Calculate the area of the outer triangle. Show all your calculations.

From the previous image (the net), the outer triangles ("The Back" and "The Front") are isosceles triangles.
The base of the triangle is $150 \text{ mm}$.
The two equal slanted sides are $215 \text{ mm}$ each.
To find the perpendicular height ($h$), we can use the Pythagorean theorem. The height divides the isosceles triangle into two right-angled triangles, where half of the base is one leg, the height is the other leg, and the slanted side is the hypotenuse.
Half of the base $= \frac{150 \text{ mm}}{2} = 75 \text{ mm}$.

Using the Pythagorean theorem ($a^2 + b^2 = c^2$):

$$h^2 + (75 \text{ mm})^2 = (215 \text{ mm})^2$$ $$h^2 + 5625 \text{ mm}^2 = 46225 \text{ mm}^2$$ $$h^2 = 46225 \text{ mm}^2 - 5625 \text{ mm}^2$$ $$h^2 = 40600 \text{ mm}^2$$ $$h = \sqrt{40600} \text{ mm}$$ $$h \approx 201.4944 \text{ mm}$$

The formula for the area of a triangle is:

$$\text{Area} = \frac{1}{2} \times \text{Base} \times \text{Perpendicular height}$$

Substitute the values for the base and the calculated perpendicular height:

$$\text{Area} = \frac{1}{2} \times 150 \text{ mm} \times \sqrt{40600} \text{ mm}$$ $$\text{Area} = 75 \text{ mm} \times \sqrt{40600} \text{ mm}$$ $$\text{Area} \approx 75 \times 201.4944 \text{ mm}^2$$ $$\text{Area} \approx 15112.08 \text{ mm}^2$$

The area of the outer triangle is approximately $\text{15112.08 mm}^2$.