Chapter 7.3, Problem 104E

(a)

To determine

To show: that the area of the rectangle is modeled by the function $A\left(\theta \right)=25\sin 2\theta$ .

Explanation of Solution

Given information:

A rectangle is to be inscribed in a semicircle of radius 5 cm as shown in the following figure.

  

Calculation:

A rectangle is to be inscribed in a semicircle of radius 5cm, as shown in the figure.

  

To find the area of the rectangle, need to length $\left(DC\right)$ of the rectangle and breadth $\left(BC\right)$ of the rectangle.

From the triangle $OBC$ ,

  $\begin{array}{c}\sin \theta =\frac{BC}{OB}\\ BC=OB\sin \theta \end{array}$

Also,

  $\begin{array}{c}\cos \theta =\frac{OC}{OB}\\ OC=OB\cos \theta \end{array}$

From the figure,

  $\begin{array}{c}DC=2OC\\ =2\left(OB\right)\cos \theta \end{array}$

The area of the rectangle $DCBE$ is $A=DC\times BC$ .

Substituting $DC=2\left(OB\right)\cos \theta$ and $BC=OB\sin \theta$ , get

  $\begin{array}{c}A=2\left(OB\right)\cos \theta \times \left(OB\right)\sin \theta \\ =2{\left(OB\right)}^2\sin \theta \cdot \cos \theta \end{array}$

Substituting $OB=5$ , get

  $\begin{array}{c}A=2\times {\left(5\right)}^2\times \sin \theta \cdot \cos \theta \\ =25\times \left(2\sin \theta \cdot \cos \theta \right)\\ =25\times 2\sin 2\theta \end{array}$

Since the area of the rectangle is the function of $\theta$ , get

  $A\left(\theta \right)=25\sin 2\theta$

(b)

To determine

To find: the largest possible area for such an inscribed rectangle.

Answer to Problem 104E

The largest possible area are $25{\text{cm}}^{\text{2}}$ .

Explanation of Solution

Given information:

A rectangle is to be inscribed in a semicircle of radius 5 cm as shown in the following figure.

  

Calculation:

The area of the rectangle is modeled by the function

  $A\left(\theta \right)=25\sin 2\theta$ .

Therefore, the area of the rectangle is the function of the angle $\theta$ .

The function $\sin \theta$ attains its maximum value at $\theta =\frac{\pi }{2}$ . Therefore, the function $\sin 2\theta$ attains its maximum value at $\theta =\frac{\pi }{4}$ .

On substituting $\theta =\frac{\pi }{4}$ is $A\left(\theta \right)$ , get the largest area as,

  $\begin{array}{c}A\left(\frac{\pi }{4}\right)=25\sin \left(2\times \frac{\pi }{4}\right)\\ =25\sin \left(\frac{\pi }{2}\right)\\ =25{\text{cm}}^{\text{2}}\end{array}$

So, the largest possible area are $25{\text{cm}}^{\text{2}}$ .

(c)

To determine

To find: the dimensions of the inscribed rectangle with the largest possible area.

Answer to Problem 104E

The length is 7 cm.

Explanation of Solution

Given information:

A rectangle is to be inscribed in a semicircle of radius 5 cm as shown in the following figure.

  

Calculation:

The largest possible area as $25{\text{cm}}^{\text{2}}$ when the $\theta =\left(\frac{\pi }{4}\right)$ .

In the first part, shown that

  $BC=OB\sin \theta$

On putting the value of $OB=5\text{cm}$ and $\theta =\left(\frac{\pi }{4}\right)$ , get the breadth $BC$ as,

  $\begin{array}{c}BC=4\sin \left(\frac{\pi }{4}\right)\\ =5\times \frac{1}{\sqrt{2}}\\ =3.5\text{cm}\end{array}$

Know that, $OC=OB\cos \theta$ .

On putting the value of $OB=5\text{cm}$ and $\theta =\left(\frac{\pi }{4}\right)$ , get

  $\begin{array}{c}OC=5\times \cos \left(\frac{\pi }{4}\right)\\ =5\times \frac{1}{\sqrt{2}}\\ =3.5\end{array}$

Thus, the length $DC$ of the rectangle is,

  $\begin{array}{c}DC=2\left(OC\right)\\ =2\times 3.5\\ =7\text{cm}\end{array}$

So, the length is 7 cm.