If the area of an equilateral triangle is increasing at the rate of $4\sqrt{3}$ cm²/sec, then the rate of increase of its perimeter when the side is 4cm, is

For an equilateral triangle with side $a$:

Area $= \dfrac{\sqrt{3}}{4}a^2$

Perimeter $= 3a$

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Given: $\dfrac{dA}{dt} = 4\sqrt{3}$ cm²/sec

Differentiating the area formula with respect to time:

$A = \dfrac{\sqrt{3}}{4}a^2$

$\dfrac{dA}{dt} = \dfrac{\sqrt{3}}{4} \times 2a \times \dfrac{da}{dt}$

$\dfrac{dA}{dt} = \dfrac{\sqrt{3}}{2} \times a \times \dfrac{da}{dt}$

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Substituting when $a = 4$ cm:

$4\sqrt{3} = \dfrac{\sqrt{3}}{2} \times 4 \times \dfrac{da}{dt}$

$4\sqrt{3} = 2\sqrt{3} \times \dfrac{da}{dt}$

$\dfrac{da}{dt} = \dfrac{4\sqrt{3}}{2\sqrt{3}}$

$\dfrac{da}{dt} = 2$ cm/sec

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Differentiating the perimeter formula with respect to time:

$P = 3a$

$\dfrac{dP}{dt} = 3 \times \dfrac{da}{dt}$

$\dfrac{dP}{dt} = 3 \times 2$

$\dfrac{dP}{dt} = 6$ cm/sec

Therefore, the rate of increase of the perimeter is 6 cm/sec.