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If p is the perimeter of rectangle Q, what is the value of p [#permalink]
05 Oct 2010, 02:11
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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.
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wininblue 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
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. Similar question: one-more-geometry-96381.html?hilit=perimeter%20rectangle#p742145Hope it helps.
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Great Method! Thank you so much Bunuel, +1
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Since it is a rectangle, the diagnal opposite side would make 90 degrees, so the diagnal = a^2 + b^2 which is equal to 10 ------ from (i)
I can nt determine the p value from the (i)
From (2), ab = 48 not sufficient. now combine both, we know that (a+b)^2 = a^2 + b^2 + 2ab .... we have all the values ..
so the ans is : C
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Perimeter = p = 2* (l+b)
To caluculate the perimeter, we need both l and b
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.
Hence C.
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Perimeter of rectangle Q ? [#permalink]
30 Jan 2011, 08:14
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.
Where is my error?
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Re: Perimeter of rectangle Q ? [#permalink]
30 Jan 2011, 08:33
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Merging similar topics. 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.
Where is my error? Check this post: one-more-geometry-96381.html#p742164You 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. Check this post for more on triangles: math-triangles-87197.htmlHope it helps.
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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 For the rectangle; if Diagonal=d length=l width=w Rephrasing: What is 2(l+w) or simply what is (l+w) 1) d=10 l^2+w^2=d^2(l+w)^2-2lw=d^2(l+w)^2=d^2+2lw=10^2+2lw=100+2lw(l+w)=sqrt{100+2lw}If we knew lw; we would have known l+w; but we don't know that Not Sufficient. 2) Area=48 lw=48 l^2+w^2=d^2(l+w)^2-2lw=d^2(l+w)^2=d^2+2lw=d^2+2*48=d^2+96(l+w)=sqrt{d^2+96}If we knew diagonal "d"; we would have known l+w; but we don't know that Not Sufficient. Using both the statements; We know lw=48 and d=10 l^2+w^2=d^2(l+w)^2-2lw=d^2(l+w)^2=d^2+2lw=10^2+2*48=100+96(l+w)=sqrt{196}=14Sufficient using both statements. Ans: "C"
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Official Guide - 12 - Question D48 [#permalink]
15 Sep 2011, 00:13
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
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Re: Official Guide - 12 - Question D48 [#permalink]
17 Sep 2011, 01:05
deephoenix wrote: 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.
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Re: Official Guide - 12 - Question D48
[#permalink]
17 Sep 2011, 01:05
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