The points A, B, C, D and E are shown in the diagram below such that ABC and AED are straight lines and BE = $\sqrt{8}$ cm.

1. Identify the opposite side and hypotenuse for angle ABE, then apply the sine ratio.

From $B(1,1)$ to $E(4,3)$, the vertical change is
$3-1=2.$
For $\angle ABE$, the opposite side is $2$ cm and the hypotenuse is $BE=\sqrt{8}$ cm.
$\sin \angle ABE=\frac{\text{opposite}}{\text{hypotenuse}}=\frac{2}{\sqrt{8}}=\frac{1}{\sqrt{2}}.$

1. Use the obtuse angle property of cosine with the sine value from part (a)(i).

From part (a)(i),
$\sin \angle ABE=\dfrac{1}{\sqrt{2}}$.

Since $\angle ABE$ is obtuse, its cosine is negative. Hence
$\cos \angle ABE=-\dfrac{1}{\sqrt{2}}.$

1. Find the length of side $AB$ from the coordinates.

Since $A(-3,1)$ and $B(1,1)$ lie on the same horizontal line,
$AB=1-(-3)=4\text{ cm}$.

2. Apply the area formula using two sides and the included angle.

$\text{Area of }\triangle ABE=\frac12(AB)(BE)\sin\angle ABE$
$=\frac12(4)(\sqrt8)\left(\frac{1}{\sqrt2}\right)$
$=2\sqrt{\frac{8}{2}}=2\sqrt4=4$

1. Find the linear scale factor from triangle ABE to triangle ACD using corresponding sides.

Since triangles $ABE$ and $ACD$ are similar, $AB$ corresponds to $AC$.

From the coordinates,
$AB=1-(-3)=4\text{ cm}$
$AC=4-(-3)=7\text{ cm}$

Linear scale factor from $\triangle ABE$ to $\triangle ACD$ is
$\frac{AC}{AB}=\frac{7}{4}.$

2. Use the square of the linear scale factor to find the area of triangle ACD.

For similar triangles,
$\frac{\text{Area of }\triangle ACD}{\text{Area of }\triangle ABE}=\left(\frac{7}{4}\right)^2.$

Using the area of $\triangle ABE$ from part (b),
$\text{Area of }\triangle ACD=4\left(\frac{7}{4}\right)^2$
$=4\times \frac{49}{16}$
$=\frac{49}{4}\text{ cm}^2.$