Question

When the area of a square is increasing four times as fast as the diagonals, what is the length of a side of the square

Answer 1

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.

$\frac{d(A)}{dt}=4(\frac{dp}{dt} )$ -------(1)

$\frac{d(A)}{dt}=\frac{d(x^2)}{dt}$

$\frac{d(A)}{dt}=2x\frac{d(x)}{dt}$

Similarly, $\frac{d(p)}{dt}=\frac{d(x\sqrt{2})}{dt}$

                      $=\sqrt{2}\frac{dx}{dt}$

Now by placing the value of $\frac{d(A)}{dt}$ and $\frac{d(p)}{dt}$ in equation (1),

$2x\frac{dx}{dt}=4\sqrt{2}\frac{dx}{dt}$

$(2x - 4\sqrt{2})\frac{dx}{dt}=0$

Since, $\frac{dx}{dt}\neq 0$

$(2x - 4\sqrt{2})=0$

x = 2√2

Therefore, length of a side of the square is 2√2.