Use the theorem of Pythagoras to answer the questions below. a. Find the perimeter and area of the triangle in the diagram below.

The legs of the right-angled triangle are 8 m and 12 m. Let $a$ be the hypotenuse.

$$a^2 = (8\text{ m})^2 + (12\text{ m})^2$$ $$a^2 = 64\text{ m}^2 + 144\text{ m}^2$$ $$a^2 = 208\text{ m}^2$$ $$a = \sqrt{208}\text{ m}$$

To simplify $\sqrt{208}$: $208 = 16 \times 13$.

$$a = \sqrt{16 \times 13}\text{ m}$$ $$a = 4\sqrt{13}\text{ m}$$

As a decimal, $a \approx 4 \times 3.60555 \text{ m} \approx 14.42\text{ m}$.

The perimeter is the sum of the lengths of all three sides.

$$\text{Perimeter} = 8\text{ m} + 12\text{ m} + 4\sqrt{13}\text{ m}$$ $$\text{Perimeter} = (20 + 4\sqrt{13})\text{ m}$$

Using the decimal approximation for $a$:

$$\text{Perimeter} \approx 20\text{ m} + 14.42\text{ m}$$ $$\text{Perimeter} \approx 34.42\text{ m}$$

The area of a right-angled triangle is given by $\frac{1}{2} \times \text{base} \times \text{height}$.
Here, the base is 12 m and the height is 8 m.

$$\text{Area} = \frac{1}{2} \times 12\text{ m} \times 8\text{ m}$$ $$\text{Area} = 6\text{ m} \times 8\text{ m}$$ $$\text{Area} = 48\text{ m}^2$$

The perimeter of the triangle is $\boxed{(20 + 4\sqrt{13})\text{ m} \approx 34.42\text{ m}}$ and the area is $\boxed{48\text{ m}^2}$.