Question

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

Answer 1

Answer:

Area of the equilateral triangle is increasing at the rate of = 5√3 m/s²

Step-by-step explanation:

To find :

dA/dt

Given that,

The sides of an equilateral triangle are increasing at the rate of 2 cm/sec

let the side of the triangle be a

since it is an equilateral triangle

so

it's area will be √3/4 a²

let it's area be A

so,

A = √3/4 a²

and,

side of the triangle is the function of time

i.e. a = f(t)

here, given,

da/dt = 2 cm/s ......(1)

now,

differenciation of its area with respect to

it's side,

dA/da

d(√3/4 a² )/da

√3/4 da²/da

= 2√3 a/4

= √3a /4

so,

dA/da = √3a/4. ....(2)

now,

Differenciation of area with respect to time

= dA/dt

dA/dt = da/dt × dA/da

from (1) da/dt = 2 cm/s

and,

from (2)

dA/da = √3 a/4

putting the values,

dA/dt = 2 × √3 a/4

= √3a/2

since a is the side of the triangle

and given its length = 10 cm

putting the value of a

dA/dt = √3 × 10/2

dA/dt = 5√3 cm/s²

so,

Area of the equilateral triangle is increasing at the rate of

= 5√3 m/s²

Answer 2

Answers—

➡10√3 cm²/sec

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Given the sides the equilateral triangle = a cm

We know —

Area of te equilateral triangle A = √3/4 a²

Differentiating w.r.t. t we get

dA/dt = √3/4 * 2a * da/dt

But it is given that da/dt = 2 cm/s

When a = 10 cm

∴ dA/dt = √3/4 ×2 ×10 ×2 = 10√3 cm²/s

∴The rate of change of area is 10√3 cm²/s