What is the area of a triangle with two sides 10 cm and 12 cm, and an included angle of 45°?

The area of the triangle is approximately 42.43 square centimetres.

To find the area of a triangle when you know two sides and the included angle, you can use the formula:
\[ \text{Area} = \frac{1}{2} \times a \times b \times \sin(C) \]
where \(a\) and \(b\) are the lengths of the sides, and \(C\) is the included angle.

In this case, the sides are 10 cm and 12 cm, and the included angle is 45°. Plugging these values into the formula, we get:
\[ \text{Area} = \frac{1}{2} \times 10 \times 12 \times \sin(45^\circ) \]

First, calculate \(\sin(45^\circ)\). The sine of 45° is \(\frac{\sqrt{2}}{2}\), which is approximately 0.7071. Now substitute this value into the formula:
\[ \text{Area} = \frac{1}{2} \times 10 \times 12 \times 0.7071 \]

Next, perform the multiplication:
\[ \text{Area} = \frac{1}{2} \times 120 \times 0.7071 \]
\[ \text{Area} = 60 \times 0.7071 \]
\[ \text{Area} \approx 42.426 \]

Rounding to two decimal places, the area of the triangle is approximately 42.43 square centimetres. This method is particularly useful in trigonometry when dealing with non-right-angled triangles.