Question

1. The sides of a triangle have lengths of 5, 8, and 6.(1/ 2). An angle bisector meets the side of length 6.(1/ 2). Find the lengths x and y.

Answer 1

Answer:

Step-by-step explanation:

i assume x+y=6(1/2)=\frac{13}{2}

$\frac{5}{8}=\frac{x}{y}=\frac{x}{\frac{13}{2}-x}=\frac{2x}{13-2x}\\65-10x=16x \\26 x=65\\2x=5\\x=\frac{5}{2}\\y=\frac{13}{2}-\frac{5}{2}=\frac{8}{2}=4$