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In triangle $ABC$, $\angle A = 30^\circ$ and $\angle B = 90^\circ$. Point $X$ is on side $\overline{AC}$ such that line segment $\overline{BX}$ bisects $\angle ABC$. If $BC = 12$, then find the area of triangle $BXA$.

 Mar 24, 2024
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avatar+128772 
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  30    A

 

                     X

 

90     B       12             C

 

Angle B = 60

AB =  12sqrt 3

AC  = 24

Since BX is a bisector...let  XC  = 24-m   and AX = m

And we have that

m / AB  = (24-m) / BC

m / (sqrt 3)  = (24 -m)

m = (24 - m) sqrt 3

m = 24sqrt 3 - msqrt 3

m+ msqrt 3 = 24sqrt 3

m ( 1 + sqrt 3)  =24sqrt 3

m = 24sqrt 3 / ( 1 + sqrt 3)  =  12sqrt3 ( sqrt 3 - 1)  = 12 [ 3 - sqrt 3 ]

 

[ BXA ]  = (1/2) AX * AB * sin 30°  = (1/4) AX * AB  =  12 [3 -sqrt 3 ] * [12 sqrt 3] / 4 =

 

3 [ 3 -sqrt 3 ] [ 12sqrt 3 ]  =  108 (sqrt 3 - 1)

 

cool cool cool

 Mar 25, 2024

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