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The area of ABE = the area of rectangle BCDE
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21 May 2019, 10:56
21 May 2019, 10:56
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Question Stats:
67% (01:45) correct
32% (01:42) wrong based on 58 sessions
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#GREpracticequestion The area of ΔABE.jpg [ 17.85 KiB | Viewed 7017 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
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24 May 2019, 06:17
24 May 2019, 06:17
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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)
