GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 17 Jul 2018, 18:23

### 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

# Arc AED above is a semicircle. EC is the height to the base AD. The ar

Author Message
TAGS:

### Hide Tags

Intern
Joined: 30 May 2014
Posts: 3
Concentration: Finance, Entrepreneurship
GMAT 1: 600 Q47 V27
GPA: 3.47
Arc AED above is a semicircle. EC is the height to the base AD. The ar [#permalink]

### Show Tags

27 Oct 2014, 09:17
2
00:00

Difficulty:

75% (hard)

Question Stats:

64% (02:44) correct 36% (03:01) wrong based on 160 sessions

### HideShow timer Statistics

Attachment:

T8910x.png [ 7.7 KiB | Viewed 3346 times ]
Arc AED above is a semicircle. EC is the height to the base AD. The area of triangle AEC is 96. If CE=12, what is the area of triangle CDE?

A. 48
B. 54
C. 72
D. 84
E. 150
SVP
Status: The Best Or Nothing
Joined: 27 Dec 2012
Posts: 1837
Location: India
Concentration: General Management, Technology
WE: Information Technology (Computer Software)
Re: Arc AED above is a semicircle. EC is the height to the base AD. The ar [#permalink]

### Show Tags

27 Oct 2014, 20:02
8
1

Area $$\triangle ECA = 96 = \frac{1}{2} * AC * 12$$

AC = 16

Arc AED is a semicircle means AD is the diameter of the circle

EA & ED are touching the end points of diameter, which means $$\angle AED = 90^{\circ}$$ (This is by property of circle)

Refer diagram below which shows various angles of both the triangles

Attachment:

T8910x.png [ 22.56 KiB | Viewed 3255 times ]

Note that $$\triangle ECA$$ and $$\triangle DCE$$ are similar triangles.

Now that they are similar, then there corresponding sides are also proportional

$$\triangle$$ ECA Base = 16 & Height = 12

$$\triangle$$ DCE Base = 12; so Height = $$12 * \frac{12}{16} = 9$$

Area $$\triangle DCE = \frac{1}{2} * 12 * 9 = 54$$
_________________

Kindly press "+1 Kudos" to appreciate

##### General Discussion
SVP
Status: The Best Or Nothing
Joined: 27 Dec 2012
Posts: 1837
Location: India
Concentration: General Management, Technology
WE: Information Technology (Computer Software)
Re: Arc AED above is a semicircle. EC is the height to the base AD. The ar [#permalink]

### Show Tags

27 Oct 2014, 20:09
1
minhphammr wrote:
Arc AED above is a semicircle. EC is the height to the base AD. The area of triangle AEC is 96. If CE=12, what is the area of triangle CDE?
A. 48
B. 54
C. 72
D. 84
E. 150

For detailed explanation, above post can be referred.

Shortcut:

Area AEC = 96; so AC = 16

Area shaded region = $$96 * \frac{12}{16} = 54$$

_________________

Kindly press "+1 Kudos" to appreciate

Current Student
Joined: 21 Aug 2014
Posts: 155
Location: United States
Concentration: Other, Operations
GMAT 1: 700 Q47 V40
GMAT 2: 690 Q44 V40
WE: Science (Pharmaceuticals and Biotech)
Re: Arc AED above is a semicircle. EC is the height to the base AD. The ar [#permalink]

### Show Tags

17 Dec 2014, 14:41
PareshGmat wrote:

Area $$\triangle ECA = 96 = \frac{1}{2} * AC * 12$$

AC = 16

Arc AED is a semicircle means AD is the diameter of the circle

EA & ED are touching the end points of diameter, which means $$\angle AED = 90^{\circ}$$ (This is by property of circle)

Refer diagram below which shows various angles of both the triangles

Attachment:
T8910x.png

Note that $$\triangle ECA$$ and $$\triangle DCE$$ are similar triangles.

Now that they are similar, then there corresponding sides are also proportional

$$\triangle$$ ECA Base = 16 & Height = 12

$$\triangle$$ DCE Base = 12; so Height = $$12 * \frac{12}{16} = 9$$

Area $$\triangle DCE = \frac{1}{2} * 12 * 9 = 54$$

Is it a general property that both triangles are going to be similar?
SVP
Status: The Best Or Nothing
Joined: 27 Dec 2012
Posts: 1837
Location: India
Concentration: General Management, Technology
WE: Information Technology (Computer Software)
Re: Arc AED above is a semicircle. EC is the height to the base AD. The ar [#permalink]

### Show Tags

17 Dec 2014, 19:13
1
nachobioteck wrote:
PareshGmat wrote:

Area $$\triangle ECA = 96 = \frac{1}{2} * AC * 12$$

AC = 16

Arc AED is a semicircle means AD is the diameter of the circle

EA & ED are touching the end points of diameter, which means $$\angle AED = 90^{\circ}$$ (This is by property of circle)

Refer diagram below which shows various angles of both the triangles

Attachment:
T8910x.png

Note that $$\triangle ECA$$ and $$\triangle DCE$$ are similar triangles.

Now that they are similar, then there corresponding sides are also proportional

$$\triangle$$ ECA Base = 16 & Height = 12

$$\triangle$$ DCE Base = 12; so Height = $$12 * \frac{12}{16} = 9$$

Area $$\triangle DCE = \frac{1}{2} * 12 * 9 = 54$$

Is it a general property that both triangles are going to be similar?

When corresponding angles are same (irrespective of dimensions), both triangles are similar

In this case, YES (Refer the diagram above for angle breakup)
_________________

Kindly press "+1 Kudos" to appreciate

Director
Joined: 25 Apr 2012
Posts: 701
Location: India
GPA: 3.21
Re: Arc AED above is a semicircle. EC is the height to the base AD. The ar [#permalink]

### Show Tags

17 Dec 2014, 21:31
minhphammr wrote:
Attachment:
T8910x.png
Arc AED above is a semicircle. EC is the height to the base AD. The area of triangle AEC is 96. If CE=12, what is the area of triangle CDE?

A. 48
B. 54
C. 72
D. 84
E. 150

The 2 triangles are similar and some relations can come handy

and CE^2=AC*CD

we know CE=12 and Area of Triangle AEC=96 so AC=96*2/12 or AC =16

So CD=CE^2/AC or CD=9

Then area of triangle CDE =1/2*9*12=54
_________________

“If you can't fly then run, if you can't run then walk, if you can't walk then crawl, but whatever you do you have to keep moving forward.”

GMAT Club Legend
Joined: 16 Oct 2010
Posts: 8124
Location: Pune, India
Re: Arc AED above is a semicircle. EC is the height to the base AD. The ar [#permalink]

### Show Tags

18 Dec 2014, 04:39
nachobioteck wrote:
Is it a general property that both triangles are going to be similar?

This post gives you some figures which should make you think about similar triangles. Check them out.
http://www.veritasprep.com/blog/2014/03 ... -the-gmat/
_________________

Karishma
Private Tutor for GMAT
Contact: bansal.karishma@gmail.com

Retired Moderator
Joined: 17 Sep 2013
Posts: 369
Concentration: Strategy, General Management
GMAT 1: 730 Q51 V38
WE: Analyst (Consulting)
Arc AED above is a semicircle. EC is the height to the base AD. The ar [#permalink]

### Show Tags

18 Dec 2014, 05:01
WoundedTiger wrote:
The 2 triangles are similar and some relations can come handy

I doubt the above 2 are true...the dimension of LHS is m while of RHS is m^2
_________________

Appreciate the efforts...KUDOS for all
Don't let an extra chromosome get you down..

Manager
Joined: 04 Oct 2013
Posts: 155
Location: India
GMAT Date: 05-23-2015
GPA: 3.45
Re: Arc AED above is a semicircle. EC is the height to the base AD. The ar [#permalink]

### Show Tags

19 Dec 2014, 05:52
Arc AED above is a semicircle. EC is the height to the base AD. The area of triangle AEC is 96. If CE=12, what is the area of triangle CDE?

A. 48
B. 54
C. 72
D. 84
E. 150

The area of triangle AEC is $$96 = \frac{1}{2}(AC)*(CE)$$ or $$AC = \frac{(96*2)}{12}=16$$

$$(EC)^2=AC*CD$$

$$CD= (12)^2/16=9$$

Area of the triangle CDE = $$\frac{1}{2}(EC)*(CD)$$ = $$\frac{(12*9)}{2}=54$$

Director
Joined: 13 Mar 2017
Posts: 610
Location: India
Concentration: General Management, Entrepreneurship
GPA: 3.8
WE: Engineering (Energy and Utilities)
Re: Arc AED above is a semicircle. EC is the height to the base AD. The ar [#permalink]

### Show Tags

06 Sep 2017, 01:01
minhphammr wrote:
Attachment:
T8910x.png
Arc AED above is a semicircle. EC is the height to the base AD. The area of triangle AEC is 96. If CE=12, what is the area of triangle CDE?

A. 48
B. 54
C. 72
D. 84
E. 150

In triangle ACE , 1/2*AC* CE = 96 -> AC = 16

Now triangle CED ~ triangle CAE
CE/CD = CA/CE
CE^2 = CA * CD (If you can remember this property of the right angles triangle, it will be highly helpful to solve the problem in a quick manner.)
12^2 = 16 * CD
CD = 9

Area CAE = 1/2 * 9 * 12 = 54

_________________

CAT 99th percentiler : VA 97.27 | DI-LR 96.84 | QA 98.04 | OA 98.95
UPSC Aspirants : Get my app UPSC Important News Reader from Play store.

MBA Social Network : WebMaggu

Appreciate by Clicking +1 Kudos ( Lets be more generous friends.)

What I believe is : "Nothing is Impossible, Even Impossible says I'm Possible" : "Stay Hungry, Stay Foolish".

Re: Arc AED above is a semicircle. EC is the height to the base AD. The ar   [#permalink] 06 Sep 2017, 01:01
Display posts from previous: Sort by

# Events & Promotions

 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®.