Answer
a) $2x\sqrt{4-{{x}^{2}}}$
b) $4x+2\sqrt{4-{{x}^{2}}}$
Work Step by Step
(a)
Consider the provided equation, $y=\sqrt{4-{{x}^{2}}}$
Let $\left( x,y \right)$ be the coordinate of point $P$
Here, the length of rectangle is twice the $x\text{-coordinate}$ and width of the rectangle is the $y\text{-coordinate}$
Consider the length of the rectangle, $l=2x$
And width of the rectangle, $w=y$
According to the formula of the area of the rectangle
$A=l\cdot w$
Substitute $2x$ for $l$ and $y$ for $w$ in the above equation
$A=2x\cdot y$
Substitute $\sqrt{4-{{x}^{2}}}$ for $y$ in the above equation
$\begin{align}
& A=2x\cdot \sqrt{4-{{x}^{2}}} \\
& =2x\sqrt{4-{{x}^{2}}}
\end{align}$
Therefore, the required area of the rectangle, $A$ with two vertices on a semi-circle of equation $y=\sqrt{4-{{x}^{2}}}$ with radius $2$ and two vertices on the $x\text{-axis}$ in terms of $x$ is, $2x\sqrt{4-{{x}^{2}}}$.
(b)
Consider the provided equation, $y=\sqrt{4-{{x}^{2}}}$
Let $\left( x,y \right)$ be the coordinate of point $P$
Here, the length of rectangle is twice the $x\text{-coordinate}$ and width of the rectangle is the $y-\text{coordinate}$
Consider the length of rectangle, $l=2x$
And width of rectangle, $w=y$
According to the formula of perimeter of rectangle
$P=2\cdot l+2\cdot w$
Substitute $2x$ for $l$ and $y$ for $w$ in the above equation
$P=2\cdot 2x+2\cdot y$
Substitute $\sqrt{4-{{x}^{2}}}$ for $y$ in the above equation
$\begin{align}
& P=4x+2\cdot \sqrt{4-{{x}^{2}}} \\
& =4x+2\sqrt{4-{{x}^{2}}}
\end{align}$
Therefore, the required perimeter of rectangle, $P$ with two vertices on a semi-circle of equation $y=\sqrt{4-{{x}^{2}}}$ with radius $2$ and two vertices on the $x\text{-axis}$ in terms of $x$ is, $4x+2\sqrt{4-{{x}^{2}}}$.
