The diagonal of a square A is (a + b) units. What is the area (in square units) of the square drawn on the diagonal of square B whose area is twice the area of A ?

$2(a + b)^2$

Let a = b = 1, then

Diagonal of square A = 2

Area of square A = \(\frac{2^2}{2}\) = 2

Area of square B = 2 × 2 = 4

Sides of square B = \(\sqrt {4}\) = 2

Diagonal of square B = 2 \(\sqrt {2}\)

Given,

Side of other square  = diagonal of square B = 2 √2

Area of other square = 2 \(\sqrt {2}\)× 2 \(\sqrt {2}\)= 8

From option 4 = 2 (a + b)²

= 2 (1 + 1)²

= 2 × 4

= 8 (satisfied)