Problem:
In right â–³ABC 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. 513​
C. 3
D. 4
E. 524​
Solution:
Since â–³ABC is a right triangle, AB=13. Also, BN=BC=5 and AM=AC=12. Thus
​BM=AB−AM=1MN=BN−BM=4​
