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We are given O = (0,0), A = (5,2) and B = (6, -8). Find [OAB], the area of triangle OAB.

 Nov 25, 2020
 #1
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We are given O = (0,0), A = (5,2) and B = (6, -8). Find [OAB], the area of triangle OAB.

 

Use Pythagorean theorem to find sides of a triangle OAB

 

Use Heron's formula to calculate the area of Δ OAB

 Nov 26, 2020
 #2
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We don't have  a right triangle here, so the Pythagorean Theorem is no good to us

 

OA =   sqrt ( 5^2 + 2^2)  = sqrt  (29)

 

OB  = sqrt ( 6^2  + 8^2)  = sqrt (100)  = 10

 

AB = sqrt [ ( 6 - 5)^2  + (-8-2)^2  ]  =  sqrt [ 1^2  + 10^2 ]  = sqrt (101)

 

Using the Law of Cosines , we have

 

[ sqrt (101)]^2  - 10^2  - [sqrt (29)]^2

______________________________   = cos (AOB)

  -2 (10)  sqrt (29)

 

101 - 100 - 29

____________  = cos (AOB)

-20sqrt (29)

 

-28

_________  =  cos (AOB)

-20sqrt (29)

 

7 / ( 5 sqrt (29) )  = cos (AOB) 

 

arccos  [  (7/(5sqrt (29)  ] ≈  74.9°

 

[ AOB ]  =  (1/2) (10) (sqrt (29)_  sin ( 74.9°)      =    26 units^2

 

 

cool cool cool

 Nov 26, 2020
 #3
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You said: "We don't have  a right triangle here, so the Pythagorean Theorem is no good to us."

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

We do have the right triangles here and they're outside the triangle OAB.

 

OA = sqrt(22 + 52) = √29

 

BO = sqrt(62 + 82) = √100

 

AB = sqrt(102 + 12) = √101

 Nov 29, 2020

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