Q99 / 180 · COMEDK 2025 (25 May Shift) · Mathematics
The two equal sides of an isosceles triangle with a fixed base 'b' cm are decreasing at the rate of $4 \mathrm{~cm} / \mathrm{sec}$. The rate at which the area is decreasing when each of the equal sides becomes equal to the base is
- Correct answerA$\dfrac{4 b}{\sqrt{3}}$
- B$\dfrac{4 b}{3}$
- C$4 \sqrt{3} b$
- D$\dfrac{2 b}{\sqrt{3}}$
Solution
Let the equal sides of the isosceles triangle be $x$ and the base be $b$. The height $h$ of the triangle is given by $h = \sqrt{x^2 - (b/2)^2} = \sqrt{x^2 - b^2/4}$.
The area $A$ of the triangle is $A = \dfrac{1}{2} \times b \times h = \dfrac{b}{2} \sqrt{x^2 - \dfrac{b^2}{4}}$.
Differentiating with respect to time $t$, we get $\dfrac{dA}{dt} = \dfrac{b}{2} \times \dfrac{1}{2\sqrt{x^2 - b^2/4}} \times 2x \dfrac{dx}{dt} = \dfrac{bx}{2\sqrt{x^2 - b^2/4}} \dfrac{dx}{dt}$.
Given $\dfrac{dx}{dt} = -4$ cm/sec and we need the rate of change when $x = b$.
Substituting $x = b$ into the expression for $\dfrac{dA}{dt}$:
$\dfrac{dA}{dt} = \dfrac{b(b)}{2\sqrt{b^2 - b^2/4}} (-4) = \dfrac{b^2}{2\sqrt{3b^2/4}} (-4) = \dfrac{b^2}{2(b\sqrt{3}/2)} (-4) = \dfrac{b^2}{b\sqrt{3}} (-4) = -\dfrac{4b}{\sqrt{3}}$.
The area is decreasing at the rate of $\dfrac{4b}{\sqrt{3}}$ cm$^2$/sec.
Answer: $\dfrac{4 b}{\sqrt{3}}$
The area $A$ of the triangle is $A = \dfrac{1}{2} \times b \times h = \dfrac{b}{2} \sqrt{x^2 - \dfrac{b^2}{4}}$.
Differentiating with respect to time $t$, we get $\dfrac{dA}{dt} = \dfrac{b}{2} \times \dfrac{1}{2\sqrt{x^2 - b^2/4}} \times 2x \dfrac{dx}{dt} = \dfrac{bx}{2\sqrt{x^2 - b^2/4}} \dfrac{dx}{dt}$.
Given $\dfrac{dx}{dt} = -4$ cm/sec and we need the rate of change when $x = b$.
Substituting $x = b$ into the expression for $\dfrac{dA}{dt}$:
$\dfrac{dA}{dt} = \dfrac{b(b)}{2\sqrt{b^2 - b^2/4}} (-4) = \dfrac{b^2}{2\sqrt{3b^2/4}} (-4) = \dfrac{b^2}{2(b\sqrt{3}/2)} (-4) = \dfrac{b^2}{b\sqrt{3}} (-4) = -\dfrac{4b}{\sqrt{3}}$.
The area is decreasing at the rate of $\dfrac{4b}{\sqrt{3}}$ cm$^2$/sec.
Answer: $\dfrac{4 b}{\sqrt{3}}$
Hence, the correct answer is option: 'A'.