I have tried some many times and failed been working on this for days still can crack it. Any help will be appreciated.

A flexible wire PQRS, of total length 12 metres, is bent into the three edged planar shape shown in the diagram attached.(Note that the arc PS is not part of the wire.) The end-segments PQ and RS of the wire are each straight and have equal length l metres, while the middle segment QR forms a circular arc. The tangent to the arc QR at Q is perpendicular to RS.
The ends P,S of the wire are placed at the edge of a disc of radius 8 metres and with centre a O,as shown. The arc QR forms part of a circle with centre at O subtends an angle x(in radians) at O. The question corncerns the area A within the disc that is enclosed between its edge and the wire,which is shown shaded.

(a)(i) Express the length of the wire PQRS in terms of l and x.Hence express l in terms of x.
(ii) Hence show that the area A can be expressed by A =f (x), where
f(x)= 8x(3-2x)(5-2x)
¬¬¬¬¬¬¬¬¬(2-x)^2
(iii) Explain why it is reasonable to choose [0,15] as the domain for f(x).

2. Hello, omkara!

A flexible wire $PQRS$ of length 12 metres, is bent into the 3-sided shape.
(Note that the arc PS is not part of the wire.)
The endsegments $PQ$ and $RS$ are each straight and have equal length $L$ m.
while the middle segment $QR$ forms a circular arc.

The ends $P,S$ of the wire are placed at the edge of a disc of radius 8 m and with centre $O.$
The arc $QR$ forms part of a circle with centre at $O$ subtends an angle $x$ radians at $O.$
The question corncerns the area A within the disc that is enclosed between its edge
and the wire, which is shown shaded.
Code:
                        P
o
8    *:::*
o::::::::
*  Q*:::A:::*
*       *:::::::
* x         ::::::::
o  *  *  *  *  o  *   o
O      8-L     R  L   S
(a) Express the length of the wire PQRS in terms of $L$ and $x$.
Hence express $L$ in terms of $x.$

We have: . $PQ + \overline{QR} + RS \:=\:12$

Then: . $L + (8-L)x + L \:=\:12\quad\Rightarrow\quad\boxed{L \;=\;\frac{12-8x}{2-x}}\;\;{\color{blue}[1]}$

(b) Hence show that the area A can be expressed by: . $A \:=\:\frac{8x(3-2x)(5-2x)}{(2-x)^2}$

$A \;=\;\text{(area sector POS)} - \text{(area sector QOR})$

$A \;=\;\frac{1}{2}8^2x - \frac{1}{2}(8-L)^2x \;=\;\frac{x}{2}L(16-L)$

Substitute [1]: . $A \;=\;\frac{x}{2}\cdot\frac{12-8x}{2-x}\left(16 - \frac{12-8x}{2-x}\right) \;=\;\frac{x(12+8x)}{2(2-x)}\cdot\frac{20-8x}{2-x}$

. . . $A \;=\;\frac{4x(3-2x)}{2(2-x)}\cdot\frac{4(5-2x)}{2-x} \quad\Rightarrow\quad\boxed{ A \;=\;\frac{8(3 - 2x)(5-2x)}{(2-x)^2}}$

(c) Explain why it is reasonable to choose $[0,\:{\bf{\color{red}1.5}}]$ as the domain for $f(x).$
I seriously doubt that angle $x$ could be 15 radians.

Look at the area function: . $A \;=\;\frac{8(3-2x)(5-2x)}{(2-x)^2}$

Since $x \geq 0\text{ and }A \geq 0$, all the factors must be positive.

We have: . $\begin{Bmatrix}3 - 2x &\geq & 0 & \Rightarrow & x & \leq & 1.5 \\ 5-2x & \geq & 0 & \Rightarrow & x & \leq & 2.5 \\ 2 - x & \geq & 0 & \Rightarrow & x & \leq & 2 \end{Bmatrix}\quad\Rightarrow\quad x \:\leq\: 1.5$

Hence, the domain is: . $x \in [0,\,1.5]$

3. ## Thanks

This is the first time I've posted on this site and really am greatfull for your time and effort. Well explained, again thank you.