Problem:
In right β³ABC with legs 5 and 12, arcs of circles are drawn, one with center A and radius 12, the other with center B and radius 5. They intersect the hypotenuse in M and N. Then MN has length
Answer Choices:
A. 2
B. 513β
C. 3
D. 4
E. 524β
Solution:
Since β³ABC is a right triangle, AB=13. Also, BN=BC=5 and AM=AC=12. Thus
βBM=ABβAM=1MN=BNβBM=4β