In the diagram, $AB$ is the diameter of the circle. AED and ABC are straight lines. $AE = 12$ cm and $EB = 16$ cm. $\angle ADC$ is a right angle.

1. Show that one pair of corresponding angles are equal right angles.

Since $AB$ is the diameter of the circle and $E$ lies on the circumference,

$\angle AEB = 90^\circ$ (angle in a semicircle).

Also, $\angle ADC = 90^\circ$ is given.

Hence,

$\angle AEB = \angle ADC$.

2. Show a second pair of corresponding angles are equal, then conclude similarity.

Since $A,E,D$ are collinear and $A,B,C$ are collinear,

$\angle EAB = \angle DAC$.

Therefore, by AA similarity,

$\triangle AEB \sim \triangle ADC$.

1. Convert the area ratio of the similar triangles to a corresponding side ratio.

Since $\triangle AEB \sim \triangle ADC$, their areas are in the ratio of the squares of their corresponding sides.

$\frac{\text{area of }\triangle AEB}{\text{area of }\triangle ADC}=\frac{4}{9}=\left(\frac{2}{3}\right)^2$

Hence the corresponding side ratio is

$\triangle AEB : \triangle ADC = 2:3.$

2. Use the side ratio to find $CD$.

From the similarity order $\triangle AEB \sim \triangle ADC$, side $EB$ corresponds to side $DC$.

$EB:DC=2:3$

Given $EB=16$ cm,

$DC=16\times \frac{3}{2}=24\text{ cm}.$

1. Find the perpendicular distance between the parallel sides $BE$ and $CD$.

From the area ratio $4:9$, the corresponding length ratio is $2:3$.

Since $AE$ corresponds to $AD$,
$AE:AD=2:3$
$12:AD=2:3$
$AD=18\text{ cm}$

Hence
$DE=AD-AE=18-12=6\text{ cm}.$

2. Use the trapezium area formula with parallel sides $BE$ and $CD$.

The parallel sides of trapezium $BCDE$ are $BE$ and $CD$.

Using $BE=16\text{ cm}$, $CD=24\text{ cm}$ from part (b), and height $DE=6\text{ cm}$,
$\text{Area of trapezium }BCDE=\frac12(16+24)(6)$
$=120\text{ cm}^2.$