In the given figre, $P Q R S$ is a quadrilateral. $Q T$ is drawn parallel to $P R$ and $Q T$ meets $S R$ produced at $T$
Prove that $a r (P Q R S) = a r (δ P S T)$

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Solution

Given :- $P R ∥ P Q T \Rightarrow P Q T R$ is a trapezium.
we know that a diagonal divides trapezium into two eqal areas .
$\Rightarrow a r (Δ P Q T) = a r (Δ P R T) = o r (P Q R) = a r (P T R) = \frac{1}{2} [a r (P Q T R)]$ .....(i)
HS,
$a r (P Q R S) \Rightarrow a r (Δ P Q R) + a r (Δ P S R)$
$\Rightarrow a r (Δ P T R) + a r (Δ P S R)$ .....[from(i)]
$\Rightarrow a r (Δ P S T)$
$= R H S$
$∴ a r (P Q R S) = a r (Δ P S T)$

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In the given figre, PQRS is a quadrilateral. QT is drawn parallel to PR and QT meets SR produced at TProve that ar(PQRS)=ar(δPST)

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In the given figre, $P Q R S$ is a quadrilateral. $Q T$ is drawn parallel to $P R$ and $Q T$ meets $S R$ produced at $T$
Prove that $a r (P Q R S) = a r (δ P S T)$