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This was posted yesterday but went unanswered. How do you find the area of triangle ABC?

 

https://web2.0calc.com/questions/hard-geometry_7

 May 12, 2020
 #1
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Let AD  = h

 

Note that 

AC  = sqrt (h^2 + 2^2)  =  sqrt (h^2 + 4)

And

AB = sqrt( h^2 + 3^2)  = sqrt ( h^2 + 9)

 

Using the Law of Cosines we have that

 

BC^2  =  AC^2  + AB^2  - 2(AC * AB) cos (45°)

 

5^2  =  (h^2 + 4) + (h^2 + 9)  - 2 sqrt [ h^4 +13h^2  + 36] sqrt (2)/2

 

25   = 2 h^2  + 13  - sqrt [ h^4 + 13h^2 + 36]  sqrt (2)

 

12 = 2h^2 -  sqrt [ h^4 + 13h^2 + 36] sqrt (2)

 

12 - 2h^2  = - sqrt [ h^4 + 13h^2 +36] sqrt (2)          square both sides

 

4h^4 - 48h^2  + 144  =  (2)[h^4 + 13h^2 + 36]

 

4h^4  - 48h^2  + 144  = 2h^4 + 26h^2  + 72

 

2h^4 - 74h^2 + 72  =  0

 

h^4 - 37h  + 36 =  0     factor

 

(h^2  - 1) ( h^2 - 36)  =  0

 

(h + 1) ( h - 1) (h + 6 ) ( h - 6)  =  0

 

The possible  solutions are   h = - 1, h = 1 , h = - 6 , h  = 6

 

We can reject the negative answers

 

And  h  cannot  be  1  because    180  - arctan (1/3) - arctan (1/2)   = BAC = 180 - pi/4  = 135°

 

If h  =    6  =  AD  then 

 

180  - arctan (6/3)  - arctan (6/2)   = BAC   = 180 - 3pi/4  =  pi/ 4  = 45°

 

So....the  area  =  (1/2)  (BC) (AD) =  (1/2)(5)(6)  =  15 units^2

 

 

cool cool cool

 May 12, 2020

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