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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

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

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

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).

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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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.

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.

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.

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

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.

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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Regards, Harsha

Note: Give me kudos if my approach is right , else help me understand where i am missing.. I want to bell the GMAT Cat

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

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.

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