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
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
The area of ABE = the area of rectangle BCDE
[#permalink]
21 May 2019, 10:56
21 May 2019, 10:56
Expert Reply
2
Bookmarks
00:00
Question Stats:
66% (01:42) correct
33% (01:21) wrong based on 45 sessions
Hide Show timer Statistics
Attachment:
#GREpracticequestion The area of ΔABE.jpg [ 17.85 KiB | Viewed 4880 times ]
The area of \(ΔABE =\) the area of rectangle \(BCDE\)
Quantity A |
Quantity B |
The ratio of side BE to side DE |
4:1 |
A)The quantity in Column A is greater.
B)The quantity in Column B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.
Re: The area of ABE = the area of rectangle BCDE
[#permalink]
24 May 2019, 06:17
24 May 2019, 06:17
1
1
Bookmarks
Triangle ABE is a right angled isosceles triangle. Therefore BE is \(2\sqrt{2}\).
Since the area of the triangle and the rectangle is same. Let us find the area of the triangle.
area of the triangle = \(\frac{1}{2} * 2 * 2 = 2\)
therefore \(2\sqrt{2} * x = 2\) where x is the length of side DE.
Therefore, x = \(2\sqrt{2} * \sqrt{2}\)
Hence the ratio is 4:1
Since the area of the triangle and the rectangle is same. Let us find the area of the triangle.
area of the triangle = \(\frac{1}{2} * 2 * 2 = 2\)
therefore \(2\sqrt{2} * x = 2\) where x is the length of side DE.
Therefore, x = \(2\sqrt{2} * \sqrt{2}\)
Hence the ratio is 4:1
Re: The area of ABE = the area of rectangle BCDE
[#permalink]
25 May 2019, 11:29
25 May 2019, 11:29
1
amorphous wrote:
Triangle ABE is a right angled isosceles triangle. Therefore BE is \(2\sqrt{2}\).
Since the area of the triangle and the rectangle is same. Let us find the area of the triangle.
area of the triangle = \(\frac{1}{2} * 2 * 2 = 2\)
therefore \(2\sqrt{2} * x = 2\) where x is the length of side DE.
Therefore, x = \(2\sqrt{2} * \sqrt{2}\)
Hence the ratio is 4:1
Since the area of the triangle and the rectangle is same. Let us find the area of the triangle.
area of the triangle = \(\frac{1}{2} * 2 * 2 = 2\)
therefore \(2\sqrt{2} * x = 2\) where x is the length of side DE.
Therefore, x = \(2\sqrt{2} * \sqrt{2}\)
Hence the ratio is 4:1
Hi,
in the highlighted part, how does x= 2[square_root]2*s? Shouldn't it be x=2/(2[square_root]2)
