A parallelogram circumscribing a circle is always a/an

Square

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Rhombus

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Rectangles

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Isosceles trapezium

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Solution

The correct option is B
Rhombus
Let ABCD be a parallelogram circumscribing a circle with centre O and touching at P, Q, R and S.

$∴ A B = C D a n d B C = A D (∵ O p p o s i t e s i d e s o f a p a r a l l e l o g r a m a r e a l w a y s e q u a l)$

Since, the lengths of tangents drawn from an external point are equal.

$\Rightarrow A P = A S . . . . (i) \Rightarrow B P = B Q . . . . (i i) \Rightarrow C R = C Q . . . (i i i) \Rightarrow D R = D S . . . (i v)$

Adding (i), (ii), (iii) and (iv), we get

AP + BP + CR + DR = AS + BQ + CQ + DS

$\Rightarrow A B + C D = A D + B C \Rightarrow A B + A B = B C + B C (∵ A B = C D a n d B C = A D) \Rightarrow A B = B C \Rightarrow A B = B C = C D = D A$

Therefore, ABCD is a rhombus.
Thus, a parallelogram circumscribing a circle is always a rhombus.

Hence the correct answer is option b.

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