Question

The two equal sides of an isosceles triangle with fixed base b are decreasing at the rate of 3 centimetre per second how fast is the area decreasing when the two equal sides are equal to the base

Answer 1

Answer:

Step-by-step explanation:

Answer 2

Pythagoras Theorem and Differentiation Helps

Step-by-step explanation:

In the isosceles triangle, if base is b, let us think x as the similar sides.

which means,Base BC is b and both equal sides AB,AC are x

We know that on x being equal to base (b), the side of triangle decreases.

i.e. $\frac {dx}{dt} = 3\ \frac {cm}{sec}$

So, is x equals b,then area is decreasing, but how fast?

i.e. $\frac {dA}{dt}$ when x =b.

For Finding Area,

• Draw a perpendicular AD to BC  

i.e. AD⊥BC

In Isosceles triangle,

perpendicular from vertex to the side bisects the side

i.e. D is the mid point of BC

BD = DC (Given BC = b)

Therefore,  

BD = DC = $\frac {b}{2}$

Now, In ΔADB  

on applying pythagoras theorem,

$AB^2 = AD^2 + BD^2$

⇒ $x^2 = AD^2 + (\frac {b}{2})^2$

⇒ $x^2 - (\frac {b}{2})^2 = AD^2$

⇒ $AD^2 = x^2 - (\frac {b}{2})^2$

AD = $\sqrt {x^2 - (\frac {b}{2})^2}$

We know that,

• Area of Isosceles triangle = $\frac {1}{2} \times base \times height$

A = $\frac {1}{2} \times b \times \frac {1}{2} \sqrt {x^2 - (\frac {b}{2})^2}$

We need,  $\frac {dA}{dt}$

Differentiating with respect to x

$\frac {dA}{dt} = \frac {1}{2} \times b \times \frac {1}{2} \frac {d(\sqrt {x^2 - (\frac {b}{2})^2})}{dt}$

⇒ $\frac {dA}{dt} = \frac {1}{2} \times b \times \frac {1}{2} \frac {d(\sqrt {x^2 - (\frac {b}{2})^2})}{dt}$

⇒ $\frac {dA}{dt} = \frac {1}{2} \times b[\frac {1}{2(\sqrt {x^2 - (\frac {b^2}{4})})} \times (\frac {dx^2}{dt}-0)]$

⇒ $\frac {dA}{dt} = \frac {1}{2} \times b[\frac {1}{2(\sqrt {x^2 - (\frac {b^2}{4})})} \times (\frac {dx^2}{dt}\times \frac {dx}{dt})]$

⇒ $\frac {dA}{dt} = \frac {1}{2} \times b[\frac {1}{2(\sqrt {x^2 - (\frac {b^2}{4})})} \times (2x \times \frac {dx}{dt})]$

⇒ $\frac {dA}{dt} = \frac {1}{2} \times b[\frac {1}{2(\sqrt {x^2 - (\frac {b^2}{4})})} \times (2x\times 3)]$

⇒ $\frac {dA}{dt} = [\frac {b}{4(\sqrt {x^2 - (\frac {b^2}{4})})} \times 6x]$

• Now, Lets find $\frac {dA}{dt}$ at x = b,

$\frac {dA}{dt} when\ x\ tends\ to\ b = [\frac {6b^2}{4(\sqrt {b^2 - (\frac {b^2}{4})})}]$

⇒ $[\frac {6b^2}{4(\sqrt {(\frac {4b^2 - b^2}{4})})}]$    

= $[\frac {6b^2}{4(\sqrt {(\frac {3b^2}{4})})}]$

= $\frac {6b^2}{\frac {4\sqrt3b}{2}}$ = $\frac {6b^2}{2\sqrt 3b}$

= $\frac {3b}{\sqrt 3}$ = $b\sqrt 3$

Since dimension of area is $\frac {cm^2}{sec}$      

$\frac {dA}{dt}$ at x = b is $b\sqrt 3$