It is currently 12 Dec 2017, 21:19

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

Events & Promotions

Events & Promotions in June
Open Detailed Calendar

An equilateral triangle ABC is inscribed in square ADEF, forming three

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

Hide Tags

3 KUDOS received
GMAT Instructor
avatar
Joined: 04 Jul 2006
Posts: 1259

Kudos [?]: 344 [3], given: 0

Location: Madrid
An equilateral triangle ABC is inscribed in square ADEF, forming three [#permalink]

Show Tags

New post 13 Apr 2007, 09:25
3
This post received
KUDOS
39
This post was
BOOKMARKED
00:00
A
B
C
D
E

Difficulty:

  75% (hard)

Question Stats:

64% (03:24) correct 36% (03:01) wrong based on 229 sessions

HideShow timer Statistics

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

Kudos [?]: 344 [3], given: 0

1 KUDOS received
Director
Director
User avatar
Joined: 14 Jan 2007
Posts: 775

Kudos [?]: 183 [1], given: 0

Re: An equilateral triangle ABC is inscribed in square ADEF, forming three [#permalink]

Show Tags

New post 13 Apr 2007, 10:54
1
This post received
KUDOS
It should be 'C'.
See the attached solution.
Attachments

Triangle.pdf [82.9 KiB]
Downloaded 757 times

To download please login or register as a user

Kudos [?]: 183 [1], given: 0

5 KUDOS received
Manager
Manager
avatar
Joined: 25 Mar 2007
Posts: 81

Kudos [?]: 9 [5], given: 0

Re: An equilateral triangle ABC is inscribed in square ADEF, forming three [#permalink]

Show Tags

New post 14 Apr 2007, 01:16
5
This post received
KUDOS
2
This post was
BOOKMARKED
My solution is very similar to yours - I got 2 as well. Maybe a bit shorter. I am attaching my solution as a jpeg file. Tell me what you think.
Attachments

SolutionTriangleProblem.jpg
SolutionTriangleProblem.jpg [ 300.68 KiB | Viewed 9391 times ]

Kudos [?]: 9 [5], given: 0

3 KUDOS received
VP
VP
avatar
Joined: 22 Nov 2007
Posts: 1078

Kudos [?]: 699 [3], given: 0

Re: An equilateral triangle ABC is inscribed in square ADEF, forming three [#permalink]

Show Tags

New post 01 Mar 2008, 23:17
3
This post received
KUDOS
5
This post was
BOOKMARKED
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. \(sqrt(5)\)

Kudos [?]: 699 [3], given: 0

1 KUDOS received
Director
Director
avatar
Joined: 01 Jan 2008
Posts: 617

Kudos [?]: 207 [1], given: 1

Re: An equilateral triangle ABC is inscribed in square ADEF, forming three [#permalink]

Show Tags

New post 04 Mar 2008, 06:42
1
This post received
KUDOS
marcodonzelli 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?

4/3
sqrt3
2
5/2
sqrt5


let's denote AB = BC = AC = 1. B belongs to DE, C belongs to EF
angle BAD = 15 degrees, EBC = 45 degrees
let's draw DH: H belongs to AB and DH is perpendicular to AB -> S(ABD) = 1/2*(AB)*(DH) = 1/2*1*(sin 15*cos15) = 1/2*(1/2)*sin(30) = 1/8 //DH = AB*sin(BAD)*cos(BDH) = AB*sin(BAD)*cos(BAD) = 1/2*sin(2*BAD)

let's draw EN: N belongs to BC and EN is perpendicular to BC -> S(BEC) = 1/2*(BC)*(EN) = 1/2*1*(sin45*cos45) = 1/2*1/sqrt(2)*1/sqrt(2)= 1/4

S(BEC)/S(ADB) = 2 -> C

Kudos [?]: 207 [1], given: 1

1 KUDOS received
Intern
Intern
avatar
Joined: 04 Mar 2008
Posts: 42

Kudos [?]: 3 [1], given: 6

Re: An equilateral triangle ABC is inscribed in square ADEF, forming three [#permalink]

Show Tags

New post 04 Mar 2008, 11:44
1
This post received
KUDOS
Let's denote AB = BC = AC = 1. B belongs to DE, C belongs to EF
angle BAD = 15 degrees, EBC = 45 degrees
Assume the length of the side of the square as y and BE = x.
Now, A(BEC)/A(ADB) =(AD* DB)/ (BE*CE) ignoring 1/2 in denominator as well as numerator
as BE = CE = x and AD = y and DB = (y-x)
if you replace values and on solving, A(BEC)/A(ADB) = x^2 / (y^2 -xy)

Now, as per pythagorus, BC = AB = 2 * sqrt x
and AB^2 = AD^2 + DB^2
put values for AB = 2 * sqrt x and AD = y and DB = y-x
solve, you'll get x^2 / (y^2 -xy) = 2

Hence, c

Kudos [?]: 3 [1], given: 6

2 KUDOS received
VP
VP
User avatar
Joined: 03 Apr 2007
Posts: 1339

Kudos [?]: 876 [2], given: 10

Reviews Badge
Re: An equilateral triangle ABC is inscribed in square ADEF, forming three [#permalink]

Show Tags

New post 04 Mar 2008, 22:06
2
This post received
KUDOS
chica wrote:
is there any easier approach without trigonometry involved ?? I am confused with drawing here.. do not get it


Let Y be length of the square
ABC is a triangle such that B intersects DE and C intersects FE (This info is from problem description)

Since it is an quilateral triangle , each interior angle of triangle ABC is 60 degrees.
Therefore angle BAD = angle CAF = 15.
Tiangle AFC is right angled. So angle ACF = 75.

Now solve angles for triangles ADB abd BCE.

Tiangle BCE turns out to be an isocles right angled triangle.


Area of BCE = BE * CE/2

let BE =x


Area of BCE = x* x/2


Area of ADB = AD* DB/2

but DB+BE=y

->DB = y-x

Area of ADB = y* (y-x)/2


Since ADB is a right angled triangle.

y^2 +(y-x)^2 = 2 * x^2

->y^2 -xy = x^2/2

Area of BCE/Area of ADB = [x* x/2]/[ y* (y-x)/2]
= x* x/y* (y-x) =2

Kudos [?]: 876 [2], given: 10

Expert Post
4 KUDOS received
CEO
CEO
User avatar
Joined: 17 Nov 2007
Posts: 3583

Kudos [?]: 4716 [4], given: 360

Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
GMAT 1: 750 Q50 V40
GMAT ToolKit User Premium Member
Re: An equilateral triangle ABC is inscribed in square ADEF, forming three [#permalink]

Show Tags

New post 07 Mar 2008, 09:46
4
This post received
KUDOS
Expert's post
Fig
Attachments

t60718.png
t60718.png [ 3.53 KiB | Viewed 9962 times ]


_________________

HOT! GMAT TOOLKIT 2 (iOS) / GMAT TOOLKIT (Android) - The OFFICIAL GMAT CLUB PREP APP, a must-have app especially if you aim at 700+ | PrepGame

Kudos [?]: 4716 [4], given: 360

CEO
CEO
User avatar
Joined: 21 Jan 2007
Posts: 2734

Kudos [?]: 1089 [0], given: 4

Location: New York City
Re: An equilateral triangle ABC is inscribed in square ADEF, forming three [#permalink]

Show Tags

New post 07 Mar 2008, 10:01
walker wrote:
Fig


Thanks, knew i could count on you, by the way, whats the program you use to draw these diagrams? i've been using adobe photoshop and it looks too crude
_________________

You tried your best and you failed miserably. The lesson is 'never try'. -Homer Simpson

Kudos [?]: 1089 [0], given: 4

Expert Post
CEO
CEO
User avatar
Joined: 17 Nov 2007
Posts: 3583

Kudos [?]: 4716 [0], given: 360

Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
GMAT 1: 750 Q50 V40
GMAT ToolKit User Premium Member
Re: An equilateral triangle ABC is inscribed in square ADEF, forming three [#permalink]

Show Tags

New post 07 Mar 2008, 10:51
bmwhype2 wrote:
by the way, whats the program you use to draw these diagrams? i've been using adobe photoshop and it looks too crude


adobe photoshop. I made template with grid.
_________________

HOT! GMAT TOOLKIT 2 (iOS) / GMAT TOOLKIT (Android) - The OFFICIAL GMAT CLUB PREP APP, a must-have app especially if you aim at 700+ | PrepGame

Kudos [?]: 4716 [0], given: 360

Expert Post
2 KUDOS received
CEO
CEO
User avatar
Joined: 17 Nov 2007
Posts: 3583

Kudos [?]: 4716 [2], given: 360

Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
GMAT 1: 750 Q50 V40
GMAT ToolKit User Premium Member
Re: An equilateral triangle ABC is inscribed in square ADEF, forming three [#permalink]

Show Tags

New post 07 Mar 2008, 11:14
2
This post received
KUDOS
Expert's post
By the way, second approach:

let x - a side of the equilateral triangle ABC
y - a side of the squre

1. The diagonal of squre = \(\sqrt{2}y = \frac{\sqrt{3}}{2}x+\frac{1}{2}x=\frac{x}{2}*(\sqrt{3}+1)\)
2. The area of squre = \(y^2=\frac{x^2}{8}*(\sqrt{3}+1)^2=\frac{x^2}{8}*(4+2\sqrt{3})=\frac{x^2}{4}*(2+sqrt{3})\)
3. The area of triangle ABC = \(\frac12*\frac{\sqrt{3}}{2}x*x=\frac{\sqrt{3}}{4}x^2\)
4. The area of triangle BEC = \(\frac12*\frac{x}{2}*x=\frac{x^2}{4}\)
5. The area of triangle ADB = 1/2 * (The area of squre - The area of triangle ABC - The area of triangle BEC) = \(\frac12*(\frac{x^2}{4}*(2+sqrt{3})-\frac{\sqrt{3}}{4}x^2 -\frac{x^2}{4})=\frac{x^2}{8}\)
6. ratio = \(\frac{\frac{x^2}{4}}{\frac{x^2}{8}}=2\)
_________________

HOT! GMAT TOOLKIT 2 (iOS) / GMAT TOOLKIT (Android) - The OFFICIAL GMAT CLUB PREP APP, a must-have app especially if you aim at 700+ | PrepGame

Kudos [?]: 4716 [2], given: 360

CEO
CEO
User avatar
Joined: 21 Jan 2007
Posts: 2734

Kudos [?]: 1089 [0], given: 4

Location: New York City
Re: An equilateral triangle ABC is inscribed in square ADEF, forming three [#permalink]

Show Tags

New post 07 Mar 2008, 11:32
goalsnr wrote:
chica wrote:
is there any easier approach without trigonometry involved ?? I am confused with drawing here.. do not get it


Let Y be length of the square
ABC is a triangle such that B intersects DE and C intersects FE (This info is from problem description)

Since it is an quilateral triangle , each interior angle of triangle ABC is 60 degrees.
Therefore angle BAD = angle CAF = 15.
Tiangle AFC is right angled. So angle ACF = 75.

Now solve angles for triangles ADB abd BCE.

Tiangle BCE turns out to be an isocles right angled triangle.


Area of BCE = BE * CE/2

let BE =x


Area of BCE = x* x/2


Area of ADB = AD* DB/2

but DB+BE=y

->DB = y-x

Area of ADB = y* (y-x)/2


Since ADB is a right angled triangle.

y^2 +(y-x)^2 = 2 * x^2

->y^2 -xy = x^2/2

Area of BCE/Area of ADB = [x* x/2]/[ y* (y-x)/2]
= x* x/y* (y-x) =2


we have the areas of both triangles, why do we need to maniipulate the second triangle with the pythg theorem?
_________________

You tried your best and you failed miserably. The lesson is 'never try'. -Homer Simpson

Kudos [?]: 1089 [0], given: 4

1 KUDOS received
VP
VP
avatar
Joined: 22 Nov 2007
Posts: 1078

Kudos [?]: 699 [1], given: 0

Re: An equilateral triangle ABC is inscribed in square ADEF, forming three [#permalink]

Show Tags

New post 08 Mar 2008, 10:14
1
This post received
KUDOS
Since |AB|=|BC|, x2 + 2xy + y2 +x2 = 2y2, so 2x2 +2xy =2x(x+y)= y2
The ratio of the area of triangle BEC to that of triangle ADB is y2/x(x+y) =2

AB and BC obviously are 2 sides of the triangle. x and y are the 2 parts of the side of the square

Kudos [?]: 699 [1], given: 0

VP
VP
User avatar
Joined: 03 Apr 2007
Posts: 1339

Kudos [?]: 876 [0], given: 10

Reviews Badge
Re: An equilateral triangle ABC is inscribed in square ADEF, forming three [#permalink]

Show Tags

New post 09 Mar 2008, 14:15
bmwhype2 wrote:
goalsnr wrote:
chica wrote:
is there any easier approach without trigonometry involved ?? I am confused with drawing here.. do not get it


Let Y be length of the square
ABC is a triangle such that B intersects DE and C intersects FE (This info is from problem description)

Since it is an quilateral triangle , each interior angle of triangle ABC is 60 degrees.
Therefore angle BAD = angle CAF = 15.
Tiangle AFC is right angled. So angle ACF = 75.

Now solve angles for triangles ADB abd BCE.

Tiangle BCE turns out to be an isocles right angled triangle.


Area of BCE = BE * CE/2

let BE =x


Area of BCE = x* x/2


Area of ADB = AD* DB/2

but DB+BE=y

->DB = y-x

Area of ADB = y* (y-x)/2


Since ADB is a right angled triangle.

y^2 +(y-x)^2 = 2 * x^2

->y^2 -xy = x^2/2

Area of BCE/Area of ADB = [x* x/2]/[ y* (y-x)/2]
= x* x/y* (y-x) =2


we have the areas of both triangles, why do we need to maniipulate the second triangle with the pythg theorem?


I used the pythg theorem to get the Area of ADB in terms of x.

Kudos [?]: 876 [0], given: 10

Senior Manager
Senior Manager
User avatar
Joined: 13 May 2013
Posts: 458

Kudos [?]: 204 [0], given: 134

Re: An equilateral triangle ABC is inscribed in square ADEF, forming three [#permalink]

Show Tags

New post 16 Dec 2013, 16:18
walker wrote:
By the way, second approach:

let x - a side of the equilateral triangle ABC
y - a side of the squre

1. The diagonal of squre = \(\sqrt{2}y = \frac{\sqrt{3}}{2}x+\frac{1}{2}x=\frac{x}{2}*(\sqrt{3}+1)\)


It seems as if you are saying the diagonal is equal to a 30:60:90 triangle. For some reason this problem is really killing me! Can someone explain?

Kudos [?]: 204 [0], given: 134

Senior Manager
Senior Manager
User avatar
Joined: 13 May 2013
Posts: 458

Kudos [?]: 204 [0], given: 134

Re: An equilateral triangle ABC is inscribed in square ADEF, forming three [#permalink]

Show Tags

New post 16 Dec 2013, 19:13
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. √(3)
C. 2
D. 5/2
E. √(5)

This is the closest I could get to a real answer:

The sides of the square = (Y) while the sides of the equilateral triangle = (S) Isosceles triangle CBE = 45:45:90 and as such, their lengths are equal to side CB=√2 so EB = EC = S/√2. If the sides of the square = y then sides DB = FC = y-(s/√2).

We know:

EB = EC = S/√2
DB = FC = y-(s/√2)

Area CBE = 1/2 (b*h)
Area = 1/2*(s/√2)*(s/√2)
Area = s^2 / 4

Area ADB = 1/2 (b*h)
Area = 1/2 (y-[s/√2])*(y)
Area = 1/2 y^2 - sy/√2
Area = (y^2)/2 - sy/√2*2
Area = (y^2)/2 - sy√2 / (√2 * √2 * 2)
Area = (y^2)/2 - sy√2 / (2 * 2)
Area = (y^2)/2 - sy√2 / (4)
Area = (2y^2)/4 - sy√2 / (4)
Area = [(2y^2) - sy√2] / (4)
Area = [y(2y - s√2)] / (4)

Difference in area:

[s^2 / 4] - [y(2y - s√2)] / (4)
[s^2 - y(2y - s√2)] / (4)

But, that's unlike ANY answer choice :-D

Kudos [?]: 204 [0], given: 134

Expert Post
4 KUDOS received
Veritas Prep GMAT Instructor
User avatar
G
Joined: 16 Oct 2010
Posts: 7791

Kudos [?]: 18108 [4], given: 236

Location: Pune, India
Re: An equilateral triangle ABC is inscribed in square ADEF, forming three [#permalink]

Show Tags

New post 16 Dec 2013, 22:07
4
This post received
KUDOS
Expert's post
WholeLottaLove 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. √(3)
C. 2
D. 5/2
E. √(5)

This is the closest I could get to a real answer:

The sides of the square = (Y) while the sides of the equilateral triangle = (S) Isosceles triangle CBE = 45:45:90 and as such, their lengths are equal to side CB=√2 so EB = EC = S/√2. If the sides of the square = y then sides DB = FC = y-(s/√2).

We know:

EB = EC = S/√2
DB = FC = y-(s/√2)

Area CBE = 1/2 (b*h)
Area = 1/2*(s/√2)*(s/√2)
Area = s^2 / 4

Area ADB = 1/2 (b*h)
Area = 1/2 (y-[s/√2])*(y)
Area = 1/2 y^2 - sy/√2
Area = (y^2)/2 - sy/√2*2
Area = (y^2)/2 - sy√2 / (√2 * √2 * 2)
Area = (y^2)/2 - sy√2 / (2 * 2)
Area = (y^2)/2 - sy√2 / (4)
Area = (2y^2)/4 - sy√2 / (4)
Area = [(2y^2) - sy√2] / (4)
Area = [y(2y - s√2)] / (4)

Difference in area:

[s^2 / 4] - [y(2y - s√2)] / (4)
[s^2 - y(2y - s√2)] / (4)

But, that's unlike ANY answer choice :-D



You need to get y in terms of s to get the ratio.
y - side of square
s - side of equilateral triangle

Note that the diagonal of a square is \(\sqrt{2}*side = \sqrt{2}*y\)

Also note that the diagonal is composed of 2 parts, the altitude of the equilateral triangle + the altitude of triangle BEC

Altitude of equilateral triangle is given by \(\sqrt{3}/2 * Side = \sqrt{3}/2 * s\)
Altitude of 45-45-90 triangle is half of the hypotenuse (which is s here). So altitude of BEC = s/2

So \(\sqrt{2}*y = \sqrt{3}/2 * s + s/2\)
So \(y = s * (\sqrt{3} + 1)/(2\sqrt{2})\)

Now, area of square = \(y^2 = s^2(\sqrt{3} + 1)^2/8\)

Area of triangle ABC = \((\sqrt{3}/4) * s^2\)

Area of triangle BEC \(= s^2/4\)

Area of triangle ADB = Area of triangle AFC \(= 1/2 * (s^2(\sqrt{3} + 1)^2/8 - (\sqrt{3}/4) * s^2 - s^2/4) = s^2/8\)

Hence the required ratio is \((s^2/4)/(s^2/8) = 2\)
_________________

Karishma
Veritas Prep | GMAT Instructor
My Blog

Get started with Veritas Prep GMAT On Demand for $199

Veritas Prep Reviews

Kudos [?]: 18108 [4], given: 236

Senior Manager
Senior Manager
User avatar
Joined: 13 May 2013
Posts: 458

Kudos [?]: 204 [0], given: 134

Re: An equilateral triangle ABC is inscribed in square ADEF, forming three [#permalink]

Show Tags

New post 17 Dec 2013, 05:14
Oh wow - the answer is so obvious to me now. Thank you for the clarification! :-D

VeritasPrepKarishma wrote:
WholeLottaLove 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. √(3)
C. 2
D. 5/2
E. √(5)

This is the closest I could get to a real answer:

The sides of the square = (Y) while the sides of the equilateral triangle = (S) Isosceles triangle CBE = 45:45:90 and as such, their lengths are equal to side CB=√2 so EB = EC = S/√2. If the sides of the square = y then sides DB = FC = y-(s/√2).

We know:

EB = EC = S/√2
DB = FC = y-(s/√2)

Area CBE = 1/2 (b*h)
Area = 1/2*(s/√2)*(s/√2)
Area = s^2 / 4

Area ADB = 1/2 (b*h)
Area = 1/2 (y-[s/√2])*(y)
Area = 1/2 y^2 - sy/√2
Area = (y^2)/2 - sy/√2*2
Area = (y^2)/2 - sy√2 / (√2 * √2 * 2)
Area = (y^2)/2 - sy√2 / (2 * 2)
Area = (y^2)/2 - sy√2 / (4)
Area = (2y^2)/4 - sy√2 / (4)
Area = [(2y^2) - sy√2] / (4)
Area = [y(2y - s√2)] / (4)

Difference in area:

[s^2 / 4] - [y(2y - s√2)] / (4)
[s^2 - y(2y - s√2)] / (4)

But, that's unlike ANY answer choice :-D



You need to get y in terms of s to get the ratio.
y - side of square
s - side of equilateral triangle

Note that the diagonal of a square is \(\sqrt{2}*side = \sqrt{2}*y\)

Also note that the diagonal is composed of 2 parts, the altitude of the equilateral triangle + the altitude of triangle BEC

Altitude of equilateral triangle is given by \(\sqrt{3}/2 * Side = \sqrt{3}/2 * s\)
Altitude of 45-45-90 triangle is half of the hypotenuse (which is s here). So altitude of BEC = s/2

So \(\sqrt{2}*y = \sqrt{3}/2 * s + s/2\)
So \(y = s * (\sqrt{3} + 1)/(2\sqrt{2})\)

Now, area of square = \(y^2 = s^2(\sqrt{3} + 1)^2/8\)

Area of triangle ABC = \((\sqrt{3}/4) * s^2\)

Area of triangle BEC \(= s^2/4\)

Area of triangle ADB = Area of triangle AFC \(= 1/2 * (s^2(\sqrt{3} + 1)^2/8 - (\sqrt{3}/4) * s^2 - s^2/4) = s^2/8\)

Hence the required ratio is \((s^2/4)/(s^2/8) = 2\)

Kudos [?]: 204 [0], given: 134

Current Student
User avatar
Joined: 06 Sep 2013
Posts: 1966

Kudos [?]: 758 [0], given: 355

Concentration: Finance
GMAT ToolKit User
Re: An equilateral triangle ABC is inscribed in square ADEF, forming three [#permalink]

Show Tags

New post 15 Jan 2014, 05:24
1
This post was
BOOKMARKED
VeritasPrepKarishma wrote:
WholeLottaLove 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. √(3)
C. 2
D. 5/2
E. √(5)

This is the closest I could get to a real answer:

The sides of the square = (Y) while the sides of the equilateral triangle = (S) Isosceles triangle CBE = 45:45:90 and as such, their lengths are equal to side CB=√2 so EB = EC = S/√2. If the sides of the square = y then sides DB = FC = y-(s/√2).

We know:

EB = EC = S/√2
DB = FC = y-(s/√2)

Area CBE = 1/2 (b*h)
Area = 1/2*(s/√2)*(s/√2)
Area = s^2 / 4

Area ADB = 1/2 (b*h)
Area = 1/2 (y-[s/√2])*(y)
Area = 1/2 y^2 - sy/√2
Area = (y^2)/2 - sy/√2*2
Area = (y^2)/2 - sy√2 / (√2 * √2 * 2)
Area = (y^2)/2 - sy√2 / (2 * 2)
Area = (y^2)/2 - sy√2 / (4)
Area = (2y^2)/4 - sy√2 / (4)
Area = [(2y^2) - sy√2] / (4)
Area = [y(2y - s√2)] / (4)

Difference in area:

[s^2 / 4] - [y(2y - s√2)] / (4)
[s^2 - y(2y - s√2)] / (4)

But, that's unlike ANY answer choice :-D



You need to get y in terms of s to get the ratio.
y - side of square
s - side of equilateral triangle

Note that the diagonal of a square is \(\sqrt{2}*side = \sqrt{2}*y\)

Also note that the diagonal is composed of 2 parts, the altitude of the equilateral triangle + the altitude of triangle BEC

Altitude of equilateral triangle is given by \(\sqrt{3}/2 * Side = \sqrt{3}/2 * s\)
Altitude of 45-45-90 triangle is half of the hypotenuse (which is s here). So altitude of BEC = s/2

So \(\sqrt{2}*y = \sqrt{3}/2 * s + s/2\)
So \(y = s * (\sqrt{3} + 1)/(2\sqrt{2})\)

Now, area of square = \(y^2 = s^2(\sqrt{3} + 1)^2/8\)

Area of triangle ABC = \((\sqrt{3}/4) * s^2\)

Area of triangle BEC \(= s^2/4\)

Area of triangle ADB = Area of triangle AFC \(= 1/2 * (s^2(\sqrt{3} + 1)^2/8 - (\sqrt{3}/4) * s^2 - s^2/4) = s^2/8\)

Hence the required ratio is \((s^2/4)/(s^2/8) = 2\)


Karishma this is an awesome explanation for this tough problem, thanks a lot

Just one small doubt, how do you know that ADB and AFC right triangles are equal? and both different from BCE, I have a bit of trouble making those sort of observations

Many thanks!
Cheers!
J :)

Kudos [?]: 758 [0], given: 355

Expert Post
1 KUDOS received
Veritas Prep GMAT Instructor
User avatar
G
Joined: 16 Oct 2010
Posts: 7791

Kudos [?]: 18108 [1], given: 236

Location: Pune, India
Re: An equilateral triangle ABC is inscribed in square ADEF, forming three [#permalink]

Show Tags

New post 15 Jan 2014, 20:16
1
This post received
KUDOS
Expert's post
jlgdr wrote:
Just one small doubt, how do you know that ADB and AFC right triangles are equal? and both different from BCE, I have a bit of trouble making those sort of observations

Many thanks!
Cheers!
J :)


Regular polygons inscribed in other regular polygons or circles make symmetrical figures. Try drawing them out. The sides/angles you think are equal will usually be equal.
Angle DAF is 90 and BAC is 60 so angles DAB and FAC will be 15 degrees each. Also AD = AF and AB = AC. So triangles ADB and AFC are congruent. On the other hand, in triangle BEC, side BE = CE so it is a 45-45-90 triangle.
_________________

Karishma
Veritas Prep | GMAT Instructor
My Blog

Get started with Veritas Prep GMAT On Demand for $199

Veritas Prep Reviews

Kudos [?]: 18108 [1], given: 236

Re: An equilateral triangle ABC is inscribed in square ADEF, forming three   [#permalink] 15 Jan 2014, 20:16

Go to page    1   2    Next  [ 23 posts ] 

Display posts from previous: Sort by

An equilateral triangle ABC is inscribed in square ADEF, forming three

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


GMAT Club MBA Forum Home| About| Terms and Conditions| GMAT Club Rules| Contact| Sitemap

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

Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.