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DS Question

by arpitad » Sun Jan 15, 2012 4:58 pm
If p is the perimeter of rectangle Q, what is the value of p?

1. Each Diagonal of rectangle Q has length 10.
2. The area of rectangle Q is 48.



The correct answer is C but I thought that A was correct and statement 1 was sufficient after I used the pythagorean triples to solve for the answer (6,8,10) triple. I'm unclear about why the rectangle can have different values for the sides if you are given that the diagonal is 10 and the only possible options for the sides are 6 and 8 since it is a right triangle.
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by shankar.ashwin » Sun Jan 15, 2012 5:29 pm
The correct answer is C but I thought that A was correct and statement 1 was sufficient after I used the pythagorean triples to solve for the answer (6,8,10) triple. I'm unclear about why the rectangle can have different values for the sides if you are given that the diagonal is 10 and the only possible options for the sides are 6 and 8 since it is a right triangle.
Not necessarily, you can form infinite right triangles which have hypotenuse as 10. 6-8-10 is just one such combination. Say in triangle ABC, angle B = 90, now A+C should be = 90 and you can frame infinite values of A and C to satisfy that
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by Anurag@Gurome » Sun Jan 15, 2012 9:52 pm
arpitad wrote:If p is the perimeter of rectangle Q, what is the value of p?

1. Each Diagonal of rectangle Q has length 10.
2. The area of rectangle Q is 48.



The correct answer is C but I thought that A was correct and statement 1 was sufficient after I used the pythagorean triples to solve for the answer (6,8,10) triple. I'm unclear about why the rectangle can have different values for the sides if you are given that the diagonal is 10 and the only possible options for the sides are 6 and 8 since it is a right triangle.
Let l = length and w = width of rectangle.
Then p = 2l + 2w

(1) Diagonal length = 10

Image

By Pythagoras Theorem, 10² = l² + w²
Now we do not know l, w, so cannot find p; NOT sufficient.

(2) Area of rectangle Q = 48 implies lw = 48
Again we don't know the values of l, w; NOT sufficient.

Combining (1) and (2), l² + w² = 100 and lw = 48
(l + w)² = l² + w² + 2lw
(l + w)² = 100 + 96 = 196
l + w = 14
2(l + w) = 2 * 14 = 28; SUFFICIENT.

The correct answer is C.
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by ronnie1985 » Sun Jan 15, 2012 10:34 pm
10^2 = a^2+b^2, there is no restriction to a n b as a,b belong to Real no set. So S1 is not sufficient
Similarly S2 is not sufficient
But together the give a = 6 b = 8 or vice-versa giving perimeter = 28
(C) is right.
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by ritzzzr » Sun Jan 15, 2012 10:53 pm
Arpitad:
you have to keep in mind while solving data sufficiency question that you dont have to find a solution to it :but it has to be unique!!!!
what you got through a triplet is right but there can be many other combinations also,dont forget decimal places values also..so you are not getting a unique solution.
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by avik.ch » Mon Jan 16, 2012 3:37 am
arpitad wrote: The correct answer is C but I thought that A was correct and statement 1 was sufficient after I used the pythagorean triples to solve for the answer (6,8,10) triple. I'm unclear about why the rectangle can have different values for the sides if you are given that the diagonal is 10 and the only possible options for the sides are 6 and 8 since it is a right triangle.
the rule you are mentioning is only applicable for integers - but you are not considering the decimal part of it.

L^2 + B^2 = C^2

this will hold true for integers in which case we called it as pythagorean triplets and in case for decimals too. Since it is the length and breadth - L and B can take decimal values as well as integer values.

lets consider this on a more broad sense :

a^n + b^n = c^n ( where n is an integer)

for N = 3,4,.... : this equation will be satisfied only for non-intergers
for N = 1 and 2 : this equation will be satisfied for both integers and non-integers.

So I hope you understand why you need the second statement.

Hope this helps !!
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