Question

Match List – I with List – II and select the correct answer using the codes given below the lists:

$\begin{array}{cc}\text{Properties}& \text{Quadrilaterals}\\ \text{In a parallelogram ABCD}& \\ \text{If}\angle D={115}^{\circ }\text{,then the measure of}\angle A\text{=}& {68}^{\circ }\\ \text{PQRS is a square such that PR and SQ intersect at O}& \\ \text{then the measure of}\angle POQ=& {58}^{\circ }\\ \text{The diagonals of a rectangle ABCD meet at O. If}\angle BOC={44}^{\circ }\text{,}& \\ \text{then the measure of}\angle OAD\text{=}& {90}^{\circ }\\ \text{If ABCD is a rectangle with}\angle BAC={32}^{\circ }\text{, then the measure of}& \\ \angle DBC\text{is =}& {65}^{\circ }\end{array}$
Choose the correct option:

• A. $\angle A={58}^{\circ },\angle POQ={65}^{\circ },\angle OAD={90}^{\circ },\angle DBC={58}^{\circ }$
• B. $\angle A={65}^{\circ },\angle POQ={68}^{\circ },\angle OAD={90}^{\circ },\angle DBC={58}^{\circ }$
• C. $\angle A={65}^{\circ },\angle POQ={90}^{\circ },\angle OAD={68}^{\circ },\angle DBC={58}^{\circ }$
• D. $\angle A={65}^{\circ },\angle POQ={90}^{\circ },\angle OAD={58}^{\circ },\angle DBC={68}^{\circ }$

Answer 1

The correct option is C $\angle A={65}^{\circ },\angle POQ={90}^{\circ },\angle OAD={68}^{\circ },\angle DBC={58}^{\circ }$

$\angle A={180}^{\circ }-{115}^{\circ }={65}^{\circ }$---- (Adjacent angles of a parallelogram are supplimentary)

$\angle POQ={90}^{\circ }$ -----(Diagonals of a square bisect each other at right angles)

$\angle AOD={44}^{\circ }$ ------ (They are vertically opposite angles.
$\angle OAD$ and $\angle ODA$ are equal as diagonals are equal and bisect each other in an rectangle.
$\therefore In△AOD$

$\angle OAD+\angle ODA+\angle AOD={180}^o$
(Sum of angles in a triangle is ${180}^{\circ }$)
$\angle OAD=\angle ODA=\frac{{180}^o-{44}^o}{2}={68}^o$

$\angle DBC={58}^{\circ }$ ------- ($\angle DAB$ and $\angle CBA$ are supplementary)