Chapter 10.7, Problem 56PS

Solution

To find: The value of angle so that area of rectangle inscribed in half circle is maximum.

The maximum area of a rectangle inscribed in a circle of radius 1 is 2 units when $\theta =45°$ .

Given data:

The rectangle ABCD inscribed in a half circle of radius 1 unit is shown in Fig.1.

  

Method/Formula used:

The area of a rectangle equals product of length and width.

Calculations:

In Fig. 1, the area $A_1$ of rectangle OCBD is

  $\begin{array}{c}{A}_1=OC\cdot CB\\ =\cos \theta \cdot \sin \theta \end{array}$

The area $A$ of rectangle ABDC equals $2A_1$ , i.e., area ABCD is

  $\begin{array}{c}A=4\cos \theta \cdot \sin \theta \left[\because \cos \theta \cdot \sin \theta =\sin 2\theta \right]\\ =2\sin 2\theta \end{array}$

Area A is maximum when $\sin 2\theta$ is maximum, i.e., $\sin 2\theta$ equals 1 (maximum value of $\sin 2\theta$ ), therefore,

  $\begin{array}{c}{A}_{\max .}=2{\left(\sin 2\theta \right)}_{\max .}\\ A_{\max .}=2\cdot 1\\ =2\end{array}$

Further,

  $\begin{array}{c}\sin 2\theta =1\\ \sin 2\theta =\sin 90°\\ 2\theta =90°\\ \theta =45°\end{array}$

Thus, the maximum area of a rectangle inscribed in a circle of radius 1 is 2 units when $\theta =45°$ .