OFFICIAL GMAT REVIEW 11th Edition Question

Kindly show courtesy to type the full question.
As per the question is concerned, you are taking one of the infinite number of possibilities. For any given line, there are infinite right triangles that can be formed, all of which will have the given line as hypotenuse...

simplyjat wrote:Kindly show courtesy to type the full question.
As per the question is concerned, you are taking one of the infinite number of possibilities. For any given line, there are infinite right triangles that can be formed, all of which will have the given line as hypotenuse...

Simplyjat is correct. If all you know is the hypotenuse, there's no way to determine the legs of the triangle.

With a hypotenuse of 10, we know that:

a^2 + b^2 = 10^2

or

a^2 + b^2 = 100

There's no way to solve for a and b with this equation.

For example, we could pick a=6, b=8 as you suggest.

We could also pick a=root50 and b=root50 (this would give us a 45/45/90 triangle with a hypotenuse of 10).

Ok here is the question:

If p is the perimeter of rectangle O, what is the value of p?
(1) Each daigonal of rectangle Q has length 10.
(2) The area of rectangle Q is 48.


So, it is given that it is a rectangle, therefore it cannot be a 45-45-90. And it definitely WILL be a right triangle. This being the case, the only values that work for this hypotenuse, since it is a right triangle, will be 6 and 8 for the sides. This is 6-8-10 is consistent with a 3-4-5 triangle. True, we don't know WHICH leg is the 6 and which is the 8 but that doesn't matter, since all we need is the perimeter.

So I am having trouble understanding why the answer isn't just A. What other values could work for the legs that would still yield a hypotenuse of 10?

Just to add to the above explanations :
A square is a rectangle, for which length and width are equal.

Sakina wrote:Ok here is the question:

If p is the perimeter of rectangle O, what is the value of p?
(1) Each daigonal of rectangle Q has length 10.
(2) The area of rectangle Q is 48.


So, it is given that it is a rectangle, therefore it cannot be a 45-45-90. And it definitely WILL be a right triangle. This being the case, the only values that work for this hypotenuse, since it is a right triangle, will be 6 and 8 for the sides. This is 6-8-10 is consistent with a 3-4-5 triangle. True, we don't know WHICH leg is the 6 and which is the 8 but that doesn't matter, since all we need is the perimeter.

So I am having trouble understanding why the answer isn't just A. What other values could work for the legs that would still yield a hypotenuse of 10?

I still don't understand why it has to be 6-8-10.

First, as mba.dude pointed out, a square IS a rectangle, it's just a special one.

Second, even if we weren't allowed to pick a square, there are infinite other possibilities.

All we have to work with is the equation:

a^2 + b^2 = 10^2

or

a^2 + b^2 = 100

So, we could pick:

(root99)^2 + (root1)^2

or

(root98)^2 + (root2)^2

and so on, and so on, and so on.

Just because

(root36)^2 + (root64)^2 happen to work out to integers doesn't mean that they're the only possible values.