Class 12th Maths Miscellaneous Exercise on Chapter 6 – Question-2

Class 12th   Class 12th Maths

Question 2

The two equal sides of an isosceles triangle with fixed base $b$ are decreasing at the rate of 3 cm per second. How fast is the area decreasing when the two equal sides are equal to the base?

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Solution

Let the triangle be isosceles with:

• Base $=b$ (constant)

• Two equal sides each $=x$ (changing with time)

Given:

$$\frac{dx}{dt}=-3\text{cm/sec}$$

We want:

$$\frac{dA}{dt}\text{ when }x=b$$

Step 1: Find the height of the triangle

For an isosceles triangle, dropping a perpendicular from the vertex to the base divides it into two equal parts:

$$h=\sqrt{x^2-{\left(\frac{b}{2}\right)}^2}$$

Step 2: Write the area formula

$$A=\frac{1}{2}\cdot b\cdot h=\frac{1}{2}b\sqrt{x^2-\frac{b^2}{4}}$$

Step 3: Differentiate with respect to time $t$

$$A=\frac{b}{2}\sqrt{x^2-\frac{b^2}{4}}$$
$$\frac{dA}{dt}=\frac{b}{2}\cdot \frac{1}{2\sqrt{x^2-\frac{b^2}{4}}}\cdot (2x)\frac{dx}{dt}$$
$$\frac{dA}{dt}=\frac{bx}{2\sqrt{x^2-\frac{b^2}{4}}}\cdot \frac{dx}{dt}$$

Step 4: Substitute $x=b$

$$\frac{dA}{dt}=\frac{b\cdot b}{2\sqrt{b^2-\frac{b^2}{4}}}\cdot (-3)$$
$$=\frac{b^2}{2\sqrt{\frac{3b^2}{4}}}\cdot (-3)$$
$$=\frac{b^2}{2\cdot \frac{b\sqrt{3}}{2}}\cdot (-3)$$
$$=\frac{b}{\sqrt{3}}\cdot (-3)$$
$$\boxed{{\displaystyle \frac{dA}{dt}=-b\sqrt{3}{\text{cm}}^2\mathrm{/}\text{sec}}}$$

Final Answer

$$\boxed{{\displaystyle \text{The area is decreasing at }b\sqrt{3}{\text{ cm}}^2\mathrm{/}\text{sec when the equal sides equal the base.}}}$$