The triangles in the figure above are equilateral and the : GMAT Problem Solving (PS)
Check GMAT Club Decision Tracker for the Latest School Decision Releases https://gmatclub.com/AppTrack

 It is currently 22 Feb 2017, 18:27

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

The triangles in the figure above are equilateral and the

Author Message
TAGS:

Hide Tags

Manager
Joined: 31 Oct 2007
Posts: 113
Location: Frankfurt, Germany
Followers: 2

Kudos [?]: 62 [4] , given: 0

The triangles in the figure above are equilateral and the [#permalink]

Show Tags

05 Apr 2008, 21:14
4
KUDOS
4
This post was
BOOKMARKED
00:00

Difficulty:

45% (medium)

Question Stats:

63% (02:25) correct 37% (01:16) wrong based on 548 sessions

HideShow timer Statistics

Attachment:

2triangles.GIF [ 1.58 KiB | Viewed 9473 times ]
The triangles in the figure above are equilateral and the ratio of the length of a side of the larger triangle to the length of a side of the smaller triangle is 2/1. If the area of the larger triangular region is K, what is the area of the shaded region in terms of K?

(A) 3/4K
(B) 2/3K
(C) 1/2K
(D) 1/3K
(E) 1/4K
[Reveal] Spoiler: OA

_________________

Jimmy Low, Frankfurt, Germany
Blog: http://mytrainmaster.wordpress.com
GMAT Malaysia: http://gmatmalaysia.blogspot.com

VP
Joined: 10 Jun 2007
Posts: 1459
Followers: 7

Kudos [?]: 263 [1] , given: 0

Re: 1000PS: Section 7 Question 16 [#permalink]

Show Tags

05 Apr 2008, 21:44
1
KUDOS
jimmylow wrote:
16. The triangles in the figure above are equilateral and the ratio of the length of a side of the larger triangle to the length of a side of the smaller triangle is 2/1. If the area of the larger triangular region is K, what is the area of the shaded region in terms of K?
(A) 3/4K
(B) 2/3K
(C) 1/2K
(D) 1/3K
(E) 1/4K

Area of shaded = Area of Large - Area of Small
Give: Area of Large = K
We need to find Area of Small in term of K.
Triangle area = 1/2 * base * height
and since a side of the large triangle is twice larger than a side of small triangle, we are dealing with (1/2) * (1/2) = 1/4 factor.
Therefore, Area of shaded = K - K/4 = 3K/4
SVP
Joined: 04 May 2006
Posts: 1925
Schools: CBS, Kellogg
Followers: 23

Kudos [?]: 1044 [0], given: 1

Re: 1000PS: Section 7 Question 16 [#permalink]

Show Tags

05 Apr 2008, 21:48
jimmylow wrote:
16. The triangles in the figure above are equilateral and the ratio of the length of a side of the larger triangle to the length of a side of the smaller triangle is 2/1. If the area of the larger triangular region is K, what is the area of the shaded region in terms of K?
(A) 3/4K
(B) 2/3K
(C) 1/2K
(D) 1/3K
(E) 1/4K

A.

the ratio tells you that the larger also is a equilateral triangle.
Length of small one is a
length of larger one is b
a =b/2

the larger arear = K= b^2*(square root(3)/4) I called square root(3)/4 S
the smaller area = a^2 *S = S*b^2 /4 = K/4

the shaded area = K -K/4 =3/4 K
_________________
Director
Joined: 14 Oct 2007
Posts: 758
Location: Oxford
Schools: Oxford'10
Followers: 15

Kudos [?]: 214 [4] , given: 8

Re: 1000PS: Section 7 Question 16 [#permalink]

Show Tags

06 Apr 2008, 07:56
4
KUDOS
an easy way to solve this would be to visualise the below triangle instead. it shows u the answer straight away without the need for any calculations
Attachments

trig.JPG [ 3.86 KiB | Viewed 9363 times ]

Intern
Joined: 01 Dec 2010
Posts: 4
Followers: 0

Kudos [?]: 5 [4] , given: 0

Re: 1000PS: Section 7 Question 16 [#permalink]

Show Tags

03 Dec 2010, 12:06
4
KUDOS
Another way to do this!
Theorem : If there are Two similar triangles with sides in Ratio : S1 : S2 - then their areas are in the ratio S1^2 : S2 ^2
=> Area of Larger : Area of Smaller = S1^2 : S2^2
=> K : As = 2^2 : 1
=> As = k/4

Therefore, Area of the shaded region : K-K/4 = 3k/4
Math Expert
Joined: 02 Sep 2009
Posts: 37083
Followers: 7243

Kudos [?]: 96337 [1] , given: 10737

Re: 1000PS: Section 7 Question 16 [#permalink]

Show Tags

04 Dec 2010, 03:07
1
KUDOS
Expert's post
2
This post was
BOOKMARKED
jimmylow wrote:
16. The triangles in the figure above are equilateral and the ratio of the length of a side of the larger triangle to the length of a side of the smaller triangle is 2/1. If the area of the larger triangular region is K, what is the area of the shaded region in terms of K?
(A) 3/4K
(B) 2/3K
(C) 1/2K
(D) 1/3K
(E) 1/4K

Yes, the property given above is very useful. It states: in two similar triangles, the ratio of their areas is the square of the ratio of their sides: $$\frac{AREA}{area}=\frac{S^2}{s^2}$$.

As both big and inscribed triangles are equilateral then they are similar, so $$\frac{AREA}{area}=\frac{S^2}{s^2}=\frac{2^2}{1^2}=4$$, so if $$AREA=K$$ then $$area=\frac{K}{4}$$ --> the area of the shaded region equals to $$area_{shaded}=K-\frac{K}{4}=\frac{3K}{4}$$.

_________________
Manager
Joined: 13 Jul 2010
Posts: 169
Followers: 1

Kudos [?]: 76 [0], given: 7

Re: 1000PS: Section 7 Question 16 [#permalink]

Show Tags

06 Dec 2010, 20:28
very useful ratio thanks guys.
Senior Manager
Joined: 06 Aug 2011
Posts: 405
Followers: 2

Kudos [?]: 197 [1] , given: 82

Re: The triangles in the figure above are equilateral and the [#permalink]

Show Tags

18 Oct 2012, 10:08
1
KUDOS
m not geting this onne..how K/4??

large triangle have area 4...how k/4?
_________________

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

Math Expert
Joined: 02 Sep 2009
Posts: 37083
Followers: 7243

Kudos [?]: 96337 [0], given: 10737

Re: The triangles in the figure above are equilateral and the [#permalink]

Show Tags

18 Oct 2012, 10:16
sanjoo wrote:
m not geting this onne..how K/4??

large triangle have area 4...how k/4?

Check here: the-triangles-in-the-figure-above-are-equilateral-and-the-62201.html#p827422

We have that $$\frac{AREA}{area}=4$$. Now, since $$AREA=K$$ then $$\frac{K}{area}=4$$ --> $$area=\frac{K}{4}$$ --> the area of the shaded region equals to $$area_{shaded}=K-\frac{K}{4}=\frac{3K}{4}$$.

Hope it's clear.
_________________
Manager
Joined: 05 Nov 2012
Posts: 71
Schools: Foster '15 (S)
GPA: 3.65
Followers: 1

Kudos [?]: 126 [1] , given: 8

Re: The triangles in the figure above are equilateral and the [#permalink]

Show Tags

09 Jun 2013, 21:14
1
KUDOS
jimmylow wrote:
Attachment:
2triangles.GIF
The triangles in the figure above are equilateral and the ratio of the length of a side of the larger triangle to the length of a side of the smaller triangle is 2/1. If the area of the larger triangular region is K, what is the area of the shaded region in terms of K?

(A) 3/4K
(B) 2/3K
(C) 1/2K
(D) 1/3K
(E) 1/4K

The easiest way to solve this one would be by picking numbers. Lets say each side of the larger triangle is 6 and given the ratio 2:1 each side of smaller triangle then is 3. Area of an equilateral triangle can be calculated using formula: $$s^2(\sqrt{3})/4$$ where s = side. So area of large triangle = k = $$9(\sqrt{3})$$ and area of small triangle = $$9(\sqrt{3})/4$$ = $$k/4$$. So the area of the smaller triangle is 1/4 the area of the large triangle. Area of shaded region=$$k-k/4$$ = $$3/4k$$
_________________

___________________________________________
Consider +1 Kudos if my post helped

Math Expert
Joined: 02 Sep 2009
Posts: 37083
Followers: 7243

Kudos [?]: 96337 [0], given: 10737

Re: The triangles in the figure above are equilateral and the [#permalink]

Show Tags

06 Mar 2014, 01:32
Bumping for review and further discussion.

_________________
GMAT Club Legend
Joined: 09 Sep 2013
Posts: 13921
Followers: 589

Kudos [?]: 167 [0], given: 0

Re: The triangles in the figure above are equilateral and the [#permalink]

Show Tags

01 Oct 2015, 22:41
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.
_________________
Director
Joined: 05 Mar 2015
Posts: 710
Followers: 8

Kudos [?]: 116 [0], given: 37

Re: The triangles in the figure above are equilateral and the [#permalink]

Show Tags

20 Jul 2016, 10:37
Bunuel wrote:
Bumping for review and further discussion.

Let side of larger eq. triangle=2X
area=sq.root3/4*(2X)^2------->sq.root3*x^2------(A)
therefore side of smaller eq. triangle=X(given ratio=2/1)
area of smaller- sq.root3*x^2/4
=sq.root3*x^2-sq.root3*x^2/4------------>sq.root3*x^2(1-1/4)
substituting from eq. (A)
Ans A
Re: The triangles in the figure above are equilateral and the   [#permalink] 20 Jul 2016, 10:37
Similar topics Replies Last post
Similar
Topics:
6 In the figure above, triangle CDE is an equilateral triangle with side 6 07 Sep 2015, 22:02
53 In the figure above, triangle ABC is equilateral, and point 23 03 Dec 2012, 03:36
3 In the figure above, equilateral triangle ABC is inscribed in the circ 4 07 May 2011, 13:26
31 In the figure above, equilateral triangle ABC is inscribed 23 17 Jul 2010, 09:33
13 In the figure above, equilateral triangle ABC is inscribed i 17 02 Mar 2009, 08:50
Display posts from previous: Sort by