Question

ABCD is a parallelogram. DK is perpendicular to BA produced at K and DM is perpendicular to BC produced at M. Find the measure of $\angle KDM.$

• A. ${200}^{\circ }$
• B. ${150}^{\circ }$
• C. ${90}^{\circ }$
• D. ${140}^{\circ }$

Answer 1

The correct option is B ${150}^{\circ }$

ABCD is a parallelogram with $\angle ADC={30}^{\circ }$

$DK\perp AK$ and $DM\perp BM$

$AK||CD$

$\therefore \angle KDC+\angle DKB={180}^{\circ }$ [co-interior angles]

$\Rightarrow \angle KDC+{90}^{\circ }={180}^{\circ }$

$\Rightarrow \angle KDC={90}^{\circ }$

$AD||BM$

$\therefore \angle ADM+\angle DMB={180}^{\circ }$ [co-interior angles]

$\Rightarrow \angle ADM+{90}^{\circ }={180}^{\circ }$

$\Rightarrow \angle ADM={90}^{\circ }$

Now, $\angle ADM=\angle ADC+\angle CDM$

$\Rightarrow {90}^{\circ }={30}^{\circ }+\angle CDM$

$\Rightarrow \angle CDM={60}^{\circ }$

$\angle KDM=\angle KDC+\angle CDM={90}^{\circ }+{60}^{\circ }\Rightarrow \angle KDM={150}^{\circ }$