Bunuel wrote:

Stiv wrote:

The size of a television screen is given as the length of the screen's diagonal. If the screens were flat, then the area of a square 21-inch screen would be how many square inches greater than the area of a square 19-inch screen?

A. 2

B. 4

C. 16

D. 38

E. 40

\(d_1=21\) and \(d_2=19\) --> \(area_{square}=\frac{d^2}{2}\) --> \(area_1-area_2=\frac{(d_1)^2}{2}-\frac{(d_2)^2}{2}=\frac{21^2-19^2}{2}=\frac{(21-19)(21+19)}{2}=\frac{2*40}{2}=40\)

Answer: E.

Just to refresh my understanding, how is the \(area_{square}=\frac{d^2}{2}\)

The side x side is simple enough. Given that a square is essentially 2 isoceles triangles, I was going with 21=\(x\sqrt{2}\)

That's where I got stuck. But I still don't get your formula...