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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

I am not able to understand how do you solve a quadratic eqn of this size and conclude that length and breadth can have 2 distinct values. Please help!

Question: P=2(a+b)=?

(1) d^2=a^2+b^2=100. Not sufficient. (2) ab=48. Not sufficient.

(1)+(2) Square P --> P^2=4(a^2+b^2+2ab), now as from (1) a^2+b^2=100 and from (2) ab=48 then P^2=4(a^2+b^2+2ab)=4(100+2*48)=4*196 --> P=\sqrt{4*196}=28. Sufficient.

Answer: C.

So you can see that it's not necessary to solve quadratic equation for a and b to get P.

1) Length of diagonal = 10 Length of a diagonal = sqrt{l^2 + b^2} = 10 Squaring both sides, we get (l^2 + b^2) = 100 This is not sufficient since we have one equation and 2 variables.

2) Area = l*b = 48 This is not sufficient since we have one equation and 2 variables.

(l+b)^2 = l^2 + b^2 + 2*l*b Substituting the values obtained in 1) and 2), we can get the value of (l+b) and thus we can calculate the perimeter.

Perimeter of rectangle Q ? [#permalink]
30 Jan 2011, 07:14

Hi,

in the book "GMAT review 12th edt.", there is diagnostic test question #48 (DS). ----- If p is the perimeter of rectangle Q, what is the value of p? 1) Each diagonal of rectangle Q has length of 10. 2) The area of rectangle Q is 48. ---- Now, the answer explanation says C is correct. However, when looking at answer 1), I know the hypotenuse of both triangles is 10. Using the Pythagorean theorem, I know that my sides are 8 and 6 -> (5:4:3) x 2.

So p = 2l + 2w = 16 + 12... hence A is sufficient to determine the value.

Re: Perimeter of rectangle Q ? [#permalink]
30 Jan 2011, 07:33

1

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demeuse81 wrote:

Hi,

in the book "GMAT review 12th edt.", there is diagnostic test question #48 (DS). ----- If p is the perimeter of rectangle Q, what is the value of p? 1) Each diagonal of rectangle Q has length of 10. 2) The area of rectangle Q is 48. ---- Now, the answer explanation says C is correct. However, when looking at answer 1), I know the hypotenuse of both triangles is 10. Using the Pythagorean theorem, I know that my sides are 8 and 6 -> (5:4:3) x 2.

So p = 2l + 2w = 16 + 12... hence A is sufficient to determine the value.

You assume with no ground for it that the lengths of the sides are integers. Knowing that hypotenuse equals to 10 DOES NOT mean that the sides of the right triangle necessarily must be in the ratio of Pythagorean triple - 6:8:10. Or in other words: if a^2+b^2=10^2 DOES NOT mean that a=6 and b=8, certainly this is one of the possibilities but definitely not the only one. In fact a^2+b^2=10^2 has infinitely many solutions for a and b and only one of them is a=6 and b=8.

For example: a=1 and b=\sqrt{99} or a=2 and b=\sqrt{96} or a=4 and b=\sqrt{84} ...

So knowing that the diagonal of a rectangle (hypotenuse) equals to one of the Pythagorean triple hypotenuse value is not sufficient to calculate the sides of this rectangle.

Hi.. am a member for some while now, however this is my 1st ever post. In Q 48 (data sufficiency) of diagnostic test of OG-12, the diagonal length of rectangle is given as 10 inches, and we need to find the perimeter. According to me, the statement is sufficient in that the sides have to be 6 and 8 inhes (using the pythagorean triple 6-8-10). However, book says it's not sufficient.

Hi.. am a member for some while now, however this is my 1st ever post. In Q 48 (data sufficiency) of diagnostic test of OG-12, the diagonal length of rectangle is given as 10 inches, and we need to find the perimeter. According to me, the statement is sufficient in that the sides have to be 6 and 8 inhes (using the pythagorean triple 6-8-10). However, book says it's not sufficient.

Can someone plz clarify or explain? Tx

OG is right. You cannot take 6-8-10 to solve this quesion. In the above posts Bunnel and fluke have explained the solution of this question using the right approach. Please refer their posts and reply back if you have any doubts.