Diagonal of square = √2a [a = side]
4√2 = √2 a
a = 4 cm
Now, area of square = a² = (4²) = 16
Side of a square whose area is 2 x 16.
a₁² = 32
⇒ a₁ = √32 ⇒a₁4√2

Correct Option: A

Diagonal of square = √2a [a = side]
4√2 = √2 a
a = 4 cm
Now, area of square = a² = (4²) = 16
Side of a square whose area is 2 x 16.
a₁² = 32
⇒ a₁ = √32 ⇒a₁4√2

Now, diagonal of new square = √2a
= √2x 4 √2
= 8 cm

Let the diagonals of the squares be 3x and 2x.
∴ Ratio of their areas = [(1/2) (3x²)] / [(1/2) (2x²)]

Correct Option: A

Let the diagonals of the squares be 3x and 2x.
∴ Ratio of their areas = [(1/2) (3x²)] / [(1/2) (2x²)] = 9/4

Required increment = 2a + [a² / 100] %

Correct Option: C

Required increment = 2a + [a² / 100] %
= 2 x 25 + [(25²)/100)] %
= 50 + (625/100)%
= 56.25%

Ratio of area of 2 squares = (ratio of perimeter of 2 squares)²

Correct Option: C

Ratio of area of 2 squares = (ratio of perimeter of 2 squares)²
= (24/12)²
= 4

Area of the field =1215/135 = 9 hec
= 90000 m² [1 hec =10000 m²]
∴ Side of the field = √90000 = 300 m

Correct Option: C

Area of the field =1215/135 = 9 hec
= 90000 m² [1 hec =10000 m²]
∴ Side of the field = √90000 = 300 m
Perimeter of the field = 4 x 300 = 1200 m

Now, cost of putting a fence around field = (1200 x 75)/100 = ₹ 900