In the figure is a right triangle ABC with AC=10cm and AB=6cm. Determine: The height AB (hint: Theorem of Pythagoras) The area of the triangular.

In a right-angled triangle, the square of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides.
Given:
Hypotenuse $AC = 10 \text{ cm}$
Side $BC = 6 \text{ cm}$
We need to find side $AB$.

$$AB^2 + BC^2 = AC^2$$

Substitute the given values:

$$AB^2 + (6 \text{ cm})^2 = (10 \text{ cm})^2$$ $$AB^2 + 36 \text{ cm}^2 = 100 \text{ cm}^2$$

Subtract $36 \text{ cm}^2$ from both sides:

$$AB^2 = 100 \text{ cm}^2 - 36 \text{ cm}^2$$ $$AB^2 = 64 \text{ cm}^2$$

Take the square root of both sides:

$$AB = \sqrt{64 \text{ cm}^2}$$ $$AB = 8 \text{ cm}$$

The height AB is $\boxed{\text{8 cm}}$.

The formula for the area of a triangle is $\frac{1}{2} \times \text{base} \times \text{height}$.
For a right-angled triangle, the two legs can be considered the base and height.
Base $BC = 6 \text{ cm}$
Height $AB = 8 \text{ cm}$ (from Step 1)

$$\text{Area} = \frac{1}{2} \times BC \times AB$$

Substitute the values:

$$\text{Area} = \frac{1}{2} \times 6 \text{ cm} \times 8 \text{ cm}$$ $$\text{Area} = \frac{1}{2} \times 48 \text{ cm}^2$$ $$\text{Area} = 24 \text{ cm}^2$$

The area of the triangular ABC is $\boxed{\text{24 cm}^2}$.