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In triangle $ABC$, we have $\angle A = 30^\circ$, $\angle B = 60^\circ$, and $\angle C=90^\circ$. Point $X$ is on $\overline{AC}$ such that $\overline{BX}$ bisects $\angle ABC$. If $AB = 12$, then what is the area of triangle $ABX$?

 Feb 11, 2020
 #1
avatar+109740 
+1

See the following image   :

 

 

AB  is the hypotenuse of a 30-60-90 right triangle

Since BC  is opposite the 30° angle it  = (1/2)(12)   = 6

And since AC is opposite the 60° angle it = 6√3

The area of triangle ABC  = (1/2)(BC)(AC)  = (1/2)(6)(6√3)  = 18√3    (1)

 

Since BX  is an angle bisector of angle B,  then angle XBC  = 30°

So triangle XBC is also a 30-60-90 right triangle

And angle CXB  = 60°

And the side opposite this angle = BC  = 60

So.....the side opposite angle XBC  = 6/√3  =  CX 

 

So....the area of triangle XBC  = (1/2) (product of the legs) = (1/2) BC * CX  = (1/2) (6) (6/√3)  = 

18 / √ 3  =   6√3   (2)

 

So....the area of  triangle   ABX  = (1)  - (2)  = 18√3 - 6√3  =   12√3  units^2

 

 

cool cool cool

 Feb 12, 2020
 #2
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+1

Thanks for the help CPhill

 

Here's another challenge for you:

 

Find the area of triangle PQR if PQ = QR = 12 and

Guest Feb 12, 2020
 #4
avatar+531 
+1

Find the area of triangle PQR if PQ = QR = 12 and

It,s impossible to solve! Not enough information. cheeky

Dragan  Feb 12, 2020
 #3
avatar+531 
+1

In triangle $ABC$, we have $\angle A = 30^\circ$, $\angle B = 60^\circ$, and $\angle C=90^\circ$. Point $X$ is on $\overline{AC}$ such that $\overline{BX}$ bisects $\angle ABC$. If $AB = 12$, then what is the area of triangle $ABX$?

 

 Area of the triangle  ABX = ( AB/2) * [( AB/2) * tan(A)] 

                                    Area   ABX ≈ 20.785 u²  indecision

 Feb 12, 2020

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