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If the diagonal and the area of a rectangle are 25 in and 168 in2, wha [#permalink]
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If the diagonal and the area of a rectangle are 25 in and 168 in2, what is the length of the rectangle? A. 21 B. 38 C. 12 D. 24 E. 25
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Originally posted by punzoo on 28 Feb 2010, 07:54.
Last edited by Bunuel on 27 Mar 2017, 12:58, edited 1 time in total.
Renamed the topic, edited the question and added the OA.



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Re: If the diagonal and the area of a rectangle are 25 in and 168 in2, wha [#permalink]
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28 Feb 2010, 09:17
punzoo wrote: If the diagonal and the area of a rectangle are 25 in and 168 in2, what is the length of the rectangle? A. 21 B. 38 C. 12 D. 24 E. 25 Ans One way is: lw = 168 l^2 + w^2 = 625 (lw)^2 = 289 so lw = 17 or w = l  17 l(l17) = 168 solving so l = 24 or 7 offcourse l cannot be ()ve hence D



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Re: If the diagonal and the area of a rectangle are 25 in and 168 in2, wha [#permalink]
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28 Feb 2010, 09:18
punzoo wrote: If the diagonal and the area of a rectangle are 25 in and 168 in2, what is the length of the rectangle? A. 21 B. 38 C. 12 D. 24 E. 25 Ans One other way is: From common right triangle proportion we know 7:24:25 just pluging these values in area formula we get correct value hence l=24 hence D.



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Re: If the diagonal and the area of a rectangle are 25 in and 168 in2, wha [#permalink]
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01 Mar 2010, 09:50
punzoo wrote: If the diagonal and the area of a rectangle are 25 in and 168 in2, what is the length of the rectangle? A. 21 B. 38 C. 12 D. 24 E. 25 Ans Ans: D let the length and breadth be l,b sqrt(l^2+b^2)=25 => l^2+b^2=625 lb = 168 eliminate B,E because 38^2 > 625 and 25^2 = 625 Option A: if l = 21, b = 168/21 = 8 , but 21^2 + 8^2 != 625 Option C: if l = 12, b = 168/12 = 14 , but 12^2 + 14^2 != 625 Option D: if l = 24, b = 168/24 = 7 , and 24^2 + 7^2 = 625
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Re: If the diagonal and the area of a rectangle are 25 in and 168 in2, wha [#permalink]
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01 Mar 2010, 13:45
The "shortcut" is knowing the Pythagorian triples. Since the hypotenuse is 25, the sides COULD be 7 and 24. Plugging them in to the area formula shows you that they are correct.



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Re: If the diagonal and the area of a rectangle are 25 in and 168 in2, wha [#permalink]
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01 Mar 2010, 13:58
given LB = 168 Also \(L^2 + B^2\)= 625 Now use \((L+B)^2  2LB\) = 625 and \((LB)^2 + 2LB\) = 625 This gives L+B = 31 and LB = 17 add them and we get L =24
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Re: If the diagonal and the area of a rectangle are 25 in and 168 in2, wha [#permalink]
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02 Mar 2010, 02:10
gurpreetsingh wrote: given
LB = 168 Also \(L^2 + B^2\)= 625
Now use \((L+B)^2  2LB\) = 625 and \((LB)^2 + 2LB\) = 625
This gives L+B = 31 and LB = 17
add them and we get L =24 Thanks for sharing this approach.



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Re: If the diagonal and the area of a rectangle are 25 in and 168 in2, wha [#permalink]
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09 Mar 2010, 21:49
L = 24 as explained above..



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Re: If the diagonal and the area of a rectangle are 25 in and 168 in2, wha [#permalink]
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11 Mar 2010, 08:01
punzoo wrote: If the diagonal and the area of a rectangle are 25 in and 168 in2, what is the length of the rectangle? A. 21 B. 38 C. 12 D. 24 E. 25 Ans l^2+b^2 = 625 and l*b = 168 (l+b)^2 = 625+336=961=>l+b=31 (lb)^2=625336=289=>lb=17 2l=48=>l=24 so d is the answer
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Re: If the diagonal and the area of a rectangle are 25 in and 168 in2, wha [#permalink]
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30 Jun 2018, 20:48
punzoo wrote: If the diagonal and the area of a rectangle are 25 in and 168 in2, what is the length of the rectangle?
A. 21 B. 38 C. 12 D. 24 E. 25 \(lb = 8*21 = 24*7\) \(l^2 + b^2 = 625\) As \(l^2\)will end with 6 and \(7^2\) will end with 9. Hence, D
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