MCQ
In Figure. the value of $\angle\text{A}+\angle\text{B}+\angle\text{C}+\angle\text{D}+\angle\text{E}+\angle\text{F} \ \text{is}:$
- A$190^\circ $
- B$540^\circ$
- ✓$360^\circ$
- D$180^\circ $
✓
Answer
Correct option: C.
$360^\circ$
As we know, sum of all the interior angles of a triangle is $180^\circ $.
$\text{In} \ \triangle\text{ABC}, \ \angle\text{A}+\angle\text{B}+\angle\text{C}=180^{\circ}$ $[\text{interior angles of }\triangle\text{ABC}].....(\text{i})$
$\text{In} \ \triangle\text{DEF}, \ \angle\text{D}+\angle\text{E}+\angle\text{F}=180^{\circ}$ $[\text{interior angles of }\triangle\text{DEF}].....(\text{ii})$
On adding Equation. $(i)$ and $(ii)$, we get.
$\angle\text{A}+\angle\text{B}+\angle\text{C}+\angle\text{D}+\angle\text{E}+\angle\text{F}\\$$=180^{\circ}+180^{\circ}$
$=360^{\circ}$
$\text{In} \ \triangle\text{ABC}, \ \angle\text{A}+\angle\text{B}+\angle\text{C}=180^{\circ}$ $[\text{interior angles of }\triangle\text{ABC}].....(\text{i})$
$\text{In} \ \triangle\text{DEF}, \ \angle\text{D}+\angle\text{E}+\angle\text{F}=180^{\circ}$ $[\text{interior angles of }\triangle\text{DEF}].....(\text{ii})$
On adding Equation. $(i)$ and $(ii)$, we get.
$\angle\text{A}+\angle\text{B}+\angle\text{C}+\angle\text{D}+\angle\text{E}+\angle\text{F}\\$$=180^{\circ}+180^{\circ}$
$=360^{\circ}$
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