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M21#15

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M21#15 [#permalink] New post 23 Nov 2008, 11:43
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Is the perimeter of a rectangle greater than 8 inches?

1. The diagonal of the rectangle is twice as long as its shorter side
2. The diagonal of the rectangle is 4 inches longer than its shorter side

[Reveal] Spoiler: OA
B

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SOLUTION: m21-73214.html#p1099791
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Re: M21#15 [#permalink] New post 23 Nov 2008, 14:21
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yup for the diagonal to be more than 4 inches (because it is 4 inches longer than the shorter side), then the longer side would have to be at least 4.
so
2*longer side must be either more than or equal to 8 by itself
so after adding in the shorter sides, it must be more than 8 inches

2 is sufficient, B correct answer
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Re: M21#15 [#permalink] New post 08 Mar 2010, 07:10
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ventivish wrote:
Is the perimeter of a rectangle greater than 8 inches?
1. The diagonal of the rectangle is twice as long as its shorter side
2. The diagonal of the rectangle is 4 inches longer than its shorter side
[Reveal] Spoiler: OA
B

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1 is not sufficient as short side is x, diagonal is 2x, long side is sqrt(3)x. Perimeter = x(2+sqrt(3)) which can be lesser than 8 or more than it.

2 is sufficient as short side is x, diagonal is x+4 so long side can be any value > 4. Perimeter > 8.
Hence B
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Re: M21#15 [#permalink] New post 08 Mar 2010, 10:29
Ans is B

1 gives b(1+1.73)>4 insufficient (depends on the value of b-breadth)

2. gives 2*sqrt(2)*(sqrt(2+b) +b> 4 if b=0 left side = 4 so for any value of b>0 this equation is true and ans is B
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Re: M21#15 [#permalink] New post 09 Mar 2010, 06:49
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B

let length=l, breadth=b, diagonal=d
From question stem we need to prove 2(l+b) >8
Stmt(1) : d = 2b
Now consider the triangle formed by length, breadth and diagonal
sum of 2 sides > 3rd side
l + b > d => l + b > 2b => l > 2b say l=3b (b is +ve)
now perimeter = 2(l+b) = 2(3b + b) = 8b . This can be greater than 8 or less than 8. Hence not sufficient

Stmt(2): d = b+4
l+b > d => l+b > b+4 => l > 4 say l=4.1
2(l+b) = 2(4.1 +b) = 8.2 + 2b... is always greater than 8. Sufficient

Hope this helps.
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Re: M21#15 [#permalink] New post 10 Mar 2010, 17:12
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the fastest way to solve this is via dk94588's approach.


given 2 sides of a triangle the 3rd side must be less than the sum of the two sides and greater than their
difference.



If x is the shorter side and x +4 is the diagnol, then the third side y is limited by:

(x+4)-x < y < (x+4) + x

i.e.

4 < y < 2x +4

the perimeter of a rectangle is 2w + 2l where w is the width and l is the length. If lets say y is the length and it is greater than 4, then 2y will be greater than 8, therefore the perimeter has to be greater than 8.



let assume that you didn't know the ranges of the sides in a triangle, you can still solve this via the Pythagorean theorem: z^2 = x^2 + y^2 , where z is the hypotenuse of a right triangle and x and y are the legs. Here our triangle is formed from two sides of the box (x & y) and the diagnol (z).

(4+x)^2 = x^2 + y^2

16 + 8x + x^2 = x^2 + y^2
16 + 8x = y^2
4(4+2x) = y^2

therefore y = sqrt(4(4+2x)) = 2*sqrt(4 +2x)

now the perimeter = 2y + 2x = 2(2*sqrt(4+2x)) + 2x = 4*sqrt(4+2x) + 2x

lets try to make x as small as possible and see if the perimeter can still be greater than 8.
x obviously cannot equal zero but lets assume its so very close to zero so we can set it to zero to see a boundary condition.

therefore the perimeter > 4*sqrt(4) = 8

sufficient.
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Re: M21#15 [#permalink] New post 07 Sep 2010, 22:41
Its B.

1) Perimeter = 2x(1+sqrt(3)) => Insuffiecient since perimeter depends on x.

2) l^2 + b^2 > (b + 4 ) ^ 2 => l > 4 So irrespective of the value of b, perimeter is > 8.
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Re: M21#15 [#permalink] New post 13 Mar 2011, 20:08
Answer is B.
If L is the length and b is the bredth. Then,
(b+4)^2=b^2+l^2
So solving the above equation:
b=L^2-16/8
Now putting the value of L:
When we will put integer value for L, first value that will give integer value for b is 8 (Hidden constraint).
b=64-16/8=6.
So, L=8;b=6; and diagonal is equal to 10.
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Re: M21#15 [#permalink] New post 14 Mar 2011, 10:19
ventivish wrote:
Is the perimeter of a rectangle greater than 8 inches?

1. The diagonal of the rectangle is twice as long as its shorter side
2. The diagonal of the rectangle is 4 inches longer than its shorter side



My answer is B.

Statement 2: We have a right triangle with sides x, x+4, and unknown. Since the length of any side of a triangle must be larger than the positive difference of the other two sides, unknown must be greater than 4.

2L + 2W = Perimeter
2(4) + 2x = Perimeter
8 + 2x = Perimeter

The perimeter must be greater than 8.
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Re: M21#15 [#permalink] New post 15 Mar 2011, 07:13
A: perimeter is [2 + 2 sqrt(3)]*x = 5.46 x - not sufficient
B: Simple concept- sum of two sides of triangle is greater than third side.
hence long side (y) + short side (x) > diagonal (x+4)
perimeter [2*(x+y)]>2x + 8. x can not be negative.

therefore ans is B
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Re: M21#15 [#permalink] New post 14 Mar 2012, 15:19
Chose C, I went too fast through the answers. Instead of analyzing B on it's own, I tried to see if applying the info in B with the info in A would get the answer and then jumped to conclusions. Oops
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Re: M21#15 [#permalink] New post 14 Mar 2012, 19:32
1) insuff... we know perimeter is goign to be 2(x) + 2 squareroot 3 x
2) suff. plug in the smallest possible value for x, 1
we know 1^2 x b^2 = 5^2
it'll still be larger than 8.
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Re: M21#15 [#permalink] New post 25 Apr 2012, 06:44
ventivish wrote:
Is the perimeter of a rectangle greater than 8 inches?

1. The diagonal of the rectangle is twice as long as its shorter side
2. The diagonal of the rectangle is 4 inches longer than its shorter side

[Reveal] Spoiler: OA
B

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from the below explanations, i am guessing the answer to have come like below

1: sum of two sides > third side in a triange... ( simple basic principle which we often forget)

so lets take l and b

so l + b > d

taking l to be shorter side

so from 1, it is said that d = 2l

so l + b > 2l

so b > l

now from perimeter => 2l + 2b
taking minimum value for b to be just 0.1 greater than b, the total perimeter could be somewhere around 4.2 or 4.3 which is less than 8
or taking double the value of l, the perimeter could have crossed 8...hence we get both YES and NO as answers

hence Option A is not the answer and thereby option D is ignored too

2) taking l as shorter side, it is given that d = l + 4
so from principle l + b > d
l + b > 4 + l
so b > 4

so 2L + 2b > 8 YES always

so answer is B
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Re: M21#15 [#permalink] New post 27 Jun 2012, 06:46
Answer is B

Considering 1) if small side = x, diagonal = 2x, long side = square root of (4x^2-x^2) = 3x^2.......nothin mentioned about the kind of no that x is and hence we cannot conclude anything

Considering 2) small side= x, diagonal=x+4, lond side by pythagoras is = square root of(16+4x)..........clearly the long side is greater than 4 and hence the perimeter is greater than 8
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Re: M21#15 [#permalink] New post 27 Jun 2012, 07:01
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Is the perimeter of a rectangle greater than 8 inches?

The question asks whether Perimeter=2(a+b)>8, or whether a+b>4, (where a and b are the length of the sides of the rectangle).

(1) The diagonal of the rectangle is twice as long as its shorter side. Clearly insufficient, we know the shape of the rectangle but not its size.

(2) The diagonal of the rectangle is 4 inches longer than its shorter side. This statement basically says that the the length of the diagonal is greater than 4 inches: d>4. Now, consider the triangle made by the diagonal and the two sides of the rectangle: since the length of any side of a triangle must be smaller than the sum of the other two sides, then we have that a+b>d, so a+b>d>4. Sufficient.

Answer: B.
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Re: M21#15 [#permalink] New post 03 Jul 2012, 04:49
B is the answer.

Rephrasing the question renders as:
Is x+y>4? Where x is the longer side of the rectangle, y is the shorter

1) Can be broken down to x=y * 3^(0.5)Not Sufficient
2) Can be broken down to x = (16+8y)^(0.5)
Clearly x>4, since y>0, therefore 8y>0.
Therefore (16+8y)^(0.5) > 4, therefore x>4. Sufficient

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Re: M21#15 [#permalink] New post 15 Mar 2013, 19:30
Perimeter of a rectangle is given by
P = 2(l+w), where, l = length of rectangle, and w= width of rectangle

for P > 8,(l+w) should greater than 4

I- Diameter of rectangle is two times the shorter side of the rectangle,

D = 2w
Also,D^2 = l^2 + w^2

From these two equations,l = d\sqrt{(3)}

We can not conclude anything about the condition (l+w) > 4, so not sufficient

II- Given, diameter of the rectangle is 4 inches more than shortest side.
So in any case longer side is greater than 4 as well.

So, in this condition, (l+w) >4, so sufficient.

Answer is B.
Re: M21#15   [#permalink] 15 Mar 2013, 19:30
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