A triangle has a 13-inch side, a 14-inch side, and a 15-inch side. To the nearest tenth
of an inch, how long is the median drawn to the 14-inch side?
2 answers
I just got the answer 14 is that right? I formed a parallelogram and added the vector lengths and multiplied it by 1/2.
Nope. You have ignored the triangle inequality.
From the law of cosines, the angle ? between the 13 and 15 sides is
14^2 = 13^2+15^2-2*13*15 cos?
? = 59.5°
So, if we let the short side be
u = 13i
The 15" side is
v = 7.62i + 12.92j
The median m is then
m = (u+v)/2 = (20.62i + 12.92j)/2
= 10.31i + 6.47j
|m| = 12.17
Or, use the handy formula found here:
http://www.cut-the-knot.org/triangle/LengthOfMedian.shtml
From the law of cosines, the angle ? between the 13 and 15 sides is
14^2 = 13^2+15^2-2*13*15 cos?
? = 59.5°
So, if we let the short side be
u = 13i
The 15" side is
v = 7.62i + 12.92j
The median m is then
m = (u+v)/2 = (20.62i + 12.92j)/2
= 10.31i + 6.47j
|m| = 12.17
Or, use the handy formula found here:
http://www.cut-the-knot.org/triangle/LengthOfMedian.shtml
1 answer