ABCD is a regular tetrahedron (right triangular pyramid) with side length 1. If M is the midpoint of line CD, then what is the area of triangle ABM?
+128460 The base of this triangle will be the altitude of an equilateral triangle with a side of 1 = sqrt (3)/2
Sides of triangle = two sides = sqrt (3)/2 + a side of 1
Semi-perimeter of triangle = [ 1 + 2 sqrt (3)/2 ] /2 = 1/2 + sqrt (3)/2 = S
Area = sqrt [ S ( S -1) ( S - sqrt (3)/2) ( S - sqrt (3)/2) ]
Area =
sqrt [ (1/2 + sqrt (3)/2) sqrt (3)/2 - 1/2) (1/2)^2 ] =
sqrt [ (-1/4 + 3/4 ) ( 1/4) ] =
(1/2)sqrt (1/2) =
(1/2) sqrt (2) /2 =
sqrt (2) / 4 units^2
