Question 6 - NCERT Exemplar - MCQs - Chapter 6 Class 12 Application of Derivatives (Term 1)

Last updated at Dec. 4, 2021 by Teachoo

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The sides of an equilateral triangle are increasing at the rate of 2 cm/sec. The rate at which the area increases, when side is 10 cm is:

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Question 6
The sides of an equilateral triangle are increasing at the rate of 2 cm/sec. The rate at which the area increases, when side is 10 cm is:
10 cm2/s (B) 3 cm2/s
(C) 10βπ cm2/s (D) 10/3 cm2/s
Let Area of equilateral triangle = A cm2
& let Side = π cm
Given that
Sides of equilateral triangle are increasing at the rate of 2 cm/sec
β΄ π π/π π = 2
We need to find rate of change of area w.r.t. side
i.e., we need to find π π¨/π π
We know that
Area of equilateral triangle = A = β3/4 π₯^2
Finding rate of change of area
Differentiating A w.r.t.x
ππ΄/ππ‘ = β3/4 (π₯^2 )β²
ππ΄/ππ‘ = β3/4 Γ (πγ(π₯γ^2))/ππ₯ Γ ππ₯/ππ‘
ππ΄/ππ‘ = β3/4 (2π₯) ππ₯/ππ‘
π π¨/π π = (βπ π)/π π π/π π
Putting π π/π π = 2, from equation (1)
ππ΄/ππ‘ = (β3 π₯)/2 Γ 2
ππ΄/ππ‘ = βπ π
Since, we have to find rate of change of area when side is 10 cm
β΄ Putting π = 10 cm in ππ΄/ππ‘
π π¨/π π = 10 βπ cm2/sec
Hence, area increases at the rate of 10 βπ cm2/sec
So, the correct answer is (C)
π π¨/π π = (βπ π)/π π π/π π
Putting π π/π π = 2, from equation (1)
ππ΄/ππ‘ = (β3 π₯)/2 Γ 2
ππ΄/ππ‘ = βπ π
Since, we have to find rate of change of area when side is 10 cm
β΄ Putting π = 10 cm in ππ΄/ππ‘
π π¨/π π = 10 βπ cm2/sec
Hence, area increases at the rate of 10 βπ cm2/sec
So, the correct answer is (C)

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