In triangle ABC, angle C is a right angle. Point M is the midpoint of AB, point N is the midpoint of AC, and point O is the midpoint of AM. The perimeter of triangle ABC is 112 and ON = 15. What is the area of triangle ABC?
+23245 Since N is the midpoint of AC and O is the midpoint of BC, NO = ½AB.
Since M is the midpoint of AB, MB = NO = 15.
Therefore, AB = 2·MB = 30.
Since the perimeter of the triangle is 112 and AB = 30, AC + BC = 82.
Since AC2 + BC2 = AB2, AC2 + BC2 = 302 = 900
Since AC2 + BC2 = 900 ---> AC2 + 2·AC·BC + BC2 = 900 + 2·AC·BC
---> (AC + BC)2 = 2·AC·BC
Since AC + BC = 52, ---> (82)2 = 2·AC·BC
6724 = 2·AC·BC
Dividing by 4, 1681 = ½·AC·BC
But, ½·AC·BC is the area of triangle(ABC).
AB is NOT equal to 30!!! AB is the hypotenuse of triangle ABC.
btw, point O is the midpoint of AM.
+118609 Thanks Gino,
You have done a lot of work there but unfortunately, you did label O incorrectly.
\(15^2=m^2+t^2-2mt\cdot cos\alpha\\ Cos\alpha = \frac{2t}{4m}=\frac{t}{2m}\\ so\\ 225=m^2+t^2-2mt \cdot \frac{t}{2m}\\ 225=m^2\\ m=15 \)
This means that the hypotenuse is 60
and the sum of the other 2 sides is 112-60 =52
As guest has already told us very loudly 
, this is impossible.
So such a triangle cannot exist.
