Question

Prove that the rectangle of maximum area inscribed in a circle is a square.

Answer 1

Hint: In this question we need to prove that the rectangle of maximum area inscribed in a circle is a square. A square has the property that all of its sides are equal. So to prove it assume some sides of the rectangle inscribed within a circle and find out the area. Use the concept of maxima and minima by differentiating the area and then cross verify by taking out the double derivative that it’s a maxima or not. This concept will help in proving the required.

Complete step-by-step answer:

Let ABCD be the rectangle inscribed in the circle with center O and radius (r).
The diagonal of the rectangle will be the diameter of the circle.
As we know diameter (d) is twice the radius.
$\Rightarrow DB=d=2r$
Since the rectangle has all the four coordinates inscribed on the circumference of the circle.
Hence let the sides of the rectangle be x and y respectively as shown in figure.
Now in triangle BCD apply Pythagoras theorem we have,
$\Rightarrow {\left(\text{hypotenuse}\right)}^2={\left(\text{perpendicular}\right)}^2+{\left(\text{base}\right)}^2$
$\Rightarrow {\left(2r\right)}^2=x^2+y^2$…………………… (1)
Now the area (A) of the rectangle is length multiplied by breadth.
$\Rightarrow A=xy$………………………. (2)
Now from equation (1) calculate the value of x
$$\begin{gathered} \Rightarrow 4r^2=x^2+y^2 \\ \Rightarrow x^2=4r^2-y^2 \end{gathered}$$
Now take square root on both sides we have,
$\Rightarrow x={\left(4r^2-y^2\right)}^{{\displaystyle \frac{1}{2}}}$……………………………….. (3)
Now substitute this value in equation (2) we have,
$\Rightarrow A=y{\left(4r^2-y^2\right)}^{{\displaystyle \frac{1}{2}}}$
Now we have to maximize the area so differentiate the area w.r.t. y and put it equal to zero we have,
${\displaystyle \frac{dA}{dy}}={\displaystyle \frac{d}{dy}}\left[y{\left(4r^2-y^2\right)}^{{\displaystyle \frac{1}{2}}}\right]=0$
Here we use product rule of differentiation $\left[{\displaystyle \frac{d}{dx}}ab=a{\displaystyle \frac{d}{dx}}b+b{\displaystyle \frac{d}{dx}}a\right]$ so using this property we have,
$$\Rightarrow y{\displaystyle \frac{d}{dy}}{\left(4r^2-y^2\right)}^{{\displaystyle \frac{1}{2}}}+{\left(4r^2-y^2\right)}^{{\displaystyle \frac{1}{2}}}{\displaystyle \frac{d}{dy}}y=0$$
Now differentiation we have,
$$\Rightarrow y{\displaystyle \frac{\left(-2y\right)}{2{\left(4r^2-y^2\right)}^{{\displaystyle \frac{1}{2}}}}}+{\left(4r^2-y^2\right)}^{{\displaystyle \frac{1}{2}}}\left(1\right)=0$$
Now simplify the above equation we have,
$$\Rightarrow +{\left(4r^2-y^2\right)}^{{\displaystyle \frac{1}{2}}}\left(1\right)=y{\displaystyle \frac{y}{{\left(4r^2-y^2\right)}^{{\displaystyle \frac{1}{2}}}}}$$
$\Rightarrow 4r^2-y^2=y^2$
$\Rightarrow 2y^2=4r^2$
$\Rightarrow y^2=2r^2$
Now take square root on both sides we have,
$\Rightarrow y=r\sqrt{2}$
Now substitute this value in equation (3) we have,
$\Rightarrow x={\left(4r^2-2r^2\right)}^{{\displaystyle \frac{1}{2}}}=\sqrt{2r^2}=r\sqrt{2}$
Hence x = y $=r\sqrt{2}$ thus it forms a square with maximum area.
So the rectangle of maximum area inscribed in a circle is a square.

Note: Whenever we face such types of problems the key concept is simply to have a diagrammatic representation of the information provided in the question as it helps to understand the basic geometry of the figure. Having a good gist of the properties of rectangle and square helps in getting the answer.