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Suppose that there is a right triangle $\triangle ABC$ such that the legs $AB$ and $BC$ have lengths $4\sqrt{2}$ and $5,$ respectively. What is the length of the median $BM$?

 Mar 10, 2024
 #1
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A

 

B               C

 

AC =  sqrt [ (4sqrt 2)^2 + 5^2  ] =  sqrt  [ (sqrt (32))^2 + 5^2 ] =  sqrt (57)

cos (ACB)  =  5/sqrt (57)

 

Law of Cosines

 

BM^2   = (AC / 2)^2  + BC^2  - 2 ( AC/2 * BC) (5/ sqrt 57)

 

BM^2  = AC^2 /  4  + BC^2  - (AC*BC) ( 5/sqrt 57)

 

BM^2  =   57 / 4  + 25 - (sqrt (57) * 5) (5  /sqrt 57)

 

BM^2  = 57/4 + 25 - 25

 

BM  =  sqrt (57)  /  2 =  AC / 2

 

cool cool cool

 Mar 10, 2024

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