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In PQR, we have P=30°, Q=60°, R=90°, Point X is on PR such that QX bisects PQR. If PQ equals 12, then what is the area of PQX?

 Nov 12, 2015

Best Answer 

 #1
avatar+104712 
+10

 

PR = 12*sin60 = 12 * [sqrt(3)/2]  =  6 * sqrt(3)

 

And we can find XR thusly :

 

XR/ sin30  = 6/ sin60

 

XR   = 6sin30/ sin60

 

XR  = 6(1/2) / [sqrt(3) / 2]

 

XR  = 3*2/ sqrt(3) = 6/sqrt(3)

 

So PX  =  PR - XR =   6*sqrt(3) - 6/sqrt(3) =  [18 - 6] / sqrt(3)  =  12/sqrt(3)

 

And the area of PQX   =  (1/2) *PX* RQ  = (1/2) (12/sqrt(3)) * (6)   =  36 / sqrt(3)  square units

 

Here's a pic :  

 

 

cool cool cool

 Nov 12, 2015
edited by CPhill  Nov 12, 2015
edited by CPhill  Nov 13, 2015
 #1
avatar+104712 
+10
Best Answer

 

PR = 12*sin60 = 12 * [sqrt(3)/2]  =  6 * sqrt(3)

 

And we can find XR thusly :

 

XR/ sin30  = 6/ sin60

 

XR   = 6sin30/ sin60

 

XR  = 6(1/2) / [sqrt(3) / 2]

 

XR  = 3*2/ sqrt(3) = 6/sqrt(3)

 

So PX  =  PR - XR =   6*sqrt(3) - 6/sqrt(3) =  [18 - 6] / sqrt(3)  =  12/sqrt(3)

 

And the area of PQX   =  (1/2) *PX* RQ  = (1/2) (12/sqrt(3)) * (6)   =  36 / sqrt(3)  square units

 

Here's a pic :  

 

 

cool cool cool

CPhill Nov 12, 2015
edited by CPhill  Nov 12, 2015
edited by CPhill  Nov 13, 2015

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