MCQ
In Fig , $\angle A+\angle B+\angle C+\angle D+\angle E+\angle F=$
- A$180^{\circ}$
- B$360^{\circ}$
- C$540^{\circ}$
- D$90^{\circ}$
✓
Answer
B. $360^{\circ}$
Using angle sum property in $\triangle^{\prime} s A B C$ and $D E F$, we obtain$
\begin{array}{ll}
& \angle A+\angle B+\angle C=180^{\circ} \text { and } \angle D+\angle E+\angle F=180^{\circ} \\
\Rightarrow \quad & \angle A+\angle B+\angle C+\angle D+\angle E+\angle F=180^{\circ}+180^{\circ}=360^{\circ}
\end{array}
$
Using angle sum property in $\triangle^{\prime} s A B C$ and $D E F$, we obtain$
\begin{array}{ll}
& \angle A+\angle B+\angle C=180^{\circ} \text { and } \angle D+\angle E+\angle F=180^{\circ} \\
\Rightarrow \quad & \angle A+\angle B+\angle C+\angle D+\angle E+\angle F=180^{\circ}+180^{\circ}=360^{\circ}
\end{array}
$
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