CUET UG 2025 — Mathematics Calculus

An equilateral triangle has sides increasing at 2 cm/sec. The rate at which the area increases when the side is 10 cm needs to be found.

For an equilateral triangle with side $a$a:

$$A = \frac{\sqrt{3}}{4}a^2$$A=43​​a2

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

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

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

The term $\frac{da}{dt}$dtda​ appears due to the chain rule, since $a$a changes with time.

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Given:

$a = 10$a=10 cm

$\frac{da}{dt} = 2$dtda​=2 cm/sec

Substituting these values:

$$\frac{dA}{dt} = \frac{\sqrt{3}}{2} \times 10 \times 2$$dtdA​=23​​×10×2

$\frac{dA}{dt} = 10\sqrt{3}$dtdA​=103​ cm²/sec

Therefore, the area increases at a rate of $10\sqrt{3}$103​ cm²/sec.