Question

AB & CD are two equal chords of a circle with centre O which intersect each other at the right angle at point P. If the perpendiculars from the centre to AB and CD meet them at M and N respectively, then MONP is a _______.

• A. square
• B. rectangle
• C. kite
• D. rhombus

Answer 1

The correct option is A

square

Two equal chords in a circle are equidistant from the center. Therefore OM=ON.............(i)

Join OP. In $△$ ONP and $△$OMP

OM = ON,

OP = OP (common arm)

$\angle$ONP = $\angle$OMP = ${90}^o$ (perpendicular from centre)

Hence the two triangles are congruent (by RHS )

$\Rightarrow$ ON = MP and ON = PN (cpct)............(ii)

From (i) and (ii) we get ON = OM = PN = PM (all four sides are equal)

Moreover all angles are ${90}^0$ . (Given that $\angle P=\angle M=\angle N={90}^o)$
$\Rightarrow \angle O={90}^o$

Therefore the figure OMNP forms a square as all sides are equal and angles are ${90}^0$.