The diagram shows a solid trapezoidal prism with four rectangular faces. AB = 10cm, BC = 12cm, DC = 6cm and CT = 16cm.

1. Find the horizontal difference between the two parallel sides of the trapezium.

Since $AB=10\text{ cm}$ and $DC=6\text{ cm}$,
$10-6=4\text{ cm}$
This forms one side of a right-angled triangle with height $BC=12\text{ cm}$.

2. Apply Pythagoras' theorem to find the slanted side $AD$.

$AD^2=12^2+4^2$
$AD^2=144+16=160$
$AD=\sqrt{160}=12.649110\ldots$
$AD=12.649\text{ cm}\quad \text{to 5 significant figures}$

1. Find the area of the two trapezoidal end faces.

Area of one trapezoidal end
$= \frac{1}{2}(10+6)(12)$
$=96\text{ cm}^2$
Area of two trapezoidal ends
$=2(96)=192\text{ cm}^2$

2. Find the total area of the four rectangular faces using the perimeter of the trapezium.

Using $AD=12.649\text{ cm}$ from part (a),
$\text{Perimeter of trapezium}=10+12+6+12.649=40.649\text{ cm}$
Area of four rectangular faces
$=40.649\times 16$
$=650.384\text{ cm}^2$

3. Add all exposed face areas to obtain the surface area of the prism.

$\text{Surface area}=192+650.384$
$=842.384\text{ cm}^2$
$\approx 842.4\text{ cm}^2$

1. Find the area of the trapezoidal cross-section.

The parallel sides of the trapezium are $AB=10\text{ cm}$ and $DC=6\text{ cm}$, and the perpendicular height is $BC=12\text{ cm}$.

$\text{Area of trapezium}=\frac12(10+6)(12)=96\text{ cm}^2$

2. Multiply the cross-sectional area by the prism length to find the volume.

The length of the prism is $CT=16\text{ cm}$.

$\text{Volume}=96\times16=1536\text{ cm}^3$