Question

Match the two columns.

$\begin{array}{cc}Column-I& Column-II\\ A)InaparallelogramABCD& P){68}^{\circ }\\ If\angle D={115}^{\circ },thenthemeasure& \\ of\angle Ais& \\ B)PQRSisasquaresuchthatPR& Q){58}^{\circ }\\ AndSQintersectatO.The& \\ Measureof\angle POQis& \\ C)Thediagonalsofarectangle& R){90}^{\circ }\\ ABCDmeetatO.If\angle BOC={44}^{\circ },& \\ Thenthemeasureof\angle OADis& \\ D)IfABCDisarectanglewith& S){65}^{\circ }\\ \angle BAC={32}^{\circ },thenthemeasureof& \\ \angle DBCis& \end{array}$

Choose the correct option:

• A. $\left(A\right)\to p;\left(B\right)\to q;\left(C\right)\to r;\left(D\right)\to s$
• B. $\left(A\right)\to s;\left(B\right)\to r;\left(C\right)\to q;\left(D\right)\to p$
• C. $\left(A\right)\to s;\left(B\right)\to r;\left(C\right)\to p;\left(D\right)\to q$
• D. $\left(A\right)\to r;\left(B\right)\to s;\left(C\right)\to q;\left(D\right)\to p$

Answer 1

The correct option is C

$\left(A\right)\to s;\left(B\right)\to r;\left(C\right)\to p;\left(D\right)\to q$

$\left(A\right)\to s;\left(B\right)\to r;\left(C\right)\to p;\left(D\right)\to q$

(A) $\angle A={180}^{\circ }-{115}^{\circ }={65}^{\circ }$

(B) $\angle POQ={90}^{\circ }$ Diagonals of a square intersect making an angle of ${90}^{\circ }$

(C) $\angle OAD={68}^{\circ }\angle AOD={44}^{\circ }$ because they are V.O.A. $\angle OAD$ and $\angle ODA$ are equal. Sum of angles in a triangle is ${180}^{\circ }$

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