M21-15

Question

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

Bunuel · Accepted Answer

Official Solution:

Is the perimeter of a rectangle greater than 8 inches? 
 
The question asks whether  Perimeter=2(a+b) \gt 8 , or whether  a+b \gt 4 , (where  a  and  b  are the lengths 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 length of the diagonal is greater than 4 inches:  d \gt 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 \gt d , so  a+b \gt d \gt 4 . Sufficient.

Answer: B

nrodriguez1 · Answer

I think this is a  high-quality  question and I agree with explanation.

dreamsatgmat · Answer

amazing question...

Biplab75009 · Answer

Really well crafted question.

Devbek · Answer

Amazing short-cut explanation. Thanks Bunuel

Sirakri · Answer

Can we not say from st1 that it is a 30:60:90 triangle and hence sufficient?

Bunuel · Answer

Yes, but do we know numerical value of any of the sides?

Sirakri · Answer

But we can answer YES or No, right? Thanks for your quick response Bunuel, you're a star

Bunuel · Answer

No, you cannot. The question asks: is the perimeter of a rectangle greater than 8 inches? Just knowing the ratio is not giving any actual lengths, so you cannot answer the question.

Amit700 · Answer

High quality question

coreyander · Answer

I think this is a  high-quality  question and I agree with explanation.

ocelot22 · Answer

Here is another way to solve this problem

(1) diag = 2 times short
suppose short =4, then perim >2*4 = 8 yes
suppose short = 1, diag =2, long = root(4-1) = root(3) and perim = 2root3 + 2 = approx 3.4 + 2 < 8 NS

(2) The diagonal is 4 inches longer than short.

label the short side x and the long side y. x^2 + y^2 = (x+4)^2 and x^2 + y^2 =x^2 +8x + 16, and y^2 = 8x + 16 and y = root(8x + 16) so min y >4, so min perim >8 so sufficient.

Bunuel · Answer

I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.

M21-15

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