Events & Promotions
|
|
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
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Apr 1612:00 PM EDT
-11:59 PM EDT
You’re first to know: Save $200 on GMAT prep starting tomorrow, 4/13. Get 6 official practice tests and study with real questions. Be ready to save!
Apr 2610:00 AM EDT
-11:00 AM EDT
The Target Test Prep course represents a quantum leap forward in GMAT preparation, a radical reinterpretation of the way that students should study. Try before you buy with a 5-day, full-access trial of the course for FREE!
Apr 2711:00 AM EDT
-12:00 PM EDT
Prefer video-based learning? The Target Test Prep OnDemand course is a one-of-a-kind video masterclass featuring 400 hours of lecture-style teaching by Scott Woodbury-Stewart, founder of Target Test Prep and one of the most accomplished GMAT instructors
Apr 2808:00 AM PDT
-11:00 AM PDT
Whether you are just beginning your MBA journey or fine-tuning your path to a 700+ score, GMAT Day is designed to give you a competitive edge.
13 Apr 2018, 01:31
Kudos
Bookmarks
D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
45%
(medium)
Question Stats:
61% (01:36) correct
39%
(01:35)
wrong
based on 79
sessions
History
Date
Time
Result
Not Attempted Yet
What is the perimeter of the triangle shown?
(1) The area of the triangle is 96 in^2.
(2) y = x + 4
Attachment:
2018-04-13_1227.png [ 5.79 KiB | Viewed 2903 times ]
13 Apr 2018, 03:21
Kudos
Bookmarks
Bunuel
1) 1/2xy= 96
xy =192
x^2 + y^2 = 20^2 = 400
(x+y)^2 = x^2+y^2 + 2xy = 400 + 2x192 = 784
(x+y)^2 = 28^2
x+y = 28
xy = 192
using trial and error x=12/16 y=16/12
perimeter can be found
2) y=x+4
X^2 + (x+4)^2 = 20^2
2x^ + 16 + 8x = 400
2x^2 + 8x - 384 = 0
x^2 + 4x - 192 = 0
x^2 + 16 x - 12 x - 192 = 0
X= 12
y = 16
(D) imo
13 Apr 2018, 05:27
Kudos
Bookmarks
Bunuel
Given in question
\(x^2+y^2=(20)^2\)
Question ask us to find parameter P,
\(P= 20+x+y\).
Either finding value of (x+y) or both x and y individually will suffice.
Statement 1 :The area of the triangle is 96 in^2.
Area =\(\frac{x*y}{2}\)
It gives\(xy= 192\).
Now using \((x+y)^2=x^2+y^2+2xy\),\(x^2+y^2=(20)^2\) and xy= 192, we can get
\((x+y)^2\)=784
Rejecting -ve value, we get \((x+y)=28\)
Perimeter can be calculated.
So Statement 1 alone is sufficient
Statement 2 :y = x + 4
Using \(x^2+y^2=(20)^2\) and y = x + 4, we get
\(x^2+(x+4)^2=(20)^2\)
\(x^2+4x-192=0\)
\(x\)= \(\frac{-4±\sqrt{4^2-4*1*(-192)}}{2*1}\)
\(x=12\) (Rejecting -ve value of x i.e -16)
Using y = x + 4, we get \(y =16\)
As we know value of x and y individually , Value of perimeter can be easily calculated.
So Statement 2 alone is Sufficient.
As Both Statement are alone sufficient , Answer is D
