Question

In the given figure, the sum of a,b,c,d,e and f is

• A.
  ${90}^{\circ }$
• B.
  ${180}^{\circ }$
• C.
  ${270}^{\circ }$
• D.
  ${360}^{\circ }$

Answer 1

The correct option is D

${360}^{\circ }$
$In\Delta ABC,\angle A+\angle B+\angle ACB={180}^{\circ }\text{[Sum of interior angles of a triangle is}{180}^{\circ }]\Rightarrow a+b+\angle ACB={180}^{\circ }\Rightarrow \angle ACB={180}^{\circ }-a-bNow,\angle ACB=\angle DCE\text{[Vertically opposite angles]}\Rightarrow \angle DCE={180}^{\circ }-a-b...(1)In\Delta GDF\angle GDF+c+d={180}^{\circ }\text{[Sum of interior angles of a triangle is}{180}^{\circ }]\Rightarrow \angle GDF={180}^{\circ }-c-d\Rightarrow \angle GDF={180}^{\circ }-c-d\Rightarrow \angle CDE=\angle GDF={180}^{\circ }-c-d...\left(2\right)\text{[Vertically opposite angles]}In\Delta HEI,\angle HEI+e+f={180}^{\circ }\text{[Sum of interior angles of a triangle is}{180}^{\circ }]\Rightarrow \angle HEI={180}^{\circ }-e-f\Rightarrow \angle DEC=\angle HEI={180}^{\circ }-e-fIn\Delta CDE,\angle CDE+\angle DCE+\angle CED={180}^{\circ }\Rightarrow ({180}^{\circ }-c-d)+({180}^{\circ }-a-b)+({180}^{\circ }-e-f)={180}^{\circ }\text{[using (1),(2)~and]~(3)}{540}^{\circ }-(a+b+c+d+e+f)={180}^{\circ }a+b+c+d+e+f={360}^{\circ }$

So, option d is correct.