It is currently 22 Nov 2017, 02:58

### 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

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# In the figure above, what is the area of triangular region B

Author Message
TAGS:

### Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 42302

Kudos [?]: 133014 [0], given: 12402

In the figure above, what is the area of triangular region B [#permalink]

### Show Tags

03 Feb 2014, 01:13
Expert's post
5
This post was
BOOKMARKED
00:00

Difficulty:

5% (low)

Question Stats:

84% (00:49) correct 16% (01:02) wrong based on 549 sessions

### HideShow timer Statistics

The Official Guide For GMAT® Quantitative Review, 2ND Edition

Attachment:

Untitled.png [ 11.64 KiB | Viewed 9614 times ]
In the figure above, what is the area of triangular region BCD ?

(A) $$4\sqrt{2}$$
(B) 8
(C) $$8\sqrt{2}$$
(D) 16
(E) $$16\sqrt{2}$$

Problem Solving
Question: 71
Category: Geometry Triangles; Area
Page: 71
Difficulty: 550

GMAT Club is introducing a new project: The Official Guide For GMAT® Quantitative Review, 2ND Edition - Quantitative Questions Project

Each week we'll be posting several questions from The Official Guide For GMAT® Quantitative Review, 2ND Edition and then after couple of days we'll provide Official Answer (OA) to them along with a slution.

We'll be glad if you participate in development of this project:
2. Please vote for the best solutions by pressing Kudos button;
3. Please vote for the questions themselves by pressing Kudos button;
4. Please share your views on difficulty level of the questions, so that we have most precise evaluation.

Thank you!
[Reveal] Spoiler: OA

_________________

Kudos [?]: 133014 [0], given: 12402

Math Expert
Joined: 02 Sep 2009
Posts: 42302

Kudos [?]: 133014 [0], given: 12402

Re: In the figure above, what is the area of triangular region B [#permalink]

### Show Tags

03 Feb 2014, 01:13
SOLUTION

In the figure above, what is the area of triangular region BCD ?

(A) $$4\sqrt{2}$$
(B) 8
(C) $$8\sqrt{2}$$
(D) 16
(E) $$16\sqrt{2}$$

Using Pythagorean theorem $$BD=\sqrt{4^2+4^2}=4\sqrt{2}$$.

The area of triangle BCD = $$\frac{1}{2}*BD*BC=\frac{1}{2}*4\sqrt{2}*4=8\sqrt{2}$$.

_________________

Kudos [?]: 133014 [0], given: 12402

Director
Joined: 25 Apr 2012
Posts: 722

Kudos [?]: 865 [1], given: 724

Location: India
GPA: 3.21
Re: In the figure above, what is the area of triangular region B [#permalink]

### Show Tags

03 Feb 2014, 03:30
1
KUDOS
Attachment:

untitled.PNG [ 15.91 KiB | Viewed 8697 times ]

In the figure above, what is the area of triangular region BCD ?

(A) $$4\sqrt{2}$$
(B) 8
(C) $$8\sqrt{2}$$
(D) 16
(E) $$16\sqrt{2}$$

Sol: In the given question Triangle ABD is isosele with sides AB=AD=4 and therefore BD=4$$\sqrt{2}$$

Area of triangle BCD is 1/2*4*4$$\sqrt{2}$$

600 level is okay
Ans is C
_________________

“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.”

Kudos [?]: 865 [1], given: 724

Senior Manager
Joined: 06 Aug 2011
Posts: 388

Kudos [?]: 238 [1], given: 82

Re: In the figure above, what is the area of triangular region B [#permalink]

### Show Tags

03 Feb 2014, 08:05
1
KUDOS
its C.

BD will be 4square root 2. (that will be base) * 4/2=8 square root 2.
_________________

Bole So Nehal.. Sat Siri Akal.. Waheguru ji help me to get 700+ score !

Kudos [?]: 238 [1], given: 82

Senior Manager
Joined: 20 Dec 2013
Posts: 267

Kudos [?]: 108 [1], given: 29

Location: India
Re: In the figure above, what is the area of triangular region B [#permalink]

### Show Tags

03 Feb 2014, 11:16
1
KUDOS
Ans C

First we calculate BD by Hypotenuse Theorem:
BD=\sqrt{4^2+4^2}=4\sqrt{2}

Area BCD=1/2 * 4\sqrt{2} * 4= 8\sqrt{2}

Kudos [?]: 108 [1], given: 29

Director
Joined: 04 Dec 2015
Posts: 696

Kudos [?]: 320 [1], given: 261

Location: India
Concentration: Technology, Strategy
Schools: ISB '19, IIMA , IIMB, XLRI
WE: Information Technology (Consulting)
In the figure above, what is the area of triangular region BCD ? [#permalink]

### Show Tags

08 Jul 2017, 01:56
1
KUDOS
parkerd wrote:
This question has already been answered, but I just needed clarification on how \sqrt{4^2 + 4^2} = 4\sqrt{2}?

(A) 4√2
(B) 8
(C) 8√2
(D) 16
(E) 16√2

$$\sqrt{4^2 + 4^2}$$

Taking $$4^2$$ common, we get;

$$\sqrt{4^2(1+1)}$$

$$\sqrt{4^2(2)}$$ ------------ ($$\sqrt{4^2} = 4$$)

$$4\sqrt{2}$$

Hope its clear now.

_________________
Please Press "+1 Kudos" to appreciate.

Kudos [?]: 320 [1], given: 261

Director
Joined: 22 May 2016
Posts: 995

Kudos [?]: 344 [0], given: 594

In the figure above, what is the area of triangular region B [#permalink]

### Show Tags

08 Jul 2017, 06:26
Bunuel wrote:
In the figure above, what is the area of triangular region BCD ?

(A) $$4\sqrt{2}$$
(B) 8
(C) $$8\sqrt{2}$$
(D) 16
(E)$$16\sqrt{2}$$

The area of right triangle BCD is $$\frac{1}{2}b*h$$, where b = BC = 4, and h is BD because height must be perpendicular to base. Angle DBC of 90° indicates exactly that.To find length of BD, the hypotenuse of triangle ABD ...

Triangle ABD is an isosceles right triangle having sides in the ratio $$x : x : x\sqrt{2}$$.

So to find BD, simply multiply 4 by $$\sqrt{2}$$ = $$4\sqrt{2}$$.

$$(\frac{1}{2})$$$$(4)$$$$(4\sqrt{2})$$
= $$8\sqrt{2}$$

Kudos [?]: 344 [0], given: 594

Manager
Joined: 30 Apr 2013
Posts: 91

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

Re: In the figure above, what is the area of triangular region B [#permalink]

### Show Tags

22 Sep 2017, 07:09
Can we not consider AD = 4 as the height of the triangle BCD?

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

Math Expert
Joined: 02 Sep 2009
Posts: 42302

Kudos [?]: 133014 [0], given: 12402

Re: In the figure above, what is the area of triangular region B [#permalink]

### Show Tags

22 Sep 2017, 07:19
santro789 wrote:
Can we not consider AD = 4 as the height of the triangle BCD?

The altitude (height) of a triangle is the perpendicular from the base to the opposite vertex. Does AD satisfy this for triangle BCD?
_________________

Kudos [?]: 133014 [0], given: 12402

Re: In the figure above, what is the area of triangular region B   [#permalink] 22 Sep 2017, 07:19
Display posts from previous: Sort by