# 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?

**Solution:**

Let ΔABC be isosceles where BC is the base of fixed length b

Also, let the length of the two equal sides of ΔABC be a.

Let us do the construction based on the given information.

Draw AD ⊥ BC.

Now, in ΔADC, by applying the Pythagoras theorem, we have:

AD = √a² - h² / 4

A = 1/2 b√a² - h²/4

The rate of change of the area with respect to time (t) is given by,

dA/dt = 1/2 h 2a / 2 (√a² - h² / 4) da/dt

= (ab / 4a² - b² √4a² - b²) da/dt

It is given that the two equal sides of the triangle are decreasing at the rate of 3 cm per second.

Therefore,

da/dt = - 3 cm/s

Hence,

dA/dt

= (- 3ab / √4a² - b²)

When, a = b we have:

dA/dt

= (- 3b² / √4a² - b²)

= (- 3b²) / √3b²

= - √3 b

Hence, if the two equal sides are equal to the base, then the area of the triangle is decreasing at the rate of √3 b cm²/s

NCERT Solutions Class 12 Maths - Chapter 6 Exercise ME Question 3

## 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?

**Summary:**

Given that the two equal sides of an isosceles triangle with fixed base b are decreasing at the rate of 3 cm per second.Hence the area of the triangle is decreasing at the rate of √3 b cm²/s

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