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In trapezoid PQRS, Base PQ is parallel to base RS. Let point X  be the intersection of diagonals PR and QS. The area of triangle PQX is 20 and the area of triangle RSX is 48.  Find the area of trapezoid PQRS

 Jun 15, 2022
 #1
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Note that triangles   PXQ  and   RXS  are similar

 

The scale factor of the sides of  PXQ to RXS =  sqrt (20  / 48) = sqrt ( 5 /12)

 

And note that sin  angle   PXS   = sin angle QXR  = sin angle RXS

 

So  the area of triangle   PXS =  (1/2) ( sqrt (5/12) XR) ( XS)  *sin (angle RXS)  =

 

sqrt (5/12) * (1/2) (XR) (XS)  sin (angle RXS) =  sqrt (5/12) [ area of triangle RXS ]

 

Similarly

 

Area of triangle  QXR  =  (1/2) (sqrt (5/12) XS) ( XR) * sin ( RXS)  =

 

sqrt (5/12)  * (1/2) ( XS) *XR) * sin (angle RXS)  =

 

sqrt (5/12)  [ area of triangle RXS ]

 

So....the total area =   20 + 48  +  2 sqrt (5/12) area (RXS)  =

 

20 + 48 +  2 sqrt (5/12) ( 48)   ≈  129.97 units ^2 

 

 

cool cool cool

 Jun 16, 2022

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