A second ant walks all the way around a triangle drawn in the dirt. Two of the dimensions of the triangle are 5 1/10 centimeters and 7 1/5 centimeters. The total distance traveled by the ant is 20 centimeters. What is the length of the third side of the triangle?

Answer 1

Let the third side be $x$. We apply LoC on the three possible triangles and get
$$x^2=1.04^2+4.2^2-2(1.04)(4.2)\cos A_1$$
$$x^2=1.04^2+3^2-2(1.04)(3)\cos A_2$$
$$x^2=3^2+4.2^2-2(3)(4.2)\cos A_3$$
where $A_1$, $A_2$, and $A_3$ are the angles opposite the sides of lengths $1.04$, $3$, and $4.2$ respectively. Thus $x^2=24.496$ and $x=\boxed{4.949\text{ cm}}$.

Answer 2

A second ant walks all the way around a triangle drawn in the dirt. Two of the dimensions of the triangle are 5 1/10 centimeters and 7 1/5 centimeters. The total distance traveled by the ant is 20 centimeters. What is the length of the third side of the triangle?

Answer 3

Let the third side of the triangle be $x$ centimeters. Since the ant walks all the way around the triangle, the sum of the lengths of the three sides of the triangle is equal to 20 centimeters. Thus, we have the equation \[5\frac{1}{10}+7\frac{1}{5}+x=20.\] Simplifying the left side of the equation, we have \[5\frac{1}{10}+7\frac{1}{5}=\frac{51}{10}+\frac{37}{5}=\frac{51}{10}+\frac{74}{10}=\frac{125}{10}=\frac{25}{2}.\] Therefore, we can rewrite the equation as \[\frac{25}{2}+x=20.\] Subtracting $\frac{25}{2}$ from both sides, we find that $x=20-\frac{25}{2} =\frac{40}{2}-\frac{25}{2}=\frac{15}{2}=\boxed{7\frac{1}{2}}$ centimeters.