### Instruction: Find the area of a triangle with side lengths 13, 17, and $12\sqrt{2}$.### Response: I notice that the side lengths are not integers, so I wonder if the triangle is right-angled.
I can check this by using the Pythagorean theorem.
If the triangle is right-angled, then the square of the longest side should equal the sum of the squares of the shorter sides.
So I calculate 13^2 + 17^2 and see if it equals $(12\sqrt{2})^2$.
I get 13^2 + 17^2 = 169 + 289 = 458, and $(12\sqrt{2})^2 = 144 \cdot 2 = 288$.
These are not equal, so the triangle is not right-angled.
This means that I cannot use the formula for the area of a right triangle, which is half the product of the lengths of the legs.
I need another way to find the area of the triangle.
I recall that there is a formula for the area of any triangle using the lengths of two sides and the angle between them.
This formula is $A = \frac{1}{2}ab \sin C$, where $a$ and $b$ are the side lengths and $C$ is the angle opposite the side of length $c$.
I don't know the angle $C$, but I can find it using the cosine rule.
The cosine rule says that $c^2 = a^2 + b^2 - 2ab \cos C$.
I can rearrange this to get $\cos C = \frac{a^2 + b^2 - c^2}{2ab}$.
In this case, I can let $a = 13$, $b = 17$, and $c = 12\sqrt{2}$.
Plugging these values into the cosine rule, I get $\cos C = \frac{13^2 + 17^2 - (12\sqrt{2})^2}{2 \cdot 13 \cdot 17}$.
Simplifying this expression, I get $\cos C = \frac{458 - 288}{442} = \frac{170}{442} \approx 0.3846$.
To find $C$, I need to take the inverse cosine of both sides.
I get $C = \cos^{-1}(0.3846) \approx 67.38^{\circ}$.
Now that I have the angle $C$, I can use the formula for the area of the triangle.
I get $A = \frac{1}{2} \cdot 13 \cdot 17 \cdot \sin(67.38^{\circ})$.
Using a calculator, I get $\sin(67.38^{\circ}) \approx 0.9218$.
Multiplying this by $\frac{1}{2} \cdot 13 \cdot 17$, I get $A \approx 103.01$.