y = 1/2, so we get 1/4 : 1/2 or 1/2....that's not an option.

Attachment:

TriangleInscribed.jpg [ 12.87 KiB | Viewed 6807 times ]

_________________

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J Allen Morris
**I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a$$. GMAT Club Premium Membership - big benefits and savings Senior Manager Joined: 07 Jan 2008 Posts: 289 Re: PS: Geometry [#permalink] ### Show Tags 17 Jul 2008, 10:39 tarek99 wrote: An equilateral triangle ABC is inscribed in square ADEF, forming three right triangles: ADB, ACF and BEC. What is the ratio of the area of triangle BEC to that of triangle ADB? a) 4/3 b) sqrt(s) c) 2 d) 5/2 e) sqrt(5) I believe the size of these triangles can vary, meaning we can increase one to decrease the size of another. So, we can't come up with right answer unless they provide us pre-configured info. No? If not, I guess I would just press Next on this one SVP Joined: 21 Jul 2006 Posts: 1510 Re: PS: Geometry [#permalink] ### Show Tags 17 Jul 2008, 10:40 durgesh79 wrote: side of triangle BCE = a side of square = (a+b) two sides of triangle ACF are b and (a+b) side of equilatrel triangle = $$a*sqrt(2)$$ sum of area of all triangles = area of square sqrt(3)/4 * 2a^2 + 1/2 a^2 + 2 * 1/2 * b(a+b) = (a+b)^2 simplify this b = (sqrt(3) - 1)/2 * a a+b = a * ( sqrt(3) + 1) / 2 area of BCE = 1/2 a^2 area of AFC = 1/2 * b * (a+b) = 1/2 * a^2 (3-1) / 4 ratio = 2: 1 answer C would you please explain how you got the side of the equilateral triangle to be time to solve : 7 minutes would you please explain how you got the side of the equilateral triangle to be $$a*sqrt(2)$$? SVP Joined: 30 Apr 2008 Posts: 1874 Location: Oklahoma City Schools: Hard Knocks Re: PS: Geometry [#permalink] ### Show Tags 17 Jul 2008, 10:43 if he makes 1 side, (the base) of BCE = a, it's a 45:45:90 triangle, so it has $$1:1:\sqrt{2}$$ sides. Therefore, one side of the triangle makes that hypotnuse and you have $$a\sqrt{2}$$ See the picture I drew above. I have the sides of BCE = y and the hypotnuse = z. He labeled his $$a$$ rather than y like I did. tarek99 wrote: durgesh79 wrote: side of triangle BCE = a side of square = (a+b) two sides of triangle ACF are b and (a+b) side of equilatrel triangle = $$a*sqrt(2)$$ sum of area of all triangles = area of square sqrt(3)/4 * 2a^2 + 1/2 a^2 + 2 * 1/2 * b(a+b) = (a+b)^2 simplify this b = (sqrt(3) - 1)/2 * a a+b = a * ( sqrt(3) + 1) / 2 area of BCE = 1/2 a^2 area of AFC = 1/2 * b * (a+b) = 1/2 * a^2 (3-1) / 4 ratio = 2: 1 answer C would you please explain how you got the side of the equilateral triangle to be time to solve : 7 minutes would you please explain how you got the side of the equilateral triangle to be $$a*sqrt(2)$$? _________________ ------------------------------------ J Allen Morris **I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a$$.

GMAT Club Premium Membership - big benefits and savings

Last edited by jallenmorris on 17 Jul 2008, 10:44, edited 1 time in total.
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Joined: 12 Jul 2008
Posts: 515
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Re: PS: Geometry [#permalink]

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17 Jul 2008, 10:43
tarek99 wrote:
durgesh79 wrote:
side of triangle BCE = a
side of square = (a+b)
two sides of triangle ACF are b and (a+b)

side of equilatrel triangle = $$a*sqrt(2)$$

sum of area of all triangles = area of square

sqrt(3)/4 * 2a^2 + 1/2 a^2 + 2 * 1/2 * b(a+b) = (a+b)^2

simplify this

b = (sqrt(3) - 1)/2 * a
a+b = a * ( sqrt(3) + 1) / 2

area of BCE = 1/2 a^2
area of AFC = 1/2 * b * (a+b)
= 1/2 * a^2 (3-1) / 4

ratio = 2: 1

would you please explain how you got the side of the equilateral triangle to be

time to solve : 7 minutes

would you please explain how you got the side of the equilateral triangle to be $$a*sqrt(2)$$?

The sides are equal, so it's a 45/45/90 triangle. You know the sides are equal because the inscribed triangle is equilateral.
SVP
Joined: 30 Apr 2008
Posts: 1874
Location: Oklahoma City
Schools: Hard Knocks
Re: PS: Geometry [#permalink]

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17 Jul 2008, 10:48
durgesh, can you label which of these goes to which triangle and how you came up with $$\frac{\sqrt{3}}{4}$$?

$$\frac{\sqrt{3}}{4}$$ * 2a^2

1/2 a^2

2 * 1/2 * b(a+b)
_________________

------------------------------------
J Allen Morris
**I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a. GMAT Club Premium Membership - big benefits and savings Manager Joined: 04 Apr 2008 Posts: 215 Location: Pune Re: PS: Geometry [#permalink] ### Show Tags 17 Jul 2008, 10:51 How are you guys taking wo sides as contant i am not getting it...........as how the triangle is alligned can vary and so is the ratio of the sides....like allen has taken BE and CE as same is there any rule or was it just an asuumption _________________ Every Problem Has a Sloution So keep working AB SVP Joined: 21 Jul 2006 Posts: 1510 Re: PS: Geometry [#permalink] ### Show Tags 17 Jul 2008, 10:54 jallenmorris wrote: if he makes 1 side, (the base) of BCE = a, it's a 45:45:90 triangle, so it has $$1:1:\sqrt{2}$$ sides. Therefore, one side of the triangle makes that hypotnuse and you have $$a\sqrt{2}$$ See the picture I drew above. I have the sides of BCE = y and the hypotnuse = z. He labeled his $$a$$ rather than y like I did. tarek99 wrote: durgesh79 wrote: side of triangle BCE = a side of square = (a+b) two sides of triangle ACF are b and (a+b) side of equilatrel triangle = $$a*sqrt(2)$$ sum of area of all triangles = area of square sqrt(3)/4 * 2a^2 + 1/2 a^2 + 2 * 1/2 * b(a+b) = (a+b)^2 simplify this b = (sqrt(3) - 1)/2 * a a+b = a * ( sqrt(3) + 1) / 2 area of BCE = 1/2 a^2 area of AFC = 1/2 * b * (a+b) = 1/2 * a^2 (3-1) / 4 ratio = 2: 1 answer C would you please explain how you got the side of the equilateral triangle to be time to solve : 7 minutes would you please explain how you got the side of the equilateral triangle to be $$a*sqrt(2)$$? but how can you decide that it's a 45:45:90 triangle? what if it's a 30:60:90 triangle? how can you be so sure which one it is? for example, if it were a 30:60:90 triangle, than the hypotenuse would be 2a, then what??? Last edited by tarek99 on 17 Jul 2008, 11:02, edited 3 times in total. Director Joined: 12 Jul 2008 Posts: 515 Schools: Wharton Re: PS: Geometry [#permalink] ### Show Tags 17 Jul 2008, 10:57 apurva1985 wrote: How are you guys taking wo sides as contant i am not getting it...........as how the triangle is alligned can vary and so is the ratio of the sides....like allen has taken BE and CE as same is there any rule or was it just an asuumption There's only one way the triangle can align if it's an equilateral triangle. Manager Joined: 04 Apr 2008 Posts: 215 Location: Pune Re: PS: Geometry [#permalink] ### Show Tags 17 Jul 2008, 11:00 That is what i was trying to say is that the answer can vary from person to person..........Plz can somebody expalin it in detail......... _________________ Every Problem Has a Sloution So keep working AB SVP Joined: 30 Apr 2008 Posts: 1874 Location: Oklahoma City Schools: Hard Knocks Re: PS: Geometry [#permalink] ### Show Tags 17 Jul 2008, 11:05 Because we know it is an equilateral triangle inscribe in a square and both the square and triangle share point A. So the sides coming from point A for the triangle towards the other sides fo the square are the same length. Look at my post above with the drawing in it. I believe that to be a rather accurate picture of this problem (although not drawn to scale or perfect angles). We know the inscribed triangle is an equialteral (also an iscoceles). In order for the 2 sides to extend out and hit the square, they must touch in the exact same spot (relative to the side) or one side of the triangle would be longer than the other and it would no longer be equilater (or iscoceles). tarek99 wrote: but how can you decide that it's a 45:45:90 triangle? what if it's a 30:60:90 triangle? how can you be so sure which one it is? _________________ ------------------------------------ J Allen Morris **I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a.

GMAT Club Premium Membership - big benefits and savings

Manager
Joined: 04 Apr 2008
Posts: 215
Location: Pune
Re: PS: Geometry [#permalink]

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17 Jul 2008, 11:17
Even i was feeling that as if you try to distribute the angles what will be the angles of the triangle..........either what i am decepting is wrong or the problem is wrong..........i am going by the figure given by allen
jallenmorris wrote:
Because we know it is an equilateral triangle inscribe in a square and both the square and triangle share point A. So the sides coming from point A for the triangle towards the other sides fo the square are the same length.

Look at my post above with the drawing in it. I believe that to be a rather accurate picture of this problem (although not drawn to scale or perfect angles).

We know the inscribed triangle is an equialteral (also an iscoceles). In order for the 2 sides to extend out and hit the square, they must touch in the exact same spot (relative to the side) or one side of the triangle would be longer than the other and it would no longer be equilater (or iscoceles).

tarek99 wrote:

but how can you decide that it's a 45:45:90 triangle? what if it's a 30:60:90 triangle? how can you be so sure which one it is?

_________________

Every Problem Has a Sloution So keep working
AB

SVP
Joined: 21 Jul 2006
Posts: 1510
Re: PS: Geometry [#permalink]

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17 Jul 2008, 11:24
jallenmorris wrote:
Because we know it is an equilateral triangle inscribe in a square and both the square and triangle share point A. So the sides coming from point A for the triangle towards the other sides fo the square are the same length.

Look at my post above with the drawing in it. I believe that to be a rather accurate picture of this problem (although not drawn to scale or perfect angles).

We know the inscribed triangle is an equialteral (also an iscoceles). In order for the 2 sides to extend out and hit the square, they must touch in the exact same spot (relative to the side) or one side of the triangle would be longer than the other and it would no longer be equilater (or iscoceles).

tarek99 wrote:

but how can you decide that it's a 45:45:90 triangle? what if it's a 30:60:90 triangle? how can you be so sure which one it is?

fine, but this only tells me that ADB and AFC are equal and that DB is the same as CF. I still can't see how we choose the triangle to be either 45:45:90 or 30:60:90. I'm really sorry for troubling you, but would you please explain this point?
thanks
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Joined: 04 Apr 2008
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Re: PS: Geometry [#permalink]

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17 Jul 2008, 11:41
tarek99 wrote:
jallenmorris wrote:
Because we know it is an equilateral triangle inscribe in a square and both the square and triangle share point A. So the sides coming from point A for the triangle towards the other sides fo the square are the same length.

Look at my post above with the drawing in it. I believe that to be a rather accurate picture of this problem (although not drawn to scale or perfect angles).

We know the inscribed triangle is an equialteral (also an iscoceles). In order for the 2 sides to extend out and hit the square, they must touch in the exact same spot (relative to the side) or one side of the triangle would be longer than the other and it would no longer be equilater (or iscoceles).

tarek99 wrote:

but how can you decide that it's a 45:45:90 triangle? what if it's a 30:60:90 triangle? how can you be so sure which one it is?

fine, but this only tells me that ADB and AFC are equal and that DB is the same as CF. I still can't see how we choose the triangle to be either 45:45:90 or 30:60:90. I'm really sorry for troubling you, but would you please explain this point?
thanks

I think i have got it in (triangle ABD) angle BAD((90-60)/2)=15,angle ADB=90,angle ABD=75, using these values in triangle BCE angle cbe will be(180-75-60=45) so from here we got it
i suppose rest can be calculated with this
_________________

Every Problem Has a Sloution So keep working
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Re: PS: Geometry [#permalink]

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17 Jul 2008, 11:49
Attachment:

TriangleExplanation.jpg [ 36.61 KiB | Viewed 6233 times ]

_________________

------------------------------------
J Allen Morris
**I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a$$. GMAT Club Premium Membership - big benefits and savings SVP Joined: 21 Jul 2006 Posts: 1510 Re: PS: Geometry [#permalink] ### Show Tags 17 Jul 2008, 11:58 perfect! that fills the gap in the puzzle! i'll now work through my calculations again to see whether i can get the final answer. thanks SVP Joined: 21 Jul 2006 Posts: 1510 Re: PS: Geometry [#permalink] ### Show Tags 17 Jul 2008, 12:23 I really appreciate it. ok, what a brutal problem this has been! hehe....i managed to get to the correct answer. How about we all try to figure out whether there could be a faster way to solve this? I honestly don't believe that the GMAC guys actually expect us to use all these endless steps to arrive to our answer. There must be a much faster or more abstract way to solve this....hmm... SVP Joined: 30 Apr 2008 Posts: 1874 Location: Oklahoma City Schools: Hard Knocks Re: PS: Geometry [#permalink] ### Show Tags 17 Jul 2008, 12:36 Here is a solution that works pretty quickly, but I'm not sure about finishing it to come up with an actual ratio of 2:1. See Picture Attached (same as other one, but this post is on page 2, so i attached it again). We're looking for BEC:ADB. Right now we have BEC = $$\frac{y^2}{2}$$ and ADB = $$1*(1-y)*0.5$$ or $$\frac{1-y}{2}$$ This is the same as $$\frac{y^2}{2}$$ divided by $$\frac{1-y}{2}$$ or $$\frac{y^2}{2}$$ * $$\frac{2}{1-y}=\frac{y^2}{1-y}$$ We know that y is a fraction because we made the entire side = 1, but how much of that side is y? Attachment: TriangleInscribed.jpg [ 16.13 KiB | Viewed 6121 times ] _________________ ------------------------------------ J Allen Morris **I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a$$.

GMAT Club Premium Membership - big benefits and savings

Re: PS: Geometry   [#permalink] 17 Jul 2008, 12:36

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# An equilateral triangle ABC is inscribed in square ADEF,

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