Question

Square ABPQ and ADRS are drawn on the sides AB and AD of a parallelogram ABCD. Which of the following options is correct?

• A. $\angle SAQ=\angle ABC$
• B. $\angle SRD=\angle ADC$
• C. $\angle SAD=\angle ABC$
• D. None of these

Answer 1

The correct option is A

$\angle SAQ=\angle ABC$

$\angle DAB+\angle ABC={180}^{\circ }$

($\because$ interior angles on the transversal AB which cuts parallel lines DA and CB)

$\angle ABC=180-\angle DAB\dots \dots \left(1\right)$

$\angle SAD={90}^{\circ }\dots \dots \left(2\right)$ ($\because$ SRDA is a square)

$\angle QAB={90}^{\circ }\dots \dots \left(3\right)$ ($\because$ ABPQ is a square)

$\angle SAD+\angle DAB+\angle QAB+\angle SAQ={360}^{\circ }$

($\because$ these are the sum of the total angles created by two intersecting lines)

${90}^{\circ }+\angle DAB+{90}^{\circ }+\angle SAQ={360}^{\circ }$

$\angle DAB+\angle SAQ={180}^{\circ }\dots \dots \left(4\right)$

$\angle SAQ={180}^{\circ }-\angle DAB\dots \dots \left(5\right)$

By (1) and (5)

$\angle SAQ=\angle ABC\dots \dots \left(6\right)$