Question

In the given five sided closed figure, find $\angle EAB+\angle ABC+\angle BCD+\angle CDE+\angle DEA$.

• A. ${520}^{\circ }$
• B. ${540}^{\circ }$
• C. ${530}^{\circ }$
• D. ${140}^{\circ }$

Answer 1

The correct option is B ${540}^{\circ }$

In $△ABC,$

$\Rightarrow$ $\angle ABC+\angle BCA+\angle CAB={180}^o$ [ Sum of angles of triangle is ${180}^o.$ ] ---( 1 )

In $△CDA,$

$\Rightarrow$ $\angle CDA+\angle DAC+\angle ACD={180}^o$ [ Sum of angles of triangle is ${180}^o.$ ] ---- ( 2 )

In $△DEA,$

$\Rightarrow$ $\angle DEA+\angle EAD+\angle ADE={180}^o$ [ Sum of angles of triangle is ${180}^o.$ ] --- ( 3 )

$\Rightarrow$ Pentagon $ABCDE=△ABC+△CDA+△DEA$

From ( 1 ), ( 2 ) and ( 3 )

$\Rightarrow$ Pentagon $ABCDE=\angle ABC+\angle BCA+\angle CAB+\angle CDA+\angle DAC+\angle ACD+\angle DEA+\angle EAD+\angle ADE$

$={180}^o+{180}^o+{180}^o$

$={540}^o$ ---- ( 4 )

$\Rightarrow$ $\angle EAB+\angle ABC+\angle BCD+\angle CDE+\angle DEA=Pentagon\left(ABCDE\right)$

From ( 4 ),

$\therefore$ $\angle EAB+\angle ABC+\angle BCD+\angle CDE+\angle DEA={540}^o.$