Question

In the figure, if $\angle A+\angle B+\angle C+\angle D+\angle E+\angle F=k\times$ right angle, then $k$ is :

• A. $2$
• B. $3$
• C. $4$
• D. $5$

Answer 1

The correct option is C $4$

$\Rightarrow$ In $△ABC$

$\Rightarrow$ $\angle A+\angle B+\angle C={180}^{\circ }$ [Sum of interior angles of triangle is ${180}^{\circ }$] --- ( 1 )

$\Rightarrow$ In $△DEF$

$\Rightarrow$ $\angle D+\angle E+\angle F={180}^{\circ }$ [Sum of interior angles of triangle is ${180}^{\circ }$] ---( 2 )

$\Rightarrow$ $\angle A+\angle B+\angle C+\angle D+\angle E+\angle F={180}^{\circ }+{180}^{\circ }$ [Adding ( 1 ) and ( 2 )]

$\Rightarrow$ $\angle A+\angle B+\angle C+\angle D+\angle E+\angle F={360}^{\circ }$

$\Rightarrow$ $\angle A+\angle B+\angle C+\angle D+\angle E+\angle F=k\times \left(rightangle\right)$.

$\therefore$ $k=\frac{{360}^o}{{90}^o}$

$\therefore$ $k=4$