Answer:
Length of a side of a square = 2√2 units
Step-by-step explanation:
Let the length of a square is 'x' units.
Therefore, Area of the square A = (Side)²
= x² square units
And by Pythagoras theorem,
(Diagonal)²= (Side 1)² + (Side 2)²
= x² + x²
= 2x²
Diagonal 'p' = x√2 units
It is given in the question that area of the square is increasing four times as fast as the diagonals.
[tex]\frac{d(A)}{dt}=4(\frac{dp}{dt} )[/tex] -------(1)
[tex]\frac{d(A)}{dt}=\frac{d(x^2)}{dt}[/tex]
[tex]\frac{d(A)}{dt}=2x\frac{d(x)}{dt}[/tex]
Similarly, [tex]\frac{d(p)}{dt}=\frac{d(x\sqrt{2})}{dt}[/tex]
[tex]=\sqrt{2}\frac{dx}{dt}[/tex]
Now by placing the value of [tex]\frac{d(A)}{dt}[/tex] and [tex]\frac{d(p)}{dt}[/tex] in equation (1),
[tex]2x\frac{dx}{dt}=4\sqrt{2}\frac{dx}{dt}[/tex]
[tex](2x - 4\sqrt{2})\frac{dx}{dt}=0[/tex]
Since, [tex]\frac{dx}{dt}\neq 0[/tex]
[tex](2x - 4\sqrt{2})=0[/tex]
x = 2√2
Therefore, length of a side of the square is 2√2.
