abuc0112 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
Here's my approach (without skipping any steps

).
Let x be the length (and width) of the square screen with diagonal 21
The area of the large screen will be x^2
Let y be the length (and width) of the square screen with diagonal 19
The area of the small screen will be y^2
Our goal is to find the value of x^2 - y^2
Large TV: If we examine the right triangle created by 2 sides (both with length x) and the diagonal, we can apply the Pythagorean Theorem to get x^2 + x^2 = 21^2
When we simplify this, we get 2x^2 = 441, which means x^2 = 441/2
Small TV: If we examine the right triangle created by 2 sides (both with length y) and the diagonal, we can apply the Pythagorean Theorem to get y^2 + y^2 = 19^2
When we simplify this, we get 2y^2 = 361,, which means y^2 = 361/2
We can now find the value of x^2 - y^2
We get x^2 - y^2 = 441/2 - 361/2 = 80/2 = 40 =
E
Cheers,
Brent