Find the areas of the following trapeziums. A) Given a trapezium with parallel sides 7cm and 9cm, and non-parallel sides 8cm.

Since it's an isosceles trapezium, we can draw perpendiculars from the ends of the shorter parallel side to the longer parallel side. This forms a rectangle in the middle and two right-angled triangles on the sides.
The length of the base of each right-angled triangle is $\frac{9 \text{ cm} - 7 \text{ cm}}{2} = \frac{2 \text{ cm}}{2} = 1 \text{ cm}$.
Using the Pythagorean theorem for one of the right-angled triangles:

$$h^2 + (1 \text{ cm})^2 = (8 \text{ cm})^2$$ $$h^2 + 1 = 64$$ $$h^2 = 64 - 1$$ $$h^2 = 63$$ $$h = \sqrt{63} \text{ cm}$$

We can simplify $\sqrt{63}$ as $\sqrt{9 \times 7} = 3\sqrt{7} \text{ cm}$.

$$\text{Area} = \frac{1}{2}(16 \text{ cm})(3\sqrt{7} \text{ cm})$$ $$\text{Area} = 8 \times 3\sqrt{7} \text{ cm}^2$$ $$\text{Area} = 24\sqrt{7} \text{ cm}^2$$

The area of the trapezium is \boxed{\text{24\sqrt{7} cm}^2}.