The diagonal of a square A is (a+b). The diagonal of a square whose area is twice the area of square A is:
Answer & explanation
Correct answer: option 3
Diagonal of a square = \(\sqrt {2}\) × s [s = side]
\(\sqrt {2}\) side = a + b
side = \(\frac{a + b}{\sqrt {2}}\)
⇒ Area of 1st square = \((\frac{a + b}{\sqrt {2}})^{2}\)
⇒ Area of 2nd square = 2 × \((\frac{a + b}{\sqrt {2}})^{2}\) = (a + b)2
⇒ Area of 2nd square = (side)2 = (a + b)2
⇒ Side of the 2nd square = a + b
⇒ Diagonal = \(\sqrt {2}\) (a + b)