Chapter 2.6, Problem 15AYU

15. Geometry A semicircle of radius r is inscribed in a rectangle so that the diameter of the semicircle is the length of the rectangle. See the figure

(a) Express the area A of the rectangle as a function of the radius r of the semicircle.

(b) Express the perimeter p of the rectangle as a function of r.

To determine

To find:

a. To express the Area $A$ of the rectangle as a function of radius $r$ of the semicircle.

Answer to Problem 15AYU

Solution:

a. $A\left(r\right)=2r^2$

Explanation of Solution

Given:

A semicircle of radius $r$ is inscribed in a rectangle, so that the diameter of the semicircle is the length of the rectangle as given below.

Calculation:

a. To express the Area $A$ of the rectangle as a function of radius $r$ of the semicircle:

Area A of the rectangle $= \text{Length}\left(l\right)*\text{ Breadth}\left(b\right)$.

Since the semicircle is inscribed inside the rectangle.

$\Rightarrow \text{Length}\left(l\right)=2r$

$\Rightarrow \text{Breadth}\left(b\right)=r$

Hence area $A\left(r\right)=lxb=2r^2$.

To determine

To find:

b. To express the Perimeter $‘p’$ of the rectangle as a function of $r$.

Answer to Problem 15AYU

Solution:

b. $p\left(r\right)=6r$

Explanation of Solution

Given:

A semicircle of radius $r$ is inscribed in a rectangle, so that the diameter of the semicircle is the length of the rectangle as given below.

Calculation:

b. To express the Perimeter $‘p’$ of the rectange as a function of $r$.

Perimeter of the rectangle $=2\left(\text{Length }+\text{ Breadth}\right)$

$=2(2r+r)$

$=6r$

Therefore, $p\left(r\right)=6r$.