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In triangle ABC, point D lies on side AC such that line segment BD bisects angle ABC. Angle A measures 30 degrees, angle C measures 90 degrees, and the length of side AC is 12 units. The task is to determine the area of triangle ABD.

 Apr 30, 2024
 #1
avatar+128772 
+1

      A 30

 

12  D

 

 

      C 90                          B

 

Angle  B  = 60

So angle BCD =  30

Side  BC =   12/sqrt 3  =  4sqrt 3

So...in triangle  BDC,, DC  =  BC / sqrt 3 =  4sqrt 3 /sqrt 3 =  4

 

Area of BDC = (1/2) BC * DC =  (1/2) (4sqrt 3) (4)  =  8 sqrt 3

 

Area of ABC  =  (1/2) (AC)(BC)  = (1/2)(12)(4sqrt 3)  =  24sqrt 3

 

So....  [ ABD ]  =   [ ABC ] - [ BDC ] =  24sqrt 3 - 8sqrt 3 =  16 sqrt 3

 

CORRECTED 

 

cool cool cool

 Apr 30, 2024
edited by CPhill  Apr 30, 2024
 #2
avatar+44 
0

That is wrong...

rtsdyhfdts  Apr 30, 2024
 #3
avatar+128772 
0

Ooops!!!....I found the area of CBD !!!

 

See my corrected answer....

 

 

cool cool cool

CPhill  Apr 30, 2024

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