+0  
 
0
2
2
avatar+867 

In triangle $ABC$, point $D$ is on side $\overline{AC}$ such that line segment $\overline{BD}$ bisects $\angle ABC$. If $\angle A = 45^\circ$, $\angle C = 45^\circ$, and $AC = 12$, then find the area of triangle $ABD$.

 Jun 22, 2024

Best Answer 

 #2
avatar+129725 
+1

A45

 

 

12/sqrt 2                   D    12

 

 

B 90           12/sqrt 2                     C 45

 

By AAS, triangle ABD  congruent to triangle CBD

 

[ ABC]  = (1/2)(12/sqrt 2)^2  = (1/2)(144/2)  = 144/4 =  36

 

[ABD]  = (1/2)[ABC]  = 18

 

cool cool cool

 Jun 22, 2024
 #1
avatar+1710 
0

The area of triangle ABD is 40*sqrt(2) + 12.

 Jun 22, 2024
 #2
avatar+129725 
+1
Best Answer

A45

 

 

12/sqrt 2                   D    12

 

 

B 90           12/sqrt 2                     C 45

 

By AAS, triangle ABD  congruent to triangle CBD

 

[ ABC]  = (1/2)(12/sqrt 2)^2  = (1/2)(144/2)  = 144/4 =  36

 

[ABD]  = (1/2)[ABC]  = 18

 

cool cool cool

CPhill Jun 22, 2024

1 Online Users

avatar