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# If the diagonal and the area of a rectangle are 25 in and 168 in2, wha

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Intern
Joined: 26 Jan 2010
Posts: 5
If the diagonal and the area of a rectangle are 25 in and 168 in2, wha  [#permalink]

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Updated on: 27 Mar 2017, 11:58
6
00:00

Difficulty:

55% (hard)

Question Stats:

73% (02:46) correct 27% (02:42) wrong based on 105 sessions

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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

Originally posted by punzoo on 28 Feb 2010, 06:54.
Last edited by Bunuel on 27 Mar 2017, 11:58, edited 1 time in total.
Renamed the topic, edited the question and added the OA.
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Joined: 01 Feb 2010
Posts: 233
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, 08:17
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
(l-w)^2 = 289 so l-w = 17 or w = l - 17
l(l-17) = 168
solving so l = 24 or -7
offcourse l cannot be (-)ve hence D
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Posts: 233
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, 08:18
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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Joined: 22 Nov 2009
Posts: 23
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, 08:50
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, 12:45
1
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, 12:58
2
given

LB = 168
Also $$L^2 + B^2$$= 625

Now use $$(L+B)^2 - 2LB$$ = 625 and $$(L-B)^2 + 2LB$$ = 625

This gives L+B = 31 and L-B = 17

add them and we get L =24
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Posts: 147
Location: Singapore
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, 01:10
gurpreetsingh wrote:
given

LB = 168
Also $$L^2 + B^2$$= 625

Now use $$(L+B)^2 - 2LB$$ = 625 and $$(L-B)^2 + 2LB$$ = 625

This gives L+B = 31 and L-B = 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, 20:49
L = 24 as explained above..
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Joined: 13 Dec 2009
Posts: 227
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, 07:01
1
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
(l-b)^2=625-336=289=>l-b=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, 19:48
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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Re: If the diagonal and the area of a rectangle are 25 in and 168 in2, wha &nbs [#permalink] 30 Jun 2018, 19:48
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