如图所示， $AC = 35\text{ cm}$， $EC = 15\text{ cm}$，且角 $ACB = \theta$ 弧度。 $ED$ 和 $AB$ 均垂直于 $DB$。

1. Find angle DCE using angles on a straight line

Since $D, C, B$ lie on a straight line,
$\angle DCE + \angle ECA + \angle ACB = \pi$
$\angle DCE + \frac{\pi}{2} + \theta = \pi$
$\angle DCE = \frac{\pi}{2} - \theta$

2. Find angle CED using the angle sum of a triangle

In $\triangle DEC$, the sum of angles is $\pi$.
$\angle CED + \angle CDE + \angle DCE = \pi$
$\angle CED + \frac{\pi}{2} + \left(\frac{\pi}{2} - \theta\right) = \pi$
$\angle CED + \pi - \theta = \pi$
$\angle CED = \theta$ (shown)

1. Express the lengths of the sides of triangle $CAB$ in terms of $\theta$.

In $\triangle CAB$:
$\sin\theta = \frac{AB}{35} \implies AB = 35\sin\theta$
$\cos\theta = \frac{CB}{35} \implies CB = 35\cos\theta$
Perimeter of $\triangle CAB = 35 + 35\sin\theta + 35\cos\theta$

2. Express the lengths of the sides of triangle $DEC$ in terms of $\theta$.

From (i), $\angle CED = \theta$.
In $\triangle DEC$:
$\sin\theta = \frac{DC}{15} \implies DC = 15\sin\theta$
$\cos\theta = \frac{ED}{15} \implies ED = 15\cos\theta$
Perimeter of $\triangle DEC = 15 + 15\sin\theta + 15\cos\theta$

3. Find the sum of the perimeters and simplify to the required form.

$P = (35 + 35\sin\theta + 35\cos\theta) + (15 + 15\sin\theta + 15\cos\theta)$
$P = 50 + 50\sin\theta + 50\cos\theta$
$P = 50\cos\theta + 50\sin\theta + 50$ (Shown)

1. Find the expression for $P$ from part (ii)

In $\triangle CAB$, $AB = 35\sin\theta$ and $BC = 35\cos\theta$.
Perimeter of $\triangle CAB = 35 + 35\sin\theta + 35\cos\theta$.
In $\triangle DEC$, $CD = 15\sin\theta$ and $ED = 15\cos\theta$.
Perimeter of $\triangle DEC = 15 + 15\sin\theta + 15\cos\theta$.
$P = (35 + 35\sin\theta + 35\cos\theta) + (15 + 15\sin\theta + 15\cos\theta)$
$P = 50\cos\theta + 50\sin\theta + 50$

2. Express $50\cos\theta + 50\sin\theta$ in the form $R\cos(\theta - \alpha)$

Let $50\cos\theta + 50\sin\theta = R\cos(\theta - \alpha) = R\cos\theta\cos\alpha + R\sin\theta\sin\alpha$.
Comparing coefficients:
$R\cos\alpha = 50$
$R\sin\alpha = 50$
$R = \sqrt{50^2 + 50^2} = \sqrt{5000} = 50\sqrt{2}$
$\tan\alpha = \frac{50}{50} = 1 \implies \alpha = \frac{\pi}{4}$
So, $P = 50\sqrt{2}\cos\left(\theta - \frac{\pi}{4}\right) + 50$

3. Find the maximum value of $P$ and the corresponding $\theta$

The maximum value of $\cos\left(\theta - \frac{\pi}{4}\right)$ is $1$.
Maximum $P = 50\sqrt{2}(1) + 50 = 50\sqrt{2} + 50$
This occurs when $\cos\left(\theta - \frac{\pi}{4}\right) = 1$
$\theta - \frac{\pi}{4} = 0 \implies \theta = \frac{\pi}{4}$

1. Express the lengths of the sides of triangles $DEC$ and $CAB$ in terms of $\theta$.

In $\triangle CAB$:
$AB = 35 \sin \theta$
$CB = 35 \cos \theta$

In $\triangle DEC$:
Since $\angle CED = \theta$,
$DC = 15 \sin \theta$
$ED = 15 \cos \theta$

2. Formulate the expression for the sum of the areas of the two triangles.

Area of $\triangle CAB = \frac{1}{2} (35 \cos \theta)(35 \sin \theta) = \frac{1225}{2} \sin \theta \cos \theta$
Area of $\triangle DEC = \frac{1}{2} (15 \sin \theta)(15 \cos \theta) = \frac{225}{2} \sin \theta \cos \theta$

$A = \frac{1225}{2} \sin \theta \cos \theta + \frac{225}{2} \sin \theta \cos \theta$
$A = \frac{1450}{2} \sin \theta \cos \theta = 725 \sin \theta \cos \theta$

3. Simplify the expression using the double angle identity.

Using the identity $\sin 2\theta = 2 \sin \theta \cos \theta$:
$A = 725 \left( \frac{1}{2} \sin 2\theta \right)$
$A = 362.5 \sin 2\theta$