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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?

 Dec 12, 2020
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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

 

 

 

cool cool cool

 Dec 12, 2020
edited by CPhill  Dec 13, 2020

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