Question

A parallelogram circumscribing a circle is always a/an

• A. Square
• B.
  Rhombus
• C.
  Rectangles
• D.
  Isosceles trapezium

Answer 1

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.

$\therefore AB=CDandBC=AD(\because Oppositesidesofaparallelogramarealwaysequal)$

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

$\Rightarrow AP=AS....\left(i\right)\Rightarrow BP=BQ....\left(ii\right)\Rightarrow CR=CQ...\left(iii\right)\Rightarrow DR=DS...\left(iv\right)$

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

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

$\Rightarrow AB+CD=AD+BC\Rightarrow AB+AB=BC+BC(\because AB=CDandBC=AD)\Rightarrow AB=BC\Rightarrow AB=BC=CD=DA$

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

Hence the correct answer is option b.