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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