Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
Home  >>  CBSE XII  >>  Math  >>  Model Papers
0 votes

Show that the rectangle of maximum area that can be inscribed in a circle is a square.

Can you answer this question?

1 Answer

0 votes
  • Perimeter=$2ab$
Step 1:
Let the length and breadth of the rectangle inscribed in a circle of radius 'a' be 'x' and 'y' respectively.
Differentiating with respect to x we get
Step 2:
Differentiating with respect to x we get
Step 3:
For $P(x)$ to be minimum $P'(x)=0$ and $P''(x)<0$
$\Rightarrow 2\bigg[1-\large\frac{x}{\sqrt{4a^2-x^2}}\bigg]$$=0$
$\Rightarrow 1-\large\frac{x}{\sqrt{4a^2-x^2}}$$=0$
$\Rightarrow 1=\large\frac{x}{\sqrt{4a^2-x^2}}$
$\Rightarrow 4a^2-x^2=x^2$
$x=\pm \sqrt 2 a$
Step 4:
$\Rightarrow P(x)$ is maximum at $x=\sqrt 2a$
Step 5:
$y=\sqrt 2a$
Thus $x=y$
$\therefore$ the rectangle is a square of side $\sqrt 2a$
answered Sep 27, 2013 by sreemathi.v
Ask Question
student study plans
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App