An angle bisector of a triangle divides the opposite side of the triangle into segments 6 cm and 8 cm long. A second side of the triangle measures 7 cm long. Find the longest and shortest possible lengths of the third side of the triangle. Round answers to the nearest tenth.
a. 7.4cm and 9.3 cm
b. 4.2 cm and 7.6 cm
c. 5.3 cm and 9.3
d. 8.3 cm and 5.8
Let $AC=6$ and $BC=8$. Let $AB=x$. Recall that the angle bisector of an angle in a triangle divides the opposite side into segments proportional to the lengths of the other two sides. Therefore, we have the following proportion: \[\frac{AC}{BC}=\frac{AB}{CB}\] Substituting $AC=6$, $BC=8$ and $AB=x$, we get: \[\frac{6}{8}=\frac{x}{8}\] Cross multiplying and simplifying, we have $8x=48$ and $x=6$. Therefore, the third side $AB$ has length $6$. Now, we can use the triangle inequality to find the range of possible side lengths for the third side. The triangle inequality states that the sum of the lengths of any two sides of a triangle must be greater than the length of the third side. Therefore, we have the following inequalities: \[AC+BC>AB\] \[AC+AB>BC\] \[BC+AB>AC\] Substituting in $AC=6$, $BC=8$ and $AB=6$, we obtain the following inequalities: \[6+8>6 \rightarrow 14>6 \quad \text{(True)}\] \[6+6>8 \rightarrow 12>8 \quad \text{(True)}\] \[8+6>6 \rightarrow 14>6 \quad \text{(True)}\] Since all the inequalities are true, the values $AC=6$, $BC=8$, and $AB=6$ do satisfy the triangle inequality. Therefore, the longest possible length for the third side is $14$ and the shortest possible length is $8$. Rounding these to the nearest tenth, we get $14.0$ and $8.0$. Hence, the answer is $\boxed{\text{(C) }5.3 \text{cm and } 9.3 \text{cm}}$.
