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. 7.4 cm and 9.3 cm
B. 8.3 cm and 5.8 cm
C. 4.2 cm and 7.6 cm
D. 5.3 cm and 9.3 cm
Let the lengths of the third side of the triangle be $x$. By the Angle Bisector Theorem, we have $\frac{6}{x}=\frac{8}{7}$, or $7\cdot6=8x$. Solving for $x$, we get $x=\frac{21}{4}$.
To find the longest and shortest possible lengths of the third side, we have to consider the extremes. The longest possible length occurs when the angle bisector coincides with the longer segment, so the third side has length $6+8=\boxed{\textbf{(A) } 7.4 \text{ cm}}$. The shortest possible length occurs when the angle bisector coincides with the shorter segment, so the third side has length $8-6=\boxed{\textbf{(B) } 5.8 \text{ cm}}$.