GMAT Question of the Day: Daily via email | Daily via Instagram New to GMAT Club? Watch this Video

It is currently 03 Apr 2020, 16:11

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

In the diagram above, ED is parallel to GH, and the circle has a diame

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

Hide Tags

Find Similar Topics 
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 62496
In the diagram above, ED is parallel to GH, and the circle has a diame  [#permalink]

Show Tags

New post 08 Mar 2015, 19:27
00:00
A
B
C
D
E

Difficulty:

  15% (low)

Question Stats:

98% (02:49) correct 3% (04:35) wrong based on 69 sessions

HideShow timer Statistics

Manager
Manager
User avatar
Joined: 23 Oct 2014
Posts: 84
Concentration: Marketing
In the diagram above, ED is parallel to GH, and the circle has a diame  [#permalink]

Show Tags

New post 08 Mar 2015, 19:40
2
1
Bunuel wrote:
Attachment:
The attachment gpp_img4.png is no longer available
In the diagram above, ED is parallel to GH, and the circle has a diameter of 13. If ED = 5 and GH = 15, what is the area of triangle FGH?

(A) 240
(B) 270
(C) 300
(D) 330
(E) 360

Kudos for a correct solution.


Angle FED = 90 degrees since FD is the diameter of the circle. Since GH is || to ED we know that angle GHF is also 90 degrees as well.

We should know at this point that the triangles are similar with all the sides of triangle GHF is three times bigger than the sides of triangle FED.

To figure out the area of triangle GHF we need to figure out the height of the right triangle FED.

FD^2 = ED^2 + FE^2
13^2 = ED^2 + 25
ED = 12

Knowing that FH = 3*FE we can plug in.

FH = 3(12)
FH = 36

Now we can get the area of right triangle GHF.

(B*H)/2 = Area
(15*36)/2 = Area
270 = Area

The answer is B.
Attachments

delete.jpg
delete.jpg [ 73.09 KiB | Viewed 2144 times ]

Manager
Manager
avatar
B
Status: single
Joined: 19 Jan 2015
Posts: 80
Location: India
GPA: 3.2
WE: Sales (Pharmaceuticals and Biotech)
Re: In the diagram above, ED is parallel to GH, and the circle has a diame  [#permalink]

Show Tags

New post 08 Mar 2015, 21:17
1
ED=5, FD=13, THEN ANGLE AT E 90DEGREE
THEN EF=12

ED PARALLEL TO GH
THEN FOLLOWING SIDES ARE EQUAL IN RATIO
EF/GF=FD/FH=ED/GH

THEN 12/GF=5/15 =>GF=36.

13/FH=1/3=>FH=39.

FH=39, GF=36 AND GH=15

RIGHT ANGLED TRIANGLE. = 1/2*36*15=270
option B sufficient
Veritas Prep GMAT Instructor
User avatar
V
Joined: 16 Oct 2010
Posts: 10240
Location: Pune, India
Re: In the diagram above, ED is parallel to GH, and the circle has a diame  [#permalink]

Show Tags

New post 08 Mar 2015, 21:28
1
Bunuel wrote:
Attachment:
gpp_img4.png
In the diagram above, ED is parallel to GH, and the circle has a diameter of 13. If ED = 5 and GH = 15, what is the area of triangle FGH?

(A) 240
(B) 270
(C) 300
(D) 330
(E) 360

Kudos for a correct solution.


Note that the triangle FED is similar to triangle FHG by AA (angles F of both triangles are vertically opposite angles and EH is a transversal to parallel lines ED and GH so angles E and H are equal because they are alternate interior angles)

Since ratio of their sides is 5:15 i.e. 1:3, ratio of their areas will be 1:9 (square of the ratio of sides)

FD = 13 and ED = 5 in right triangle FED so this is a 5-12-13 triangle. Area = (1/2)*5*12 = 30

Area of triangle FGH = 9*30 = 270

Answer (B)
_________________
Karishma
Veritas Prep GMAT Instructor

Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >
Manager
Manager
User avatar
Status: A mind once opened never loses..!
Joined: 05 Mar 2015
Posts: 186
Location: India
MISSION : 800
WE: Design (Manufacturing)
GMAT ToolKit User
Re: In the diagram above, ED is parallel to GH, and the circle has a diame  [#permalink]

Show Tags

New post 08 Mar 2015, 23:26
2
Hi

This question test the similar triangles concept

As GF // ED
Therefore Angle marked in the diagram are equal and by AAA Triangles GHF & FED are similar
Now as we know for a pair of similar triangles the ratio of the corresponding sides is equal

Therefore GH/ED = HF/FE
Now HF = 36

Area of triangle = 1/2*15*36 = 270
Attachments

File comment: Solution
gpp_img4.png
gpp_img4.png [ 21.63 KiB | Viewed 2080 times ]

SVP
SVP
User avatar
Status: The Best Or Nothing
Joined: 27 Dec 2012
Posts: 1713
Location: India
Concentration: General Management, Technology
WE: Information Technology (Computer Software)
Re: In the diagram above, ED is parallel to GH, and the circle has a diame  [#permalink]

Show Tags

New post 09 Mar 2015, 01:35
1
Answer = b = 270

Right triangle DEF is similar to right triangle GHF; Area in the ratio 1:3

Area of right triangle GHF \(= \frac{1}{2} (12*3) * 15 = 270\)
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 62496
Re: In the diagram above, ED is parallel to GH, and the circle has a diame  [#permalink]

Show Tags

New post 15 Mar 2015, 19:30
Bunuel wrote:
Image
In the diagram above, ED is parallel to GH, and the circle has a diameter of 13. If ED = 5 and GH = 15, what is the area of triangle FGH?

(A) 240
(B) 270
(C) 300
(D) 330
(E) 360

Kudos for a correct solution.


MAGOOSH OFFICIAL SOLUTION:

Let’s begin by focusing on triangle FED. The angle ∠E spans a diameter, so ∠E = 90°. Thus, triangle FED is a right triangle with hypotenuse FD = 13 and leg ED = 5. It will save you a tremendous amount of calculations here if you already have memorized the 5-12-13 Pythagorean Triplet. Thus, FE = 12. Area = (0.5)bh = (0.5)(12)(5) = 30.

Because ED and GH are parallel, all the angles are equal, and the two triangles are similar. From ED = 5 to GH = 15 we scale up by a scale factor of k = 3. Lengths are multiplied by the scale factor, and areas are multiplied by the scale factor squared, k^2 = 9. 30*9 = 270 is the area of FGH.

Answer = (B)
_________________
Non-Human User
User avatar
Joined: 09 Sep 2013
Posts: 14464
Re: In the diagram above, ED is parallel to GH, and the circle has a diame  [#permalink]

Show Tags

New post 09 May 2016, 22:32
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________
GMAT Club Bot
Re: In the diagram above, ED is parallel to GH, and the circle has a diame   [#permalink] 09 May 2016, 22:32
Display posts from previous: Sort by

In the diagram above, ED is parallel to GH, and the circle has a diame

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





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