Question

Diagonals bisect each other at ${90}^{\circ }$.

• A. rhombus
• B. quadrilateral
• C. circle
• D. trapezium

Answer 1

Answer: (A)

rhombus

Quadrilateral is a four sided polygon and sum of the four angles of a quadrilateral is ${360}^o$.

It can have square, rectangle, parallelogram, and all polygons of four sides.

Circle do not have diagonals.

Trapezium can have angles of any measure and only two opposite sides are parallel. So, the diagonals may or may not bisect each other at ${90}^o$.

Let $ABCD$ be any rhombus.

Name the point of intersection of diagonals as $O$

$AD\parallel BC$ and $AB\parallel CD$ ...... (Opposite sides)

$\angle A=\angle C$ and $\angle B=\angle D$ ......... (Opposite angles)

Also, $m\left(\angle A\right)+m\left(\angle B\right)={180}^o$ ......... $\left(1\right)$

In $△ABO$,

$\angle OAB+\angle ABO+\angle AOB={180}^o$ ......... $\left(2\right)$

$\angle OAB=\frac{\angle A}{2}$ and $\angle ABO=\frac{\angle B}{2}$

From $\left(2\right)$,

$\frac{\angle A}{2}+\frac{\angle B}{2}+\angle AOB={180}^o$

$⟹{90}^o+\angle AOB={180}^o$ ........... (From $\left(1\right)$)

$⟹\angle AOB={90}^o$

Similarly, $\angle AOD=\angle DOC=\angle BOC={90}^o$

$\therefore$ Diagonals bisect each other at ${90}^o$.

Hence, diagonals of rhombus bisect each other at ${90}^o$.