Bunuel wrote:

In the diagram above, if the area of triangle LNO is 32, then what is the area of triangle LMN?

A. 24

B. \(24 \sqrt{2}\)

C. \(24 \sqrt{3}\)

D. 32

E. 48

Even though the figure looks like a square, that can NEVER be assumed. It explicitly states that the figure is not drawn to scale (something we should assume all the time unless it is stated otherwise), and we must ONLY work with the information given.

Examining triangle LNO, we are given that the

\(area= \frac{base * height}{2} = 32\)

Since we have a right angle at O, we can consider the base to be LO and the height to be ON. We are given the height = 8, so the base, LO, is \(\frac{area*2}{base} = \frac{32*2}{8}=8\)

So, triangle LNO is an isosceles right triangle, and therefore the hypotenuse, LN, is \(8\sqrt{2}\)

Now looking at triangle LMN, we have the height given as 6, and the base LN calculated as \(8\sqrt{2}\), so

\(area=\frac{6*8\sqrt{2}}{2} = 24\sqrt{2}\)

Answer: B

_________________

Dave de Koos

GMAT aficionado