\[\begin{array}{1 1}(1)1.5 cm^{2}/sec&(2)\frac{1}{2\pi}\\(3)4\pi&(4)\frac{\pi}{3}\end{array}\]

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Let a be the length of the side , d the length of the diagonal and A be the area of the square at time t.

$A=a^2$

$d^2=a^2+a^2= 2a^2$

$d= \sqrt 2 a$

$\large\frac{d}{dt} $$(d) =\sqrt 2 \large\frac{da}{dt}$

$0.1 = \sqrt {2} \large\frac{da}{dt}$

$\large\frac{dA}{dt} =\large\frac{da}{dt}$

$A= a^2 \large\frac{15}{\sqrt 2} \frac{0.1}{\sqrt 2}$

$\qquad= 2a$

$\qquad= 2 \times \large\frac{da}{at} \times \large\frac{0.1}{\sqrt 2}$

$=>1.5 cm^{2}/sec$

Hence 1 is the correct answer.

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