Multiple Choice

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,

A$$ P \cap Q \cap R \neq \phi $$
Correct Answer
B$$ P \cup Q \cup R = R$$
C$$ P \cup Q \cup R = P$$
Correct Answer
D$$P \cup Q = R \cup P$$
Correct Answer

Solution

$$P=$$The set of points inside the square
$$Q=$$The set of points inside the triangle
$$R=$$The set of points inside the circle

Then
$$P\cap Q\cap R \neq \phi $$ [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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