|
|
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.
18 Jun 2020, 03:27
Kudos
Bookmarks
B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
85%
(hard)
Question Stats:
47% (02:36) correct
53%
(02:57)
wrong
based on 47
sessions
History
Date
Time
Result
Not Attempted Yet
Rectangle ABCD, shown above, measures 4 cm by 16 and is folded about line MN so that point C coincides with point A, as shown below.
Label the new location of C is C’ and D as D’. What is the area of the resulting pentagon \(ABMND’\)?
A. 32
B. 47
C. 49
D. 51
E. 53
Are You Up For the Challenge: 700 Level Questions
Attachment:
2020-06-18_1415.png [ 20.24 KiB | Viewed 4118 times ]
Attachment:
2020-06-18_1415_001.png [ 33.44 KiB | Viewed 4141 times ]
18 Jun 2020, 04:44
Kudos
Bookmarks
AN^2 = D'N^2 + AD'^2
=> x^2 = (16-x)^2 + 4^2
=> x = 17/2
Area of the resultant figure = Ar. of ABCD - Ar. of AMN
=> A = 4*16 - 0.5*4*17/2 = 64 - 17 = 47
Ans: B
Attachments
Untitled.png [ 32.76 KiB | Viewed 4639 times ]
General Discussion
yashikaaggarwal
Senior Moderator - Masters Forum
Joined: 19 Jan 2020
Last visit: 29 Mar 2026
Posts: 3,088
Given Kudos: 1,510
Location: India
Schools: Cranfield (A) Warwick MiM "23 (M)
GPA: 4
WE:Analyst (Internet and New Media)
18 Jun 2020, 05:52
Kudos
Bookmarks
AM = 1/2 of AC(Diagonal)
AM = 1/2*√16^2+4^2
AM = 1/2*√256+16
AM = 1/2*√272
AM = 1/2*4√17
AM = 2√17 = 8.25 = AN (∆AMN is an isosceles triangle)
Area of Pentagon ABMND'= Area of triangle ABM + triangle AD'N + triangle AMN
∆AD'N = ∆ABM (right Angle triangle)
BM = D'N = 16-8.25 = 7.75
Area of ∆AD'N+∆ABM = 2*1/2*7.75*4 => 31 -----------(1)
Draw a line perpendicular on AD from M suc, namely P such that PMN is an right Angle triangle.
PM = 4
PN = 16-7.75-7.75 = 0.5
MN = √4^2+0.5^2
MN = √16.25
MN = 4.03
Area of isosceles triangle AMN whose sides are 8.25,8.25 and 4.03
=> √S(S-A)(S-B)(S-C)
=> S = 8.25+8.25+4.03/2
=> 20.53/2 => 10.265
=> √10.265(10.265-8.25)(10.265-8.25)(10.265-4.03)
=> √10.265(2.015)(2.015)(6.235)
=> √10.265(2.015)(2.015)(6.235)
=> √259.86 => 16.12 or 16 ----(2)
Adding all 3 areas = 31+16 = 47
Answer is 47
Posted from my mobile device
AM = 1/2*√16^2+4^2
AM = 1/2*√256+16
AM = 1/2*√272
AM = 1/2*4√17
AM = 2√17 = 8.25 = AN (∆AMN is an isosceles triangle)
Area of Pentagon ABMND'= Area of triangle ABM + triangle AD'N + triangle AMN
∆AD'N = ∆ABM (right Angle triangle)
BM = D'N = 16-8.25 = 7.75
Area of ∆AD'N+∆ABM = 2*1/2*7.75*4 => 31 -----------(1)
Draw a line perpendicular on AD from M suc, namely P such that PMN is an right Angle triangle.
PM = 4
PN = 16-7.75-7.75 = 0.5
MN = √4^2+0.5^2
MN = √16.25
MN = 4.03
Area of isosceles triangle AMN whose sides are 8.25,8.25 and 4.03
=> √S(S-A)(S-B)(S-C)
=> S = 8.25+8.25+4.03/2
=> 20.53/2 => 10.265
=> √10.265(10.265-8.25)(10.265-8.25)(10.265-4.03)
=> √10.265(2.015)(2.015)(6.235)
=> √10.265(2.015)(2.015)(6.235)
=> √259.86 => 16.12 or 16 ----(2)
Adding all 3 areas = 31+16 = 47
Answer is 47
Posted from my mobile device
