Bunuel 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

If the size of a square television screen is given by its diagonal, we need side lengths to calculate area.

The relationship between the a square's side and its diagonal, d, is given by

\(s\sqrt{2} = d\)

\(s = \frac{d}{\sqrt{2}}\)

The side of the 21-inch size television (d = 21), therefore, is

\(\frac{21}{\sqrt{2}}\). Square that to find area:

\((\frac{21}{\sqrt{2}}\) * \(\frac{21}{\sqrt{2}})\) = \(\frac{21*21}{2}\) = \(\frac{441}{2}\)

The side of the 19-inch size television (d = 19) is

\(\frac{19}{\sqrt{2}}\). Square that to find area:

\((\frac{19}{\sqrt{2}}\) * \(\frac{19}{\sqrt{2}})\) = \(\frac{19*19}{2}\) = \(\frac{361}{2}\)

Difference in area between larger and smaller, in square inches:

\((\frac{441}{2} - \frac{361}{2}) =\frac{80}{2} = 40\)

Answer E

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