A very quick DS question

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feeling_alive: Newbie | Next Rank: 10 Posts; Posts: 6; Joined: Mon Feb 27, 2012 8:46 pm

by feeling_alive » Mon Apr 02, 2012 11:45 am

If p is the perimeter of rectangle X,
what is the value of p?

1. Each diagonal of X has a length of 5.
2. The area of X is 12.

I don't get why statement 1 is insufficient. If the diagonal is of length 5 that means the hypotenuse of the triangle 5. But when a rectangle is cut through the diagonal doesn't that mean it will have two 45 degree angles and 1 90 degree angles for each triangle. This makes a 45-45-90 triangle and since I already know one side which is 5(opposite the 90 degrees) I can just solve for the other legs of the triangle using the ratio of 1:1:root2. Thus making this statement sufficient, could someone please clarify the logic in my understanding?

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Source: Beat The GMAT — Data Sufficiency | Mon Apr 02, 2012 11:45 am

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by Bill@VeritasPrep » Mon Apr 02, 2012 12:30 pm

The diagonal will only create two 45-45-90 triangles if the rectangle is also a square.

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ziko: Junior | Next Rank: 30 Posts; Posts: 17; Joined: Wed Feb 29, 2012 1:58 am

by ziko » Tue Apr 03, 2012 10:16 pm

You cannot say that the angles are 45-45-90 they could be any depending on the sides of the rectangle, so i guess you misunderstand it with quadrant.
In my opinion answer is C.
1st statment tells us diagonal of X has a length of 5 but no info on sides - not suff.
2nd statement, Let say sides of the rectangle are a and b, so a*b=12 - not sufficient, since there are a number of combinations (1*12, 2*6, 3*4)
Combining them: according to on the pifagorian teorem a^2+b^2=25, from the statement 2 the only combination which suits is 3 and 4.

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Anurag@Gurome: GMAT Instructor; Posts: 3835; Joined: Fri Apr 02, 2010 10:00 pmFollowed by: 523 members; Location: Milpitas, CA; Thanked: 1854 times; GMAT Score:770

by Anurag@Gurome » Tue Apr 03, 2012 10:23 pm

feeling_alive wrote:If p is the perimeter of rectangle X,
what is the value of p?

1. Each diagonal of X has a length of 5.
2. The area of X is 12.

I don't get why statement 1 is insufficient. If the diagonal is of length 5 that means the hypotenuse of the triangle 5. But when a rectangle is cut through the diagonal doesn't that mean it will have two 45 degree angles and 1 90 degree angles for each triangle. This makes a 45-45-90 triangle and since I already know one side which is 5(opposite the 90 degrees) I can just solve for the other legs of the triangle using the ratio of 1:1:root2. Thus making this statement sufficient, could someone please clarify the logic in my understanding?

p is the perimeter of rectangle X implies p = 2(l + w), where l and w are the length and width of rectangle respectively.

(1) Each diagonal of X has a length of 5 implies if d is the length of the rectangle, then dÂ² = lÂ² + wÂ² = 5Â² = 25; NOT sufficient.

(2) The area of X is 12 implies lw = 12; NOT sufficient.

Combining (1) and (2), p = 2(l + w) implies pÂ² = 4(l + w)Â² = 4(lÂ² + 2wl + wÂ²)
From statement 1, lÂ² + wÂ² = 25 and from statement 2, lw = 12
So, pÂ² = 4(25 + 2 * 12) = 4 * 49 implies p = 2 * 7 = 14; SUFFICIENT.

The correct answer is C.

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Pharo: Senior | Next Rank: 100 Posts; Posts: 61; Joined: Mon Jan 18, 2010 11:09 pm; Thanked: 8 times

by Pharo » Tue Apr 03, 2012 11:00 pm

Look at it this way dude.

perimeter = 2 (w + l); So you know you need w and l to find the solution.

1. gives you w^2 + l^2 = 25; this one does not give you w and l but gives you something you can work with.

perimeter^2 = (2(w+l))^2 ; expand this
= 4 (w^2 + l^2 + 2wl) ; sub w^2 + l^2 = 25
= 4 (25 + 2wl)
= 100 + 8wl
With statement 1 you no longer need w and l but now you need wl. Since you do not have this value this statement is insufficient.

2. gives you wl.

So, combining both you can solve this.

Hope this helps

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