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]
  3. Since the \triangle POQ is similar to the \triangle ROS
    \begin{align} & \therefore \dfrac{ PO }{ RO } = \dfrac{ OQ }{ OS } = \dfrac{ PQ }{ RS } = \dfrac{ OM }{ ON } = \dfrac{ 2 }{ 3 } && \left[ \because \dfrac{ PQ }{ RS } = \dfrac{ 10 }{ 15 } = \dfrac{ 2 }{ 3 } \right] \end{align}
  4. \begin{align} & \text{ Since } \dfrac{ OM }{ ON } = \dfrac{ 2 }{ 3 } \\ & \therefore OM = \dfrac{ 2 }{ 3 } \times ON \\ & \text{ and } OM + ON = MN \\ & \implies \dfrac{ 2 }{ 3 } \times ON + ON = MN \\ & \implies \dfrac{ 5 }{ 3 } ON = MN \\ & \implies ON = \dfrac{ 3 }{ 5 } MN && \ldots (1) \end{align}
  5. Now, in \triangle PSR
    \begin{align} & \text{Area of } \triangle PSR = \text{ Area of } \triangle POS + \text{ Area of } \triangle SOR \\ \implies & \dfrac{ 1 }{ 2 } \times PA \times SR = 30 + \dfrac{ 1 }{ 2 } \times ON \times SR \space\space\space [\text{ Given, area of triangle } POS = 30 ] \\ \implies & \dfrac{ 1 }{ 2 } \times PA \times 15 = 30 + \dfrac{ 1 }{ 2 } \times ON \times 15 \space\space\space [ \text{ Given, } SR = 15 ] \\ \implies & \dfrac{ 1 }{ 2 } \times MN \times 15 = 30 + \dfrac{ 1 }{ 2 } \times ON \times 15 \space\space\space [ MN = PA \text{ height of the trapezium }] \\ \implies & MN = 4 + ON \\ \implies & MN = 4 + \dfrac{ 3 MN } { 5 } \space \left[ \text{ From (1) } \right] \\ \implies & MN = 10 \end{align}
  6. \begin{align} \text{ Area of the trapezium } PQRS & = \dfrac{ 1 }{ 2 } \times (PQ + SR) \times MN \\ & = \dfrac{ 1 }{ 2 } \times ( 10 + 15 ) \times 10 \\ & = 125 \end{align}

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