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AB is the diameter of a circle and C is any point on the circle. Show that the area of triangle ABC is maximum, when it is an isosceles triangle.

 

 

 

 
 
 
 
 

Answers (1)

Let 'r' be the radius of the circle then,
AB= 2r\left ( AB \; is\; diameter \right )        
   
Let BC = x units
we know that angle substracted by diametric in a circle in right angle
\therefore < C= 90^{\circ}
Then, AC= \sqrt{\left ( AB \right )^{2}-\left ( BC \right )^{2}}\Rightarrow AC= \sqrt{\left ( 2r \right )^{2}-\left ( x \right )^{2}}= \sqrt{4r^{2}-x^{2}}---(i)
Now, area of \triangle ABC
                 A= \frac{1}{2}\left ( AC \right )\left ( BC \right )
                     = \frac{1}{2}\sqrt{4r^{2}-x^{2}} \times x
Differentaiating A w.r.t x , we get
\Rightarrow \frac{dA}{dx}= \frac{1}{2}\left [ \sqrt{4r^{2}-x^{2}}+x\cdot \frac{1}{2\sqrt{4r^{2}-x^{2}}}\frac{d}{dx}\left ( 4r^{2}-x^{2} \right ) \right ]
\Rightarrow \frac{dA}{dx}= \frac{1}{2}\left \{ \sqrt{4r^{2}-x^{2}}-\frac{x^{2}}{\sqrt{4r^{2}-x^{2}}} \right \}
= \frac{1}{2} \left \{ \frac{4r^{2}-x^{2}-x^{2}}{\sqrt{4r^{2}-x^{2}}} \right \}
\Rightarrow \frac{dA}{dx}= \frac{1}{2} \left \{ \frac{4r^{2}-2x^{2}}{\sqrt{4r^{2}-x^{2}}} \right \}
The centrical number of x are given by \frac{dA}{dx}= 0
\Rightarrow \frac{1}{2}\left \{ \frac{4r^{2}-2x^{2}}{\sqrt{4r^{2}-x^{2}}} \right \}= 0
\Rightarrow 4r^{2}-2x^{2}= 0 \Rightarrow 4r^{2}= 2x^{2}
=x= \sqrt{2r}
Now, \frac{dA}{dx}= \frac{1}{2}\left \{ \frac{4r^{2}-2x^{2}}{\sqrt{4r^{2}-x^{2}}} \right \}
Again differentiating w.r.t x we get
\frac{d^{2}A}{dx^{2}}= \frac{1}{2}\left \{ \left ( -4x \right )\frac{1}{\sqrt{4r^{2}-x^{2}}}+\left ( 4r^{2}-2x^{2} \right )\left ( \frac{-1}{2} \right )\left ( 4r^{2}-x^{2} \right )^{\frac{-3}{2}}\frac{d}{dx} \left ( 4r^{2}-x^{2} \right )\right \}
\frac{d^{2}A}{dx^{2}}= \frac{1}{2}\left \{ \frac{-4x}{\sqrt{4r^{2}-x^{2}}}+\frac{x\left ( 4r^{2}-2x^{2} \right )}{\left ( 4r^{2}-x^{2} \right )^{\frac{3}{2}}} \right \}
=\left ( \frac{d^{2}A}{dx^{2}} \right )_{x= \sqrt{2r}}= \frac{1}{2}\left \{ \frac{-4\left ( \sqrt{2r} \right )}{\sqrt{4r^{2}-2r^{2}}}+\frac{\sqrt{2}r\left ( 4r^{2}-4r^{2} \right )}{\left ( 4r^{2}-2r^{2} \right )^{\frac{3}{2}}} \right \}
=\frac{-2\sqrt{2r}}{\sqrt{2}r}= -2< 0
Thus , A is maximum when x=\sqrt{2}r
Putting x=\sqrt{2}r in (i)
AC= \sqrt{4r^{2}-\left ( \sqrt{2}r \right )^{2}}
\therefore AC= \sqrt{2}r  & \therefore BC= AC= \sqrt{2}r
Hence A is maximum when the triangle is isosceles Hence proved

Posted by

Ravindra Pindel

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