In the diagram, BE intersects AD at F. AF = 7 cm and FD = 4 cm. $\angle ABF$, $\angle ACD$ and $\angle DEF$ are right angles. BCDF is a trapezium and $\theta$ is an acute angle.

1. Find the lengths of $BF$ and $AB$ using triangle $ABF$.

In right-angled $\triangle ABF$,
$\sin\theta = \frac{BF}{AF} \implies BF = 7\sin\theta$
$\cos\theta = \frac{AB}{AF} \implies AB = 7\cos\theta$

2. Find the lengths of $CD$ and $AC$ using triangle $ACD$.

The total length $AD = AF + FD = 7 + 4 = 11$ cm.
In right-angled $\triangle ACD$,
$\sin\theta = \frac{CD}{AD} \implies CD = 11\sin\theta$
$\cos\theta = \frac{AC}{AD} \implies AC = 11\cos\theta$

3. Find the length of $BC$.

$BC = AC - AB = 11\cos\theta - 7\cos\theta = 4\cos\theta$

4. Calculate the perimeter of the trapezium $BCDF$.

Perimeter $P = BC + CD + DF + BF$
$P = 4\cos\theta + 11\sin\theta + 4 + 7\sin\theta$
$P = 4 + 18\sin\theta + 4\cos\theta$ (Shown)

1. Identify the constant $a$ and set up the R-formula for the trigonometric terms

From $P = 4 + 18\sin\theta + 4\cos\theta$, we can see that $a = 4$.
We need to express $18\sin\theta + 4\cos\theta$ in the form $R\sin(\theta + \alpha)$.
$R\sin(\theta + \alpha) = R\sin\theta\cos\alpha + R\cos\theta\sin\alpha$
Comparing coefficients:
$R\cos\alpha = 18$
$R\sin\alpha = 4$

2. Calculate the value of $R$

$R = \sqrt{18^2 + 4^2}$
$R = \sqrt{324 + 16} = \sqrt{340}$ (or $2\sqrt{85}$)

3. Calculate the value of $\alpha$

$\tan\alpha = \frac{4}{18} = \frac{2}{9}$
$\alpha = \tan^{-1}\left(\frac{2}{9}\right) \approx 12.5288^\circ \approx 12.5^\circ$ (to 1 d.p.)

4. Write the final expression for $P$

$P = 4 + \sqrt{340}\sin(\theta + 12.5^\circ)$

1. Determine the constants $R$ and $\alpha$ from part (b)

From part (a), $P = 4 + 18\sin\theta + 4\cos\theta$.
Using the $R$-formula, $18\sin\theta + 4\cos\theta = R\sin(\theta + \alpha)$.
$R = \sqrt{18^2 + 4^2} = \sqrt{340} = 2\sqrt{85}$
$\tan\alpha = \frac{4}{18} \implies \alpha = \tan^{-1}\left(\frac{2}{9}\right) \approx 12.5288^\circ$
So, $P = 4 + 2\sqrt{85}\sin(\theta + 12.5288^\circ)$.

2. Find the maximum exact value of $P$

The maximum value of $\sin(\theta + \alpha)$ is $1$.
Maximum exact value of $P = 4 + 2\sqrt{85}(1) = 4 + 2\sqrt{85}$.

3. Find the corresponding value of $\theta$

For maximum $P$, $\sin(\theta + 12.5288^\circ) = 1$.
$\theta + 12.5288^\circ = 90^\circ$
$\theta = 90^\circ - 12.5288^\circ = 77.4712^\circ \approx 77.5^\circ$ (to 1 d.p.)

1. Substitute $P = 11$ into the $R$-formula expression from part (b)

From part (b), $P = 4 + \sqrt{340}\sin(\theta + 12.53^\circ)$.
$4 + \sqrt{340}\sin(\theta + 12.5288^\circ) = 11$
$\sqrt{340}\sin(\theta + 12.5288^\circ) = 7$

2. Isolate the sine term and find the basic angle

$\sin(\theta + 12.5288^\circ) = \frac{7}{\sqrt{340}}$
$\text{Basic angle} = \sin^{-1}\left(\frac{7}{\sqrt{340}}\right) \approx 22.309^\circ$

3. Solve for $\theta$, considering the given domain

Since $\theta$ is an acute angle, we take the first quadrant solution:
$\theta + 12.5288^\circ = 22.309^\circ$
$\theta = 22.309^\circ - 12.5288^\circ = 9.7802^\circ \approx 9.78^\circ$

1. State the maximum value of the perimeter $P$ from earlier parts

From part (c), the maximum exact value of $P$ is $4 + \sqrt{18^2 + 4^2} = 4 + \sqrt{340}$.

2. Compare the maximum value with 28 to draw a conclusion

Since $\sqrt{340} < \sqrt{400} = 20$, the maximum value of $P$ is strictly less than $4 + 20 = 24$.
Therefore, $4 + \sqrt{340} < 28$.
Thus, the perimeter cannot be 28 cm.