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(B) Let $x$ be the length of the diagonal of the square and $A$ be its area.Given that the rate of change of the diagonal is $\frac{dx}{dt} = 0.5 	ex

Vedclass · Accepted Answer

$10 \sqrt{2} 	ext{ cm}^2/	ext{sec}$

Vedclass · Answer

(B) Let $x$ be the length of the diagonal of the square and $A$ be its area.Given that the rate of change of the diagonal is $\frac{dx}{dt} = 0.5 	ext{ cm/sec}$.The area of a square in terms of its diagonal $x$ is given by $A = \frac{x^2}{2}$.Differentiating both sides with respect to time $t$,we get:$\frac{dA}{dt} = \frac{d}{dt} \left( \frac{x^2}{2} ight) = \frac{2x}{2} \cdot \frac{dx}{dt} = x \cdot \frac{dx}{dt}$.Given that the area $A = 400 	ext{ cm}^2$,we find the diagonal $x$ using $A = \frac{x^2}{2}$:$400 = \frac{x^2}{2} \implies x^2 = 800 \implies x = \sqrt{800} = 20\sqrt{2} 	ext{ cm}$.Now,substitute $x = 20\sqrt{2}$ and $\frac{dx}{dt} = 0.5$ into the expression for $\frac{dA}{dt}$:$\frac{dA}{dt} = (20\sqrt{2}) \cdot (0.5) = 10\sqrt{2} 	ext{ cm}^2/	ext{sec}$.

The diagonal of a square is changing at the rate of $0.5 \text{ cm/sec}$. Then the rate of change of area when the area is $400 \text{ cm}^2$ is equal to

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