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

It is currently 05 Dec 2019, 09:48

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 figure above, what is the area of the shaded region?

  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: 59561
In the figure above, what is the area of the shaded region?  [#permalink]

Show Tags

New post 05 Sep 2017, 23:53
1
4
00:00
A
B
C
D
E

Difficulty:

  35% (medium)

Question Stats:

72% (01:40) correct 28% (01:42) wrong based on 108 sessions

HideShow timer Statistics

Image
In the figure above, what is the area of the shaded region?

(A) 5/36
(B) 5/18
(C) 1/3
(D) 5/12
(E) 5/6

Attachment:
2017-09-06_1052.png
2017-09-06_1052.png [ 5.16 KiB | Viewed 6134 times ]
Current Student
User avatar
P
Joined: 02 Jul 2017
Posts: 280
Concentration: Entrepreneurship, Technology
GMAT 1: 730 Q50 V38
GMAT ToolKit User
In the figure above, what is the area of the shaded region?  [#permalink]

Show Tags

New post Updated on: 06 Sep 2017, 00:26
2
We have to find area of shaded region.

area of shaded region = Area of big triangle - area of small triangle.
ie area of shaded region = Area( Triangle ABC) - Area( Triangle ADE) <- Refer attached figure

As both triangles are right triangle and both have an angle common ( Angle A)
=>Big triangle is similar to small triangle
=> triangle ABC ~ triangle ADE

Area formula for 2 similar triangles =>
\(\frac{Ar(ABC)}{Ar(ADE)} = \frac{AB^2}{AD^2} = \frac{BC^2}{DE^2} = \frac{AC^2}{AE^2}\)

Ar(ABC) = 1/2 *1*1 = 1/2
BC=1
DE=2/3

=>\(\frac{Ar(ABC)}{Ar(ADE)} = \frac{1^2}{(2/3)^2}\)
=> Ar(ADE)= Ar(ABC)* 4/9 = 1/2 * 4/9 = 2/9

Area of shaded region = Ar(ABC) - Ar(ADE) = 1/2 -2/9 =5/18

Answer: B
Attachments

Shaded area.jpg
Shaded area.jpg [ 25.54 KiB | Viewed 4771 times ]


Originally posted by Nikkb on 06 Sep 2017, 00:12.
Last edited by Nikkb on 06 Sep 2017, 00:26, edited 1 time in total.
Current Student
User avatar
P
Joined: 18 Aug 2016
Posts: 594
Concentration: Strategy, Technology
GMAT 1: 630 Q47 V29
GMAT 2: 740 Q51 V38
GMAT ToolKit User Reviews Badge
Re: In the figure above, what is the area of the shaded region?  [#permalink]

Show Tags

New post 06 Sep 2017, 00:22
Bunuel wrote:
Image
In the figure above, what is the area of the shaded region?

(A) 5/36
(B) 5/18
(C) 1/3
(D) 5/12
(E) 5/6

Attachment:
2017-09-06_1052.png

Divide the shaded region into one triangle on the top with (1-2/3) = 1/3 height and x as base
area of triangle will be 1/3x
area of rectangle with 2/3 length and x width will be 2/3x
area of shaded region = 5/6x
5/6x + (1-x)*1/3 = 1/2
3x/6 = 1/6
x=1/3
5/6*1/3 = 5/18
B
_________________
We must try to achieve the best within us


Thanks
Luckisnoexcuse
Senior PS Moderator
User avatar
V
Joined: 26 Feb 2016
Posts: 3300
Location: India
GPA: 3.12
In the figure above, what is the area of the shaded region?  [#permalink]

Show Tags

New post 06 Sep 2017, 00:41
Attachment:
2017-09-06_1052.png
2017-09-06_1052.png [ 6.09 KiB | Viewed 4749 times ]


In the figure, ABC is a right angled isosceles triangle, as AB = BC = 1
The triangle ABC is similar to ADE(AA similarity)

It has been given that DE = \(\frac{2}{3}\). Similarly, AD = \(\frac{2}{3}\)
(sides opposite equal angles are equal)

Area of triangle ABC = \(\frac{1}{2}*1*1 = \frac{1}{2}\)
Area of triangle ADE = \(\frac{1}{2}*(\frac{2}{3})^2 = \frac{2}{9}\)

Therefore, are of shaded region is \(\frac{1}{2} - \frac{2}{9} = \frac{(9-4)}{18} = \frac{5}{18}\)(Option B)
_________________
You've got what it takes, but it will take everything you've got
Director
Director
User avatar
D
Affiliations: IIT Dhanbad
Joined: 13 Mar 2017
Posts: 730
Location: India
Concentration: General Management, Entrepreneurship
GPA: 3.8
WE: Engineering (Energy and Utilities)
Re: In the figure above, what is the area of the shaded region?  [#permalink]

Show Tags

New post 06 Sep 2017, 03:52
Bunuel wrote:
Image
In the figure above, what is the area of the shaded region?

(A) 5/36
(B) 5/18
(C) 1/3
(D) 5/12
(E) 5/6

Attachment:
2017-09-06_1052.png


Since the unshaded triangle and the whole triangle are similar .
Ratio of their area = (2/3)^2 = 4/9
Ratio of area of shaded region to full triangle = 1-4/9 = 5/9

Area of complete triangle = 1/2*1*1 = 1/2
Area of shaded region = 5/9 *1/2 = 5/18

Answer B
Senior SC Moderator
avatar
V
Joined: 22 May 2016
Posts: 3723
In the figure above, what is the area of the shaded region?  [#permalink]

Show Tags

New post 06 Sep 2017, 14:28
pushpitkc wrote:
Attachment:
2017-09-06_1052.png


In the figure, ABC is a right angled isosceles triangle, as AB = BC = 1
The triangle ABC is similar to ADE(AA similarity)

It has been given that DE = \(\frac{2}{3}\). Similarly, AD = \(\frac{2}{3}\)
(sides opposite equal angles are equal)

Area of triangle ABC = \(\frac{1}{2}*1*1 = \frac{1}{2}\)
Area of triangle ADE = \(\frac{1}{2}*(\frac{2}{3})^2 = \frac{2}{9}\)

Therefore, are of shaded region is \(\frac{1}{2} - \frac{2}{9} = \frac{(9-4)}{18} = \frac{5}{18}\)(Option B)

(E) 5/6

Attachment:
2017-09-06_1052.png

pushpitkc , (and the rest of the posters, all of whom used triangles' similarity or assumed right angles at some point), I don't think we know we have two right angles.

We are not told that the two vertical lines are parallel.

If the two vertical lines were parallel, then the transversal which forms the base of both triangles, and which would cut the parallel lines, would create two congruent right angles.

We know from the diagram's mark that one angle is a right angle. That angle and the unmarked angle at the smaller triangle's base would be corresponding angles created by parallel lines cut by a transversal, and would be congruent.

In that case, the small and large triangles would share a right angle and a vertex angle, and by AA property of triangles, the triangles would be similar.

We are not told that the shorter vertical line is perpendicular to the horizontal line. Same result as above, i.e., if short vertical line were perpendicular to base, triangles would be similar.

Nor does the figure display a right angle mark where the shorter vertical line meets the base to create a small triangle with one side of length \(\frac{2}{3}\).

It seems all we know is that the two triangles share one angle, at the vertex on the left. The short vertical line might not be vertical. I'm not sure how we can conclude that the triangles are similar.

Am I missing something?
_________________
SC Butler has resumed! Get two SC questions to practice, whose links you can find by date, here.

Never doubt that a small group of thoughtful, committed citizens can change the world; indeed, it's the only thing that ever has -- Margaret Mead
Director
Director
User avatar
D
Affiliations: IIT Dhanbad
Joined: 13 Mar 2017
Posts: 730
Location: India
Concentration: General Management, Entrepreneurship
GPA: 3.8
WE: Engineering (Energy and Utilities)
Re: In the figure above, what is the area of the shaded region?  [#permalink]

Show Tags

New post 07 Sep 2017, 00:10
1
genxer123 wrote:
pushpitkc wrote:
Attachment:
2017-09-06_1052.png


In the figure, ABC is a right angled isosceles triangle, as AB = BC = 1
The triangle ABC is similar to ADE(AA similarity)

It has been given that DE = \(\frac{2}{3}\). Similarly, AD = \(\frac{2}{3}\)
(sides opposite equal angles are equal)

Area of triangle ABC = \(\frac{1}{2}*1*1 = \frac{1}{2}\)
Area of triangle ADE = \(\frac{1}{2}*(\frac{2}{3})^2 = \frac{2}{9}\)

Therefore, are of shaded region is \(\frac{1}{2} - \frac{2}{9} = \frac{(9-4)}{18} = \frac{5}{18}\)(Option B)

(E) 5/6

Attachment:
2017-09-06_1052.png

pushpitkc , (and the rest of the posters, all of whom used triangles' similarity or assumed right angles at some point), I don't think we know we have two right angles.

We are not told that the two vertical lines are parallel.

If the two vertical lines were parallel, then the transversal which forms the base of both triangles, and which would cut the parallel lines, would create two congruent right angles.

We know from the diagram's mark that one angle is a right angle. That angle and the unmarked angle at the smaller triangle's base would be corresponding angles created by parallel lines cut by a transversal, and would be congruent.

In that case, the small and large triangles would share a right angle and a vertex angle, and by AA property of triangles, the triangles would be similar.

We are not told that the shorter vertical line is perpendicular to the horizontal line. Same result as above, i.e., if short vertical line were perpendicular to base, triangles would be similar.

Nor does the figure display a right angle mark where the shorter vertical line meets the base to create a small triangle with one side of length \(\frac{2}{3}\).

It seems all we know is that the two triangles share one angle, at the vertex on the left. The short vertical line might not be vertical. I'm not sure how we can conclude that the triangles are similar.

Am I missing something?


genxer123 , You are 100% correct. There is no mention of shorter line being parallel to the larger line in the shaded region.
We just assumed it to be parallel and solved the question which is not at all acceptable on GMAT. Also there were option available, so we just solved it. Had it been a data sufficiency question, then the situation could have created a real problem. So the question should say that smaller line is parallel to longer line...
Target Test Prep Representative
User avatar
V
Status: Founder & CEO
Affiliations: Target Test Prep
Joined: 14 Oct 2015
Posts: 8605
Location: United States (CA)
Re: In the figure above, what is the area of the shaded region?  [#permalink]

Show Tags

New post 11 Sep 2017, 11:02
Bunuel wrote:
Image
In the figure above, what is the area of the shaded region?

(A) 5/36
(B) 5/18
(C) 1/3
(D) 5/12
(E) 5/6

Attachment:
2017-09-06_1052.png


We see that we have two similar triangles. The ratio of the side length of the larger triangle to the smaller triangle is 1/(2/3). Since the base and height of the larger triangle is 1, the base and height of the smaller triangle is 2/3.

Thus, the area of the larger triangle is 1 x 1 x 1/2 = ½ and the area of the smaller triangle is 2/3 x 2/3 x 1/2 = 4/18.

So, the area of the shaded region is 1/2 - 4/18 = 9/18 - 4/18 = 5/18.

Answer: B
_________________

Scott Woodbury-Stewart

Founder and CEO

Scott@TargetTestPrep.com
TTP - Target Test Prep Logo
122 Reviews

5-star rated online GMAT quant
self study course

See why Target Test Prep is the top rated GMAT quant course on GMAT Club. Read Our Reviews

If you find one of my posts helpful, please take a moment to click on the "Kudos" button.

Non-Human User
User avatar
Joined: 09 Sep 2013
Posts: 13710
Re: In the figure above, what is the area of the shaded region?  [#permalink]

Show Tags

New post 19 Sep 2018, 10:22
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 figure above, what is the area of the shaded region?   [#permalink] 19 Sep 2018, 10:22
Display posts from previous: Sort by

In the figure above, what is the area of the shaded region?

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





cron

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