Question

$\square ABCD$ is a rhombus. If it is inscribed in $ \odot (0,r)$, then $\square ABCD$ is
A. Square
B. Rectangle
C. Trapezium
D. None of the above

Answer 1

Hint: We will first draw the required figure and then divide the rhombus in two triangles by drawing a diagonal. We will then prove these triangles congruent and use angles equality to prove that one of the angle is ${90^ \circ }$

Complete step-by-step answer:
Let us draw the required figure:-

               

In the figure given below O is the center of the circle and ABCD is the rhombus inside it. And we have joined A and C to form two triangles, ABC and ACD.
Now, consider $\vartriangle ABC$ and $\vartriangle ADC$:-
AB = AD (sides of a rhombus are equal)
BC = DC (sides of a rhombus are equal)
AC = AC (Common)
Hence, we have $\vartriangle ABC \cong \vartriangle ADC$ (By SSS property)
SSS stands for "side, side, side" and means that we have two triangles with all three sides equal. If three sides of one triangle are equal to three sides of another triangle, the triangles are congruent.
Therefore, $\angle BAC = \angle DAC$ ……….(1)
$\angle BCA = \angle DCA$ ………(2) (By CPCT)
CPCT stands for Corresponding parts of congruent triangles. CPCT theorem states that if two or more triangles which are congruent to each other are taken then the corresponding angles and the sides of the triangles are also congruent to each other.
Since, we know that $\angle A + \angle C = {180^ \circ }$ because opposite angles of a rhombus are supplementary.
We can rewrite it as:-
$\angle BAC + \angle DAC + \angle BCA + \angle DCA = {180^ \circ }$
Now, using (1) and (2), we can rewrite it as:-
$2\angle BAC + 2\angle BCA = {180^ \circ }$
$ \Rightarrow 2(\angle BAC + \angle BCA) = {180^ \circ }$
$ \Rightarrow \angle BAC + \angle BCA = {90^ \circ }$ ………….(3)
Now, consider the $\vartriangle ABC$:-
$ \Rightarrow \angle BAC + \angle BCA + \angle B = {180^ \circ }$ (Sum of interior angles of a triangle)
Now, using (3), we will get:-
$\angle B = {90^ \circ }$
Hence, since all the sides of the quadrilateral are equal and one of the angles is ${90^ \circ }$, hence, we have a square.

So, the correct answer is “Option A”.

Note: The students must notice that if initially they are not given a rhombus but a parallelogram, then they will reach to the conclusion of a rectangle.
You must note that “Every square is a rhombus, but every rhombus is not a square” and “Every rectangle is a parallelogram but every parallelogram is not a rectangle”.