question_answer

\[AB\]is a diameter of a circle and\[C\]is any point on the circumference of the circle, then

A)  The area of \[\Delta ABC\] is maximum when it is isosceles

B)         The area of \[\Delta ABC\] is minimum when it is isosceles

C)         The perimeter of \[\Delta ABC\] is minimum when it is isosceles

D)         None of these

Correct Answer: A

Solution :

Area of\[\Delta ABC\],\[A=\frac{1}{2}\times \sqrt{{{d}^{2}}-{{x}^{2}}}\] for max/min\[\frac{dA}{dx}=0\] \[\Rightarrow \frac{1}{2}\sqrt{{{d}^{2}}-{{x}^{2}}}+\frac{1}{2}\times \left\{ \frac{-2x}{2\sqrt{{{d}^{2}}-{{x}^{2}}}} \right\}=0\] \[\Rightarrow \]                                       \[\frac{{{d}^{2}}-{{x}^{2}}-{{x}^{2}}}{2\sqrt{{{d}^{2}}-{{x}^{2}}}}=0\] \[\Rightarrow \]                                                           \[x=\frac{d}{\sqrt{2}}\] Also for \[x<\frac{d}{\sqrt{2}},\,\,\frac{dA}{dx}>0\] and for \[x>\frac{d}{\sqrt{2}},\,\,\frac{dA}{dx}<0\] Hence;\[x=\frac{d}{\sqrt{2}}\]is the point of maxima. \[\therefore \]\[A\]is max at\[x=\frac{d}{\sqrt{2}}\] the area is maximum\[y=\frac{d}{\sqrt{2}}\] \[\therefore \]when\[\Delta \]is isosceles,