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?
Area of square A = $$\frac{(diagonal)^2}{2}$$ =Â $$\frac{(a + b)^2}{2}$$
Area of square B = 2 $$\times$$ area of square A = 2$$\times \frac{(a + b)^2}{2} = (a + b)^2$$
Side of B = a + b
Diagonal of B = $$\sqrt{2} side$$ = $$\sqrt{2}(a+b)$$
Area (in square units) of the square drawn on the diagonal of square B = $$(side)^2$$ =Â $$(\sqrt{2}(a+b))^2 = 2(a + b)^2$$
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