1966 AHSME Problems/Problem 11

Problem

The sides of triangle $BAC$ are in the ratio $2:3:4$ . $BD$ is the angle-bisector drawn to the shortest side $AC$ , dividing it into segments $AD$ and $CD$ . If the length of $AC$ is $10$ , then the length of the longer segment of $AC$ is:

$\text{(A)} \ 3\frac{1}{2} \qquad \text{(B)} \ 5 \qquad \text{(C)} \ 5\frac{5}{7} \qquad \text{(D)} \ 6 \qquad \text{(E)} \ 7\frac{1}{2}$

Solution

1966 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 10	Followed by Problem 12
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

[[Category:]] The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

Page

Toolbox

Search

1966 AHSME Problems/Problem 11

Problem

Solution

See also