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?^@


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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 ^@
    ^@\begin{align} & \angle POQ = \angle ROS && [\text{ Vertically opposite angle }] \\ & \angle QPO = \angle SRO && [ \text{ Interior alternate angles since } PQ || SR ] \\ & \angle OQP = \angle OSR && [ \text{ Interior alternate angles since } PQ || SR ] \\ \therefore \space & \triangle POQ \sim \triangle ROS \end{align}^@
  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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