Last visit was: 22 Dec 2024, 00:30 It is currently 22 Dec 2024, 00:30

Close

GRE Prep 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 GRE 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
avatar
Intern
Intern
Joined: 04 Feb 2018
Posts: 9
Own Kudos [?]: 43 [37]
Given Kudos: 0
Send PM
Most Helpful Expert Reply
Verbal Expert
Joined: 18 Apr 2015
Posts: 30448
Own Kudos [?]: 36808 [14]
Given Kudos: 26096
Send PM
Most Helpful Community Reply
avatar
Retired Moderator
Joined: 16 Oct 2019
Posts: 63
Own Kudos [?]: 175 [13]
Given Kudos: 21
Send PM
General Discussion
avatar
Intern
Intern
Joined: 24 Jan 2018
Posts: 31
Own Kudos [?]: 9 [3]
Given Kudos: 0
Send PM
Re: Quadrilateral sides comparison [#permalink]
2
1
Bookmarks
I think the answer is C.

With the figure, I’d do this:

e^2= a^2+d^2 at the same time e^2=b^2+c^2 so, we can equal both equations:
a^2+d^2 = b^2+c^2
a^2+d^2 – c^2= b^2

Second, I'll compare the two options:
Quantity A
b^2 + 2ad

Quantity B
a^2+2ad+d^2-c^2

We can subtract “2ad” in both equations
Quantity A
b^2

Quantity B
a^2+d^2-c^2

We already know that “b^2= a^2+d^2 – c^2”

Please let me know if it is ok. It’s just my guess.
avatar
Intern
Intern
Joined: 04 Feb 2018
Posts: 9
Own Kudos [?]: 43 [0]
Given Kudos: 0
Send PM
Re: Quadrilateral sides comparison [#permalink]
Linamrs wrote:
I think the answer is C.

With the figure, I’d do this:

e^2= a^2+d^2 at the same time e^2=b^2+c^2 so, we can equal both equations:
a^2+d^2 = b^2+c^2
a^2+d^2 – c^2= b^2

Second, I'll compare the two options:
Quantity A
b^2 + 2ad

Quantity B
a^2+2ad+d^2-c^2

We can subtract “2ad” in both equations
Quantity A
b^2

Quantity B
a^2+d^2-c^2

We already know that “b^2= a^2+d^2 – c^2”

Please let me know if it is ok. It’s just my guess.


I dont think you can use pythagorean theorem, always look out for that. I used to fall for it many times.
Try without the theorem, your method is close to mine. I got B another friend D. Still not sure.
User avatar
Sherpa Prep Representative
Joined: 15 Jan 2018
Posts: 147
Own Kudos [?]: 363 [6]
Given Kudos: 0
Send PM
Re: Quadrilateral sides comparison [#permalink]
6
Expert Reply
Tricky problem. There are a few hurdles here. The first is that the quantities look pretty complicated. I'd try to simplify if possible. Notice that Quantity B contains (a + d)^2, which can be rewritten as a^2 + 2ad + d^2. Then we can subtract the 2ad from both sides, and shuffle the leftovers till we have b^2 + c^2 and a^2 + d^2. See the first picture for the algebra.

Next, we should recognize b and c as the sides of the rightmost triangle and a and d as the sides of the leftmost. They have the same opposite side, but the rightmost triangle has a 92° angle, while the leftmost has an 88° angle. If the angle were 90° then we would of course know that b^2 + c^2 would be the same as e^2. But what does it mean when the angle is either bigger or smaller than 90°? Easiest to just draw a picture and exaggerate the difference to see what it means. See the 2nd picture. Doing so shows, without any math, that the two sides on either side of the smaller angle must be larger, so their squares must also be larger. Thus, B is the answer.
Attachments

IMG_20180208_161942.jpg
IMG_20180208_161942.jpg [ 1.06 MiB | Viewed 14583 times ]

IMG_20180208_162251.jpg
IMG_20180208_162251.jpg [ 729.8 KiB | Viewed 14570 times ]

avatar
Intern
Intern
Joined: 04 Feb 2018
Posts: 9
Own Kudos [?]: 43 [2]
Given Kudos: 0
Send PM
Re: Quadrilateral sides comparison [#permalink]
2
SherpaPrep wrote:
Tricky problem. There are a few hurdles here. The first is that the quantities look pretty complicated. I'd try to simplify if possible. Notice that Quantity B contains (a + d)^2, which can be rewritten as a^2 + 2ad + d^2. Then we can subtract the 2ad from both sides, and shuffle the leftovers till we have b^2 + c^2 and a^2 + d^2. See the first picture for the algebra.

Next, we should recognize b and c as the sides of the rightmost triangle and a and d as the sides of the leftmost. They have the same opposite side, but the rightmost triangle has a 92° angle, while the leftmost has an 88° angle. If the angle were 90° then we would of course know that b^2 + c^2 would be the same as e^2. But what does it mean when the angle is either bigger or smaller than 90°? Easiest to just draw a picture and exaggerate the difference to see what it means. See the 2nd picture. Doing so shows, without any math, that the two sides on either side of the smaller angle must be larger, so their squares must also be larger. Thus, B is the answer.


Thanks I reached a similar conclusion but Im still not convinced. Can we have another method?


This is a friend method. I am more convinced of his method but not 100% sure.
Attachments

9f641fef-78e5-439a-98d5-3e097136f221.jpg
9f641fef-78e5-439a-98d5-3e097136f221.jpg [ 63.25 KiB | Viewed 14463 times ]

avatar
Intern
Intern
Joined: 04 Feb 2018
Posts: 9
Own Kudos [?]: 43 [0]
Given Kudos: 0
Send PM
Re: Quadrilateral sides comparison [#permalink]
The problem with the above method is that we cant really tell from the manipulation.

Look


- decrease + increase 0 no change, these are the subscripts meaning.
Attachments

8a31cfb8-18f8-46d6-8cf5-93b71dfeb01b.jpg
8a31cfb8-18f8-46d6-8cf5-93b71dfeb01b.jpg [ 44.98 KiB | Viewed 14480 times ]

Verbal Expert
Joined: 18 Apr 2015
Posts: 30448
Own Kudos [?]: 36808 [0]
Given Kudos: 26096
Send PM
Re: Quadrilateral sides comparison [#permalink]
Expert Reply
Please guys, could you edit your answers ??

Even though the discussion is quite interesting and useful per se, it is unuseful for the other students.
Who is trying to search for explanations will not find them due to the screenshots.

Whenever you do wanna a discussion unfold, is a good way to write all about as text.

The same is true when you post a new one question: the rules of the board forbid the use of screenshots.

Please refer to this https://gre.myprepclub.com/forum/rules-for ... -1083.html
and this https://gre.myprepclub.com/forum/qq-how-to ... -2357.html

Next time a similar topic will be locked.

Thank you for your collaboration.

regards
User avatar
Sherpa Prep Representative
Joined: 15 Jan 2018
Posts: 147
Own Kudos [?]: 363 [0]
Given Kudos: 0
Send PM
Re: Quadrilateral sides comparison [#permalink]
3
Expert Reply
@Carcass-
Sorry about the pictures.

@trunks14-
Here's a simple way to imagine it: for any right triangle, the two legs squared will equal the hypotenuse squared. But what if you kept the two legs the same length and squeezed down the hypotenuse to something smaller? Then we know that the two legs squared will now be larger than the new hypotenuse squared, and we also know that the angle must now be less than 90°.

Similarly, if we widen the two legs, the hypotenuse would now have to be larger, but the angle now exceeds 90°.

This is actually a general rule:
a^2 + b^2 < c^2 when the angle is less than 90°
and
a^2 + b^2 > c^2 when the angle is greater than 90°

So from these thought experiments we know that in the original question the two sides of the triangle squared on either side of the 88° angle must be larger than the two sides of the triangle squared on either side of the 92° angle. So again the answer is B.

A picture would've been much easier and less wordy but that's not allowed, haha.
Verbal Expert
Joined: 18 Apr 2015
Posts: 30448
Own Kudos [?]: 36808 [0]
Given Kudos: 26096
Send PM
Re: Quadrilateral sides comparison [#permalink]
Expert Reply
No need to say sorry :)

It is just a matter to keep the board efficient.

regards
avatar
Intern
Intern
Joined: 05 Oct 2017
Posts: 10
Own Kudos [?]: 9 [0]
Given Kudos: 0
Send PM
Re: Quadrilateral sides comparison [#permalink]
SherpaPrep wrote:
@Carcass-
Sorry about the pictures.

@trunks14-
Here's a simple way to imagine it: for any right triangle, the two legs squared will equal the hypotenuse squared. But what if you kept the two legs the same length and squeezed down the hypotenuse to something smaller? Then we know that the two legs squared will now be larger than the new hypotenuse squared, and we also know that the angle must now be less than 90°.

Similarly, if we widen the two legs, the hypotenuse would now have to be larger, but the angle now exceeds 90°.

This is actually a general rule:
a^2 + b^2 < c^2 when the angle is less than 90°
and
a^2 + b^2 > c^2 when the angle is greater than 90°

So from these thought experiments we know that in the original question the two sides of the triangle squared on either side of the 88° angle must be larger than the two sides of the triangle squared on either side of the 92° angle. So again the answer is B.

A picture would've been much easier and less wordy but that's not allowed, haha.



I think need correction

For a right triangle: a2+b2=c2a2+b2=c2.
For an acute (a triangle that has all angles less than 90°) triangle: a2+b2>c2
For an obtuse (a triangle that has an angle greater than 90°) triangle: a2+b2<c2
avatar
Director
Director
Joined: 09 Nov 2018
Posts: 505
Own Kudos [?]: 133 [0]
Given Kudos: 0
Send PM
Re: Quadrilateral sides comparison [#permalink]
Ans B
avatar
Manager
Manager
Joined: 08 Dec 2018
Posts: 94
Own Kudos [?]: 70 [0]
Given Kudos: 0
Send PM
Re: Quadrilateral sides comparison [#permalink]
1
AE wrote:
Ans B

Since, (angle b+ angle c )<90 degree. So, (b squared + C squared)<e squared. Similarly, ( angle a+angle d)>90 degree. So, ( a squared + d squared)>e squared. therefore, B is greater.
avatar
Intern
Intern
Joined: 24 Feb 2018
Posts: 13
Own Kudos [?]: 14 [0]
Given Kudos: 0
Send PM
Re: Which is greater b^2+2ad or (a+d)^2 - c^2 [#permalink]
1
Try to look at this problem in a very simple way.

You have 2 triangles, one angle for each is given.

Triangle 1 with sides A and D and an angle 88 will have a greater angle at its disposition when compares to the other triangle with sides B and C and an angle of 92.

this inherently means, A+D > B+C

from the equations given, simplify and get it to this form and voila, you have an answer! ------seeeeeeee!!!!(C)
Verbal Expert
Joined: 18 Apr 2015
Posts: 30448
Own Kudos [?]: 36808 [0]
Given Kudos: 26096
Send PM
Re: B^2+2AD or (A+D)^2-C^2 [#permalink]
Expert Reply
Post A Detailed Correct Solution For The Above Questions And Get A Kudos.
Question From Our New Project: GRE Quant Challenge Questions Daily - NEW EDITION!
User avatar
Senior Manager
Senior Manager
Joined: 10 Feb 2020
Posts: 496
Own Kudos [?]: 354 [1]
Given Kudos: 299
Send PM
Re: B^2+2AD or (A+D)^2-C^2 [#permalink]
1
I didnt understand the solution
avatar
Retired Moderator
Joined: 16 Oct 2019
Posts: 63
Own Kudos [?]: 175 [5]
Given Kudos: 21
Send PM
Re: B^2+2AD or (A+D)^2-C^2 [#permalink]
4
1
Bookmarks
Assume the ∠mno = 90°
So, in right angle triangle mno; \(mo^2=mn^2+no^2\) --(1)

But, in the given condition, ∠mno > 90°, this implies the length of side mo(i.e. E) is greater compare to the length of mo at ∠mno = 90°
with ∠mno > 90°, \(mo^2>mn^2+no^2\) or \(E^2>B^2+C^2\) --(2)

In the given condition, ∠mpo < 90°, this implies the length of side mo(i.e. E) is less compared to the length of mo at ∠mpo = 90°
with ∠mpo < 90°, \(mo^2<mp^2+po^2\) or \(E^2<A^2+D^2\) --(3)

from (2) & (3); \(B^2+C^2<E^2<A^2+D^2\)

\(B^2+C^2<A^2+D^2\)

\(B^2+C^2<(A+D)^2-2AD\); {\((A+D)^2=A^2+D^2-2AD\)}

\(B^2+2AD<(A+D)^2-C^2\)

Hope this helps!!
Attachments

#greprepclub B^2+2AD or (A+D)^2-C^2.jpg
#greprepclub B^2+2AD or (A+D)^2-C^2.jpg [ 13.26 KiB | Viewed 8638 times ]

avatar
Manager
Manager
Joined: 22 Jan 2020
Posts: 120
Own Kudos [?]: 241 [2]
Given Kudos: 10
Send PM
Re: Which is greater b^2+2ad or (a+d)^2 - c^2 [#permalink]
2
Quantity A: b^2 +2ad
Quantity B: (a+d)^2 -c^2

Quantity A: b^2 +2ad
Quantity B: a^2 + 2ad +d^2 -c^2

Quantity A: b^2
Quantity B: a^2 + d^2 -c^2

Quantity A: b^2 + c^2
Quantity B: a^2 + d^2



Attachment:
stretchTriangle.jpg
stretchTriangle.jpg [ 101.15 KiB | Viewed 12100 times ]

Attachment:
stretchTriangle2.jpg
stretchTriangle2.jpg [ 104.78 KiB | Viewed 12106 times ]


I'm attaching two images a user posted for another question that shows how when one of the sides of a triangle is fixed increasing the distance of the vertex opposite to the fixed side decreases the angle and decreasing the distance of the vertex to the fixed side increases the angle.


This is the case we have here.
Side "e" has a fixed length.

(angle between b and c = 92) > (angle between a and d = 88)

Therefore a and d combined are longer than b and c
(a and d are more "Stretched out")


Quantity A: b^2 + c^2
Quantity B: a^2 + d^2

B>A

Final Answer: B

Originally posted by chacinluis on 17 Aug 2020, 19:19.
Last edited by chacinluis on 18 Aug 2020, 11:05, edited 1 time in total.
Verbal Expert
Joined: 18 Apr 2015
Posts: 30448
Own Kudos [?]: 36808 [0]
Given Kudos: 26096
Send PM
Re: Which is greater b^2+2ad or (a+d)^2 - c^2 [#permalink]
Expert Reply
GREAT explanation 8-)
Prep Club for GRE Bot
Re: Which is greater b^2+2ad or (a+d)^2 - c^2 [#permalink]
 1   2   
Moderators:
GRE Instructor
88 posts
GRE Forum Moderator
37 posts
Moderator
1115 posts
GRE Instructor
234 posts

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