CBSE Board Examination 2025 · Section C · 3 marks

Q26: Rate of increase of the area of an equilateral triangle of side 15 cm

Board 2025Ch 6 · Application of Derivatives

The question

The side of an equilateral triangle is increasing at the rate of 3 cm/s. At what rate its area increasing when the side of the triangle is 15 cm ?

Exam answer

Let $a$ be the side and $A$ the area of the triangle at time $t$ ⁠. Then $\frac{da}{dt} = 3$ cm/s.

Why?

A rate of change with time is a derivative with respect to $t$ ⁠. "The side is increasing at 3 cm/s" means $\frac{da}{dt} = +3$ ⁠. It is positive because the side is growing.

We are asked for $\frac{dA}{dt}$ ⁠, so we need a formula linking $A$ and $a$ ⁠.

A = \frac{\sqrt3}{4}a^2

Why?

Area of a triangle with two sides $a$ ⁠, $a$ and the angle $60^\circ$ between them:

\begin{aligned} A &= \frac12\cdot a\cdot a\sin60^\circ \\ &= \frac12 a^2\cdot\frac{\sqrt3}{2} = \frac{\sqrt3}{4}a^2 \end{aligned}

\begin{aligned} \frac{dA}{dt} &= \frac{\sqrt3}{4}\cdot2a\frac{da}{dt} \\ &= \frac{\sqrt3}{2}a\frac{da}{dt} \end{aligned}

Why?

Differentiate both sides with respect to $t$ ⁠. The side $a$ changes with $t$ ⁠, so by the chain rule

\frac{d}{dt}(a^2) = 2a\frac{da}{dt}

Then $\frac{\sqrt3}{4}\cdot2 = \frac{\sqrt3}{2}$ ⁠.

At $a = 15$ cm,

\frac{dA}{dt} = \frac{\sqrt3}{2}\times15\times3 = \frac{45\sqrt3}{2}

Why?

Put in $a = 15$ and $\frac{da}{dt} = 3$ only after differentiating.

\begin{aligned} \frac{dA}{dt} &= \frac{\sqrt3}{2}(15)(3) \\ 15 \times 3 &= 45 \\ \frac{\sqrt3}{2}\times45 &= \frac{45\sqrt3}{2} \end{aligned}

Watch out: if you put $a = 15$ first, $A$ becomes a constant and its derivative would wrongly be $0$ ⁠.

The units are those of area per unit time: cm²/s.

$\therefore$ The area is increasing at the rate of $\dfrac{45\sqrt3}{2}$ cm²/s.

Where this comes from

Chapter 6, Application of Derivatives: Rate of change: area of an equilateral triangle.

Question type: Rates in time

How to spot it: click to see

Type 2. Rates in time: how fast is it changing now?

Spot it: words like "per second", "how fast", "at the rate of", and one moment, such as "when the radius is 10 cm".

Here: The side grows at 3 cm/s, so $\frac{da}{dt} = 3$ ⁠, and the moment is a side of 15 cm. Differentiate $A = \frac{\sqrt3}{4}a^2$ with respect to $t$ first, then put in $a = 15$ ⁠.

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