The sides of an equilateral triangle are increasing at the rate of 5 cm/sec. The rate at which the area increases when the side is 20 cm, is

The sides of an equilateral triangle are increasing at the rate of 5 cm/sec, with the current side length at 20 cm.

The area of an equilateral triangle with side $a$a is given by:

$$A = \dfrac{\sqrt{3}}{4} \times a^2$$A=43​​×a2

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Differentiating both sides with respect to time $t$t:

$$\dfrac{dA}{dt} = \dfrac{\sqrt{3}}{4} \times \dfrac{d(a^2)}{dt}$$dtdA​=43​​×dtd(a2)​

$$\dfrac{dA}{dt} = \dfrac{\sqrt{3}}{4} \times 2a \times \dfrac{da}{dt}$$dtdA​=43​​×2a×dtda​

$$\dfrac{dA}{dt} = \dfrac{\sqrt{3}}{2} \times a \times \dfrac{da}{dt}$$dtdA​=23​​×a×dtda​

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Given:
$a = 20$a=20 cm
$\dfrac{da}{dt} = 5$dtda​=5 cm/sec

Substituting the values:

$$\dfrac{dA}{dt} = \dfrac{\sqrt{3}}{2} \times 20 \times 5$$dtdA​=23​​×20×5

$$\dfrac{dA}{dt} = \dfrac{\sqrt{3}}{2} \times 100$$dtdA​=23​​×100

$\dfrac{dA}{dt} = 50\sqrt{3}$dtdA​=503​ cm²/sec

Therefore, the rate at which the area increases is $50\sqrt{3}$503​ cm²/sec.