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FInd the area of the triangle with sides sqrt(13), sqrt(61), sqrt(80).

 Dec 27, 2020
 #1
avatar+114325 
0

Using the Law of Cosines

 

80 - 13 - 61   = -2(sqrt (793)  cos x

 

6 / [ -2 sqrt (793) ]   = cos x =   -3 /sqrt (793)

 

sin x   = sqrt  [ 1 - (-3/ sqrt (793))^2  ]   =  sqrt [ 1  - 9/793 ] =  sqrt [ 784/793  ]  = 28/ sqrt (793)

 

Area  =  (1/2) ( 13) (61) sin x   = 

 

(1/2) (793) (28/ sqrt (793) )   =

 

14sqrt (793)  units^2

 

 

cool cool cool

 Dec 27, 2020
 #2
avatar+859 
+3

Find the area of the triangle with sides sqrt(13), sqrt(61), sqrt(80).

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The triangle area, using Heron's formula, is 14 square units. smiley

 Dec 27, 2020

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