Question

If all line segments are straight, in the given figure, then the sum of the angles at the corners marked in the diagram is?

• A. ${360}^o$
• B. ${450}^o$
• C. ${540}^o$
• D. ${630}^o$

Answer 1

The correct option is A ${360}^o$

We have to find the $\angle A+\angle B+\angle C+\angle D+\angle E+\angle F+\angle G=?$

$\Rightarrow$ We have sum of interier angles of triangle ${180}^{\circ }$

$\therefore$ from $\Delta AHI,\Delta BIJ,\Delta CJK,\Delta DLK,\Delta ELH,\Delta FMN,\Delta GNH$ adding all angles of the given triangles , we have,

or $\angle A+\angle AHI+\angle AIH+\angle B+\angle BIJ+\angle BJI+\angle C+\angle CJK+\angle CKJ$$+\angle D+\angle DKL+\angle DLK+\angle E+\angle ELH+\angle EML+\angle F+\angle FMN+\angle FNM$$+\angle G+\angle GNH+\angle GHN=180+180+180+180+180+180+180$

or, $\angle A+\angle B+\angle C+\angle D+\angle E+\angle F+\angle G=7\times 180-\left[\right(\angle AIH+\angle BIJ)+(\angle BJI+\angle CJK)$$+(\angle CKJ+\angle DKL)+(\angle DLK+\angle ELM)+(\angle EML+\angle FMN)$$+(\angle FNM+\angle GNH)+(\angle GHN+\angle AHI)]$

or, $\angle A+\angle B+\angle C+\angle D+\angle E+\angle F+\angle G=(7\times {180}^{\circ })-\left[14\right(180{)}^{\circ }-2($ sum of angles of polygon $HIJKLMNH\left)\right]$

also, sum of interior angle angle of polygon $=(n-2){180}^{\circ }$

where $n=$ sides of polygon

$\therefore \angle A+\angle B+\angle C+\angle D+\angle E+\angle F+\angle G=7\times 180-14\times 180+2\times 5\times {180}^{\circ }$

$=17\times 180-14\times 180$

$=3\times {180}^{\circ }$

$={540}^{\circ }$