+0  
 
0
477
4
avatar+195 

In isosceles right triangle $ABC$, shown here, $AC=BC$. Point $X$ is on side $BC$ such that $CX=6$ and $XB=12$, and $Y$ is on side $AB$ such that $\overline{XY}\perp\overline{AB}$. What is the ratio of the area of triangle $BXY$ to the area of triangle $ABC$?

 Feb 28, 2021
 #1
avatar
0

[BXY]/[ABC] = 1/27

 Mar 1, 2021
 #2
avatar+195 
+1

I actually got 2/9😎

QuestionMachine  Mar 1, 2021
 #3
avatar
+1

Where was my head? cheeky You are absolutely right !!!     [BXY]/[ABC] = 36/162 = 2/9

Guest Mar 1, 2021
 #4
avatar+129852 
+1

Note that  triangle  BYX  is similar  to triangle BCA

 

AC  =  BC  =   12 +  6   = 18

 

BA  =  sqrt  (18^2  + 18^2)  = sqrt (2 *18^2)  =   18 sqrt (2) 

 

Scale  factor  of   BYX  to  BCA  =  12/  (18 sqrt (2))  =  2/ (3 sqrt (2))  =  sqrt (2)  / 3

 

Ratio  of  area of BYX  to  BCA =   (scale  factor)^2  =   2/9

 

Just  as  QM  found  !!!! 

 

 

cool cool cool

 Mar 1, 2021

0 Online Users