Let $P$ be the set of points inside the square, $Q$ be the set of points inside the triangle and $R$ be the set of points inside the circle. If the triangle and circle intersect each other and are contained in the square then,

P \cap Q \cap R \neq ϕ

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P \cup Q \cup R = R

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P \cup Q \cup R = P

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P \cup Q = R \cup P

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Solution

The correct options are
A $P \cap Q \cap R \neq ϕ$
C $P \cup Q \cup R = P$
D $P \cup Q = R \cup P$
$P =$ The set of points inside the square
$Q =$ The set of points inside the triangle
$R =$ The set of points inside the circle

Then
$P \cap Q \cap R \neq ϕ$ [Since the set $Q$ and $R$ intersect each others and are contained in the square.]

And $P \cup Q \cup R = P$ [Since the union of $P$ and $Q$ and $R$ is the set of all points inside the square.]

$P \cup Q = R \cup P$

$P = P$ [Since the set of points in the triangle and the circle are contained in the square.]

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Let P be the set of points inside the square, Q be the set of points inside the triangle and R be the set of points inside the circle. If the triangle and circle intersect each other and are contained in the square then,

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