Question

Question 22

In the given figure, the value of
$\angle A+\angle B+\angle C+\angle D+\angle E+\angle F$ is

a) ${190}^{\circ }$
b) ${540}^{\circ }$
c) ${360}^{\circ }$
d) ${180}^{\circ }$

Answer 1

As we know, sum of all the interior angles of a triangle is ${180}^{\circ }$
In $\Delta ABC,\angle A+\angle B+\angle C={180}^{\circ }$ [interior angles of $\Delta ABC$] ..... (i)
In $\Delta DEF,\angle D+\angle E+\angle F={180}^{\circ }$ [interior angles of $\Delta DEF$] ..... (ii)
On adding Eqs. (i) and (ii), we get
$\angle A+\angle B+\angle C+\angle D+\angle E+\angle F={180}^{\circ }+{180}^{\circ }={360}^{\circ }$