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
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