Let side of bigger square is x m. And, side of smaller square is y m.
According to given condition,
x² + y² = 260 ...(i)
And 4x – 4y = 24
⇒ 4(x – y) = 24
⇒ x – y = 6
⇒ x = 6 + y ...(ii)
Putting the value of (ii) in (i), we get
(6 + y) 2 + y² = 260
⇒ 36 + y² + 12y + y² = 260
⇒ 2y² + 12y – 224 = 0
⇒ y² + 6y – 112 = 0
⇒ y² + 14y – 5y – 112 = 0
⇒ y(y + 14) –8(y + 14) = 0
⇒ (y – 8) (y + 14) = 0
⇒ y–8 = 0, y + 14 = 0
⇒ y = 8, y = – 14
Putting the value of y in (ii), we get
x = 6 + y
= 6 + 8 = 14.
Hence, sides of squares are 14 cm and 8 cm.

Let the digits at tens and units place of the number be x and y respectively. Then, Number = 10x + 7
It is given that,
⇒ 10x + y = 4(x + y)
and 10x + y = 3xy
⇒ 6x – 3y = 0 and 10x + y = 3y
⇒ y – 2x and 10x + y = 3xy
⇒ 10x + 2x = 3x X 2x
⇒ 6x² – 12x = 0
⇒ 6x (x – 2) = 0
⇒ x = 0 or x = 2
Since the given number is a two-digit number. So, its tens digit cannot be zero.
∴ x = 2
⇒ y = 2 x 2 = 4 [∵ = 2x]
Hence,
required number = 10x+y = 10 x 2 + 4 = 24.1