Question

In a quadrilateral ABCD, $\angle {\text{A = 90}}^\circ $ and ${\text{AB = BC = CD = DA}}$, then ABCD is a?
${\text{A}}{\text{.}}$Parallelogram
${\text{B}}{\text{.}}$Rectangle
${\text{C}}{\text{.}}$Square
${\text{D}}{\text{.}}$None of these

Answer 1

Hint: We know that any quadrilateral having four sides equal is called a rhombus. So using this property we can solve this problem.

Complete step-by-step answer:
So it is given in question that all sides of quadrilateral are equal i.e. ${\text{AB = BC = CD = DA}}$
therefore we can call it a rhombus now.
However one of the properties of rhombus is that the measure of its angles can be anything other than 90 degree.
Since it is given in question that one angle of quadrilateral ABCD i.e. $\angle {\text{A = 90}}^\circ $
Therefore all remaining angles will automatically be 90 degrees as the sides of the quadrilateral are equal.
Hence the quadrilateral ABCD which we called rhombus above becomes a square now.
Then ABCD is a square.

Note: In this question first we considered that the given quadrilateral is a rhombus but as one of the angles of the quadrilateral is 90 degree therefore rest of the angles will also be 90 degree therefore a quadrilateral with all sides equal and whose all angles are 90 degree becomes a square.