If the length of the diagonal of a square is increasing at the rate of 0.2 cm/sec, then the rate of increase of its area when its side is `30//sqrt(2)` cm, isA. `3 cm^(2)//sec`B. `(6)/(sqrt(2))cm^(2)//sec`C. `3sqrt(2)cm^(2)//sec`D. `6cm^(2)//sec`

If the length of the diagonal of a square is increasing at the rate of

1.	If the length of the diagonal of a square is increasing at the rate of 0.2 cm/sec, then the rate of increase of its area when its side is `30//sqrt(2)` cm, isA. `3 cm^(2)//sec`B. `(6)/(sqrt(2))cm^(2)//sec`C. `3sqrt(2)cm^(2)//sec`D. `6cm^(2)//sec`
Answer» Correct Answer - D Let at any time t the length of the diagonal be x cm. Then, each side = `(x)/(sqrt(2))` cm. We have, `(dx)/(dt) = 0.2 cm//sec` Let A be the area of the squar. Then, `A=((x)/(sqrt(2)))^(2)=(1)/(2)x^(2)` `implies (dA)/(dt)=x(dx)/(dt)` `implies ((dA)/(dt))_((x)/(sqrt(2))=(30)/(sqrt(2)))=30xx0.2 = 6cm^(2)//sec`

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If the length of the diagonal of a square is increasing at the rate of 0.2 cm/sec, then the rate of increase of its area when its side is `30//sqrt(2)` cm, isA. `3 cm^(2)//sec`B. `(6)/(sqrt(2))cm^(2)//sec`C. `3sqrt(2)cm^(2)//sec`D. `6cm^(2)//sec`

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