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looks easy but may be tough

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Re: looks easy but may be tough

by yezz » Tue Aug 18, 2009 5:40 pm
ern5231 wrote:Does the rectangle have an area less than 30?
(1) Perimeter = 20
(2) Diagonal < 10
OA given is A. But I feel it should be D. What do you say?
to maximize an area of a rectangle make it a square ( a square is a special type of rectangle which in turn is a special type of parallelogram.

from one

square(rectangle) parimeter = 20 thus side = 5

area of the square with side 5 = 25<30...suff

from 2

diagonal <10, worst case scenario is diagonal = 10

thus the 2 sides of the rectangle is 5, 5sqrt3

area = 2*5/2*5sqrt3 = 25sqrt3>30

however the diagonal can be = 4

area 2*4/2*4sqrt3 = 16sqrt3<30

thus insuff

A is the answer
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by varunkh70 » Tue Aug 18, 2009 9:51 pm
oh..I did not know this concept..Does the diagonal of a rectangle turn it into two 30-60-90 triangles?
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by PussInBoots » Wed Aug 19, 2009 4:45 pm
Nope
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by sakurle » Thu Aug 20, 2009 7:11 am
Let us take statement 2.
Diagonal < 10, worse situation Diagonal = 10, in that case sum of 2 sides, a+b > 10, (sum of any 1 sides is always greater than 3rd)

In this case a+b can be 12, a=5, b=7 axb = 35. On other hand diaginal can be 5 in which case a+b > 5, and axb < 30 Therefore insuff.
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by navalpike » Thu Aug 20, 2009 8:08 am
I have never heard of this concept. How does simply providing the diagonal turn it into a 30 60 90?
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by sreak1089 » Thu Aug 20, 2009 8:12 am
but it does turn it into twp 45-45-90 triangles right?
varunkh70 wrote:oh..I did not know this concept..Does the diagonal of a rectangle turn it into two 30-60-90 triangles?
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by sakurle » Thu Aug 20, 2009 9:24 am
Nope it does not turn into 45-45-90 either. If you go by 2. then if diagonal < 10 worse case diagonal = 10; then a+b sum of 2 sides has to be greater than 10. Assuming 5, 7 the product is 35 > 30.

On the other hand if diagonal <5, the sides can be 3,4 where area <30. Therefore IMO insuff. Answer is A.
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