+0  
 
+1
65
1
avatar

Find the ratio of the area of \(\triangle BCX\) to the area of \(\triangle ACX\) in the diagram if CX bisects \(\angle ACB\). Express your answer as a common fraction.

 Dec 14, 2018
 #1
avatar+98034 
+1

Let AX = M

Let BX = 24 - M

 

Since  ACB is bisected, we have this relationship

 

M / 30 = (24 - M) / 27     cross-multiply

 

27M =  30( 24 - M)     simplify

 

27M = 720 - 30M     add 30M to both side

 

57M = 720

 

M =  720 / 57   =   240 / 19   = AX

 

Then     24 - M   =    24 - 240/19  =   216/19  =   BX

 

Since each triangle is under the same height, the ratio of their areas is just the ratio of their bases

 

So

 

area BCX / Area ACX  =  ( 216 /19) / (240 / 19)   =  216 /240  = 9 / 10

 

 

cool cool cool

 Dec 14, 2018

29 Online Users

avatar
avatar
avatar
avatar
avatar