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. 4.2 cm and 7.6
b. 8.3 cm and 5.8
c. 7.4 cm and 9.3
d. 5.3 cm and 9.3
Let the triangle be $\triangle ABC$ with $AB=7$, $AC=8$, and $BC=x$. Let the angle bisector of $\angle B$ intersect $AC$ at $D$.
By the Angle Bisector Theorem, we have $\frac{BD}{DC} = \frac{AB}{AC} = \frac{7}{8}$. Therefore, $BD = \frac{7}{15} \cdot 8 = \frac{56}{15}$. We know that $AD = AC - CD = 8 - \frac{56}{15} = \frac{112}{15}$. Then, by the Triangle Inequality, we have $AD + BD > AB$, or $\frac{112}{15} + \frac{56}{15} > 7$. Simplifying, we have $\frac{168}{15} > 7$, which is true.
Similarly, we have $AD + DC > AC$, or $\frac{112}{15} + \frac{56}{15} > 8$, which simplifies to $\frac{168}{15} > \frac{120}{15}$, which is true.
Therefore, we have $AD + DB > AB$ and $AD + DC > AC$, so our inequalities are valid.
By the Triangle Inequality, we also have $BC + AC > AB$, or $x + 8 > 7$, which simplifies to $x > -1$. Since lengths cannot be negative, the shortest possible length for the third side is $0$.
For the longest possible length, we have $AC + AB > BC$, or $8 + 7 > x$, which simplifies to $x < 15$.
Therefore, the shortest possible length is $0$, and the longest possible length is $15$.
Rounded to the nearest tenth, our answers are $\boxed{\textbf{(d) } 5.3 \text{ cm and } 9.3 \text{ cm}}$.