alinomoto wrote:

This is from the

OG12, but I am confused about something:

48. 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.

Now, the obvious choice is

, but I figured that is the trap answer and I guessed

. Isn't it true that the hypotenuse of a right angled triangle, if 10, will have sides 6-8? Or is this not always the case? I know that if the triangle is not Right angled, then it is not the case;

but here: if the diagonal is 10, and since it is a rectangle, the 2 sides should be 6-8 and thus we can find the perimeter. Of course this is the same answer that you get by substituting values and using 2).

But OA says that we cannot know the sides from 1) or 2) and thus we need 1) and 2).

Some explanation please.

Given:

\(2*(l+b)=p\)

S1: \(l^2+b^2=10^2\) (Pythagoras theorem) With this, we can't even solve l and b, so no chance of getting the p.

S2:\(l*b=48\). With this, we can't even solve l and b, so no chance of getting the p.

What we need to solve is p, which is

\(p=2*(l+b)\)

Note that \((l+b)^2=l^2+b^2+2*l*b\)

Using S1 and S2:

\((l+b)^2=10^2+2*48\),

Thus \((l+b)\) can be found.

Which means \(p\) can be found.

Therefore the answer should be "C"