+1439 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 $BXA$.
+290 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
Answer: 124.71
+129895 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
