Im not sure if this is a faster method but it helped to solve the question in 30 sec

Let the side of the bigger square be x and smaller square be y

we can apply Pythagoras theorem to calculate the area, Take a look at the figure

Since it is a square, the 2 sides of the triangle will be same, so applying the Pythagoras theorem, we get

x^2 + x^2 = diagonal^2, inputting values of the diagonals as 21 and 19, we get

2x^2 = 21^2 - (1)

2y^2 = 19^2 - (2)

no we know that square of the 2 squares is x^2 and y^2 receptively so by (1) - (2), we get

x^2 - y^2 = 21^2/2 - 19^2/2

x^2 - y^2 = (441 - 361)/2

x^2 - y^2 = 80/2

x^2 - y^2 = 40

so the answer is E

diagonal 1 = x sqrt(2) = 21 Hence x = 21/sqrt(2)
diagonal 2 = y sqrt(2) = 19 Hence y = 19/sqrt(2)

Area 1 - Area 2 = Solution
[21/sqrt(2)]² - [19/sqrt(2)]²
21²/2 - 19²/2
(21² - 19²)/2
(21-19)(21+19)/2
(2)(40)/2
40

Correct Answer [spoiler](E)[/spoiler]