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Re: One more geometry [#permalink]
26 Jun 2010, 07:24

1

This post received KUDOS

Expert's post

I think the answer is C.

Statement 1: Area = 60

This implies that \(l*b = 60\) where l and b are length and breadth respectively. This could mean a million possibilities. So insufficient and we are down to BCE.

Statement 2: Length of diagonal = 13.

This implies that \(\sqrt{l^2 + b^2} = 13\)

So at this point, I am a little confused about one tiny detail.

Scenario 1: We can assume that all sides of the rectangle are integers, in which case Statement alone is sufficient since the only two integers whose sum of squares is 13 are 12 and 5. So we are done.

Scenario 2: We cannot assume that all sides are integers, in which case statement two is insufficient and we use statement 1 with statement 2 to figure out the answer.

b = \(\frac{60}{l}\)

Substituting that into statement 2, we get:

\(\sqrt{l^2 + (60/l)^2} = 13\)

This can be solved for a value of l and hence we get C as our answer. What's the OA?

Last edited by whiplash2411 on 26 Jun 2010, 19:54, edited 1 time in total.

Re: One more geometry [#permalink]
26 Jun 2010, 08:23

Expert's post

Hussain15 wrote:

What is the perimeter of rectangle R?

(1) The area of rectangle R is 60.

(2) The length of a diagonal of rectangle R is 13.

I wanna discuss this one in detail.

Let the sides be \(x\) and \(y\). Question: \(P=2(x+y)=?\)

(1) \(area=xy=60\). Clearly insufficient. (2) \(x^2+y^2=13^2=169\). Also insufficient.

(1)+(2) Square P --> \(P^2=4(x+y)^2=4(x^2+2xy+y^2)\) as from (1) \(xy=60\) and from (2) \(x^2+y^2=169\), then \(P^2=4(x^2+2xy+y^2)=4(169+120)=4*289\) --> \(P=\sqrt{4*289}=2*17=34\). Sufficient.

Re: One more geometry [#permalink]
26 Jun 2010, 09:21

OA is "C", but why??? It should be "B". We know the 13-12-5 principle. If hypotenuse is 13, then the other two sides can be determined. Why not "B"?????

Re: One more geometry [#permalink]
26 Jun 2010, 09:27

Expert's post

Hussain15 wrote:

OA is "C", but why??? It should be "B". We know the 13-12-5 principle. If hypotenuse is 13, then the other two sides can be determined. Why not "B"?????

What about \(13-\frac{13}{\sqrt{2}}-\frac{13}{\sqrt{2}}\) ? _________________

Re: One more geometry [#permalink]
26 Jun 2010, 09:45

3

This post received KUDOS

Expert's post

Hussain15 wrote:

OA is "C", but why??? It should be "B". We know the 13-12-5 principle. If hypotenuse is 13, then the other two sides can be determined. Why not "B"?????

Posted from my mobile device

5-12-13 is a Pythagorean triple. What is Pythagorean triple?

Triangles that have their sides in the ratio of whole numbers (integers) are called Pythagorean Triples. The smallest one is 3:4:5. If you multiply the sides by any number, the result will still be a right triangle whose sides are in the ratio 3:4:5. For example 6, 8, and 10.

A Pythagorean triple consists of three positive integers \(a\), \(b\), and \(c\), such that \(a^2 + b^2 = c^2\). Such a triple is commonly written \((a, b, c)\), and a well-known example is \((3, 4, 5)\). If \((a, b, c)\) is a Pythagorean triple, then so is \((ka, kb, kc)\) for any positive integer \(k\). There are 16 primitive Pythagorean triples with c ≤ 100: (3, 4, 5)(5, 12, 13)(7, 24, 25)(8, 15, 17) (9, 40, 41) (11, 60, 61) (12, 35, 37) (13, 84, 85) (16, 63, 65) (20, 21, 29) (28, 45, 53) (33, 56, 65) (36, 77, 85) (39, 80, 89) (48, 55, 73) (65, 72, 97).

BUT:

Knowing that hypotenuse equals to 13 DOES NOT mean that the sides of the right triangle necessarily must be in the ratio of Pythagorean triple - 5:12:13. Or in other words: if \(a^2+b^2=13^2\) DOES NOT mean that \(a=5\) and \(b=12\), certainly this is one of the possibilities but definitely not the only one. In fact \(a^2+b^2=13^2\) has infinitely many solutions for \(a\) and \(b\) and only one of them is \(a=5\) and \(b=12\).

For example: \(a=1\) and \(b=\sqrt{168}\) or \(a=2\) and \(b=\sqrt{165}\) or \(a=4\) and \(b=\sqrt{153}\) or \(a=2.5\) and \(b=\sqrt{162.75}\)...

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.

Re: One more geometry [#permalink]
26 Jun 2010, 09:46

No this is wrong. a^2+b^2 = 13^2 holds true for many values of (a,b)

You cannot deduce a and b to be 12,5

what if a^2 = 6 and b^=7, it was not even given that they have integral value. Even if it was given a and b are integers you still have to think whether any integral pair a,b exists which is unique.

no doubt 2+2 = 4, but you can not deduce 4 = 2+2 only...it can be 3+1....This reasoning will help you in CR - cause and effect. _________________

Re: One more geometry [#permalink]
26 Jun 2010, 16:24

Hussain15 wrote:

OA is "C", but why??? It should be "B". We know the 13-12-5 principle. If hypotenuse is 13, then the other two sides can be determined. Why not "B"?????

Posted from my mobile device

ans c I am no Math Genius ..but if you deduce this a^2 +b^2=C^2..then it doesn't mean that triplet a,b, c are pythgorean triplet.

But if a right angle triangle is given ..then yes a,b,c satisfy pythagoras theorem and if c is given then you can find a and b.

Re: One more geometry [#permalink]
28 Jun 2010, 06:37

Hussain15 wrote:

OA is "C", but why??? It should be "B". We know the 13-12-5 principle. If hypotenuse is 13, then the other two sides can be determined. Why not "B"?????

Posted from my mobile device

Dude,

Not Every Red car is Ferrari ...

Just kidding...

So every triplet is not 13-12-5 _________________

GGG (Gym / GMAT / Girl) -- Be Serious

Its your duty to post OA afterwards; some one must be waiting for that...

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