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1966 AHSME Problems/Problem 11

Revision as of 20:56, 14 September 2014 by Timneh (talk | contribs) (Created page with "== Problem == The sides of triangle <math>BAC</math> are in the ratio <math>2:3:4</math>. <math>BD</math> is the angle-bisector drawn to the shortest side <math>AC</math>, dividi...")
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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

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