The size of a television screen is given as the length

Hello Gmat_mission.

Let's take a look at your question.

We know that the size of a television screen is given as the length of the screen’s diagonal.

Now, if d is the diagonal of a square, then we get $$d^2=l^2+l^2=2l^2\ \ \Rightarrow\ \ l^2=\frac{d^2}{2}$$ where "l" is the length of the side of the square. Also, we know that the area of a square is equal to l^2, therefore we get: $$Area=\frac{d^2}{2}.$$ Now, the area of a square 21-inch screen is $$Area=\frac{21^2}{2}$$ and the area of a square 19-inch screen is $$Area=\frac{19^2}{2}.$$ To know how many inches greater is the first screen with respect to the second one we have to subtract both areas as follows: $$Area_1-Area_2=\frac{21^2}{2}-\frac{19^2}{2}=\frac{21^2-19^2}{2}=\frac{\left(21-19\right)\left(21+19\right)}{2}$$ $$=\frac{\left(2\right)\left(40\right)}{2}=40.$$ So, the correct answer is the option E.

I hope this explanation may help you.

Gmat_mission 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

We can save a little time if we know the following:

The area of a rhombus with diagonals d� and d₂ = (d�d₂)/2.
A square is a rhombus with 4 equal angles.
In a square, the two diagonals are equal.
Thus, the area of a square with diagonal d = d²/2.

a² - b² = (a+b)(a-b).

Thus:
Big TV - Little TV = 21²/2 - 19²/2 = (1/2)(21² - 19²) = (1/2)(21+19)(21-19) = 40.

The correct answer is E.