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A rectangle has length 2x and width x. If each diagonal of t
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28 Jan 2020, 16:42
28 Jan 2020, 16:42
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A rectangle has length 2x and width x. If each diagonal of the rectangle has length d, what is the area of the rectangle, in terms of d?
A. \(\frac{2}{5} d\)
B. \(\frac{5}{2} d\)
C. \(\frac{4}{25} d^2\)
D. \(\frac{2}{5} d^2\)
E. \(\frac{2}{3} d^2\)
Kudos for the right answer and explanation
A. \(\frac{2}{5} d\)
B. \(\frac{5}{2} d\)
C. \(\frac{4}{25} d^2\)
D. \(\frac{2}{5} d^2\)
E. \(\frac{2}{3} d^2\)
Kudos for the right answer and explanation
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Re: A rectangle has length 2x and width x. If each diagonal of t
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29 Jan 2020, 04:30
29 Jan 2020, 04:30
1
Carcass wrote:
A rectangle has length 2x and width x. If each diagonal of the rectangle has length d, what is the area of the rectangle, in terms of d?
A. \(\frac{2}{5} d\)
B. \(\frac{5}{2} d\)
C. \(\frac{4}{25} d^2\)
D. \(\frac{2}{5} d^2\)
E. \(\frac{2}{3} d^2\)
Kudos for the right answer and explanation
A. \(\frac{2}{5} d\)
B. \(\frac{5}{2} d\)
C. \(\frac{4}{25} d^2\)
D. \(\frac{2}{5} d^2\)
E. \(\frac{2}{3} d^2\)
Kudos for the right answer and explanation
Since the sides of the rectangle are 2x and x, we have:
Diagonal = \(d = \sqrt{(2x)^2 + x^2} = x\sqrt{5}\)
Area of the rectangle = \((2x)*(x) = 2x^2\)
We have:
\(d^2 = 5x^2\)
\(=> x^2 = d^2/5\)
=> Area \(= 2x^2 = 2d^2/5\)
Answer D
Re: A rectangle has length 2x and width x. If each diagonal of t
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03 Mar 2022, 08:11
03 Mar 2022, 08:11
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Thanks to another GRE Prep Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).
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