A Trapezoid PQRS has two parallel bases PQ and RS. The diagonals of the Trapezoid intersect at a point O. Suppose that PQ=10,SR=15, and the area of the triangle POS is 30. What is the area of the trapezoid PQRS?


Answer:

125

Step by Step Explanation:
  1. Given, a trapezoid PQRS,PQ=10,SR=15, and the area of triangle POS=30, where O is the point of intersection of the diagonals of the trapezium.
    Let's draw the given trapezium and a perpendicular line MN through the point O.
    P Q M N A O S R
  2. In triangle POQ and ROS
    POQ=ROS[ Vertically opposite angle ]QPO=SRO[ Interior alternate angles since PQ||SR]OQP=OSR[ Interior alternate angles since PQ||SR] POQROS
  3. Since the POQ is similar to the ROS
    PORO=OQOS=PQRS=OMON=23[PQRS=1015=23]
  4.  Since OMON=23OM=23×ON and OM+ON=MN23×ON+ON=MN53ON=MNON=35MN(1)
  5. Now, in PSR
    Area of PSR= Area of POS+ Area of SOR12×PA×SR=30+12×ON×SR   [ Given, area of triangle POS=30]12×PA×15=30+12×ON×15   [ Given, SR=15]12×MN×15=30+12×ON×15   [MN=PA height of the trapezium ]MN=4+ONMN=4+3MN5 [ From (1) ]MN=10
  6.  Area of the trapezium PQRS=12×(PQ+SR)×MN=12×(10+15)×10=125

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