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 ?
Answer & explanation
Correct answer: option 4
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)