|
|
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.
31 Jan 2016, 16:43
Kudos
Bookmarks
B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
75%
(hard)
Question Stats:
62% (03:00) correct
38%
(03:19)
wrong
based on 195
sessions
History
Date
Time
Result
Not Attempted Yet
Would anyone please be so kind to help me with this question? I have no idea how to tackle it. I literally spent hours trying to solve the question.
The rhombus (AFCE) is inscribed in rectangle (ABCD). The length of BF = DE.
If the rectangle has a width of 20 yards and a length of 25 yards, what would be the total length (the perimeter) of a fence along the sides defined by AFCE?
A. 80 yards
B. 82 yards
C. 84 yards
D. 85 yards
E. 90 yards
The rhombus (AFCE) is inscribed in rectangle (ABCD). The length of BF = DE.
If the rectangle has a width of 20 yards and a length of 25 yards, what would be the total length (the perimeter) of a fence along the sides defined by AFCE?
A. 80 yards
B. 82 yards
C. 84 yards
D. 85 yards
E. 90 yards
Attachments
gmat.JPG [ 11.11 KiB | Viewed 14975 times ]
31 Jan 2016, 19:15
Kudos
Bookmarks
The definition of a rhombus states that every side is of equal length. This definition is the key to solving this question.
1) We state that the hypotenuse of either side triangle is x. For this solution, I will be working with the left triangle.
2) We state that the bottom and top sides of the rhombus are also x, due to the definition of a rhombus.
3) We will define the bottom of the left triangle as y. That is, for the rectangle bottom itself, we can define it as 25 = y + x, where x is a side for the rhombus and y is the bottom part of the left triangle.
4) From (1), we know that x = \(\sqrt{20^2 + y^2}\).
5) From (3), we know x = 25 - y.
6) From (4) and (5), we have the equation \(\sqrt{20^2 + y^2} = 25 - y\)
7) Solving for y, we get 4.5.
8) 25 - 4.5 = 20.5 = x
9) Since all sides of the rhombus are the same, 20.5 * 4 = 41 * 2 = 82.
The answer is B, 82
1) We state that the hypotenuse of either side triangle is x. For this solution, I will be working with the left triangle.
2) We state that the bottom and top sides of the rhombus are also x, due to the definition of a rhombus.
3) We will define the bottom of the left triangle as y. That is, for the rectangle bottom itself, we can define it as 25 = y + x, where x is a side for the rhombus and y is the bottom part of the left triangle.
4) From (1), we know that x = \(\sqrt{20^2 + y^2}\).
5) From (3), we know x = 25 - y.
6) From (4) and (5), we have the equation \(\sqrt{20^2 + y^2} = 25 - y\)
7) Solving for y, we get 4.5.
8) 25 - 4.5 = 20.5 = x
9) Since all sides of the rhombus are the same, 20.5 * 4 = 41 * 2 = 82.
The answer is B, 82
02 Feb 2016, 06:23
Kudos
Bookmarks
saiesta
Hi,
the method explained above is the correct method..
But another method which you could use is the POE..
now the perimeter of rectangle is 20+25+20+25=90..
so perimeter of rhombus has to be less than 90.. elimiate E..
Also perimeter of the rhombus has to be greater than 20+20+20+20=80, as the side of rhombus is the hypotenuse , which has to be more than the side of the same triangle, 20
eliminate A..
lets see the remaining choices..
A rhombus has all equal sides, so each side = perimeter/4
B. 82 yards... side= 20.5
C. 84 yards..side=21
D. 85 yards..side=21.25
now this side = hypotenuse osf a right angle triangle whose one side is 20..
lets see if hyp=20.5 and one side = 20, what is the second side...
\(\sqrt{20.5^2-20^2}\)=\(\sqrt{(20.5-20)(20.5+20)}\)
...\(\sqrt{.5*40.5}\)=\(\sqrt{.5*.5*81}\)=4.5..
this exactly fits in as other side of rhombus..
25=4.5=20.5..
