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# help with geometry

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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 $BXA$.

Jan 11, 2024

#1
+289
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Well, we can see that this is a special right triangle, a 30-60-90 triangle, where m angle A = 30, m angle C = 60, and m angle B = 90.

The side lengths are in the ratio 1:sqrt3:2, which in our case is 12:12 x sqrt3:24

12 x sqrt 3 = 20.7846096908 or approx 20.8 as shown in the graph below:

ALERT:     Sorry, the top point is meant to be A, the bottom left one B, and the bottom right one C.

We can multiply 12x 20.7846 x 1/2 and get:

124.7076

or approximately

124.71

Jan 11, 2024
#2
+129690
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A

12sqrt 3          X

B                12                  C

Hypotenuse  = 12 * 2   = 24

Let AX =  x

Let CX =  24 -x

Since BX is an angle  bisector

AX / AB =  CX / AB

x / (12sqrt 3)  = (24 - x)  /12

x /sqrt 3  =  24 - x

x =  (24 - x) sqrt 3

x = 24sqrt 3  -  sqrt (3)x

x + sqrt 3  = 24 sqrt 3

x ( 1 + sqrt 3) =  24 sqrt (3)

x =  [ 24sqrt (3) ] / [ 1 + sqrt 3]

x = [24 sqrt 3 ] [1 - sqrt 3] / [ 1 -3]

x = -12 [ 1 - sqrt 3]   =  12 [ sqrt (3) - 1]  = AX

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

36 [sqrt (3) ] [ sqrt (3)   - 1]  =

36 ( 3 - sqrt 3 )   ≈ 45.65

Jan 12, 2024
edited by CPhill  Jan 12, 2024
edited by CPhill  Jan 12, 2024