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

Hint: Here we will proceed by taking three triangles from which the pentagon is made. Then by using the sum of all interior angles of a triangle i.e. ${180^ \circ }$, we will add all the angles and we will get the required sum of the angles.

Complete step-by-step answer:
Firstly, we will take three triangles from the pentagon-
$
  \vartriangle ABE \\
  \vartriangle EBD \\
  \vartriangle DBC \\
 $
Now we will use the property of the triangle i.e. sum of all interior angles of a triangle is ${180^ \circ }$.
Since the pentagon has 3 triangles,
The sum of all its angles$ = 3 \times {180^ \circ } = {540^ \circ }$
  So $\angle $ EAB + $\angle $ ABC + $\angle $ BCD + $\angle $ CDE + $\angle $ DEA = ${540^ \circ }$
$\therefore $ Option B is correct.

Note: There are also other ways by which we can solve this problem as if we use angle sum property of pentagon i.e. S = $\left( {n - 2} \right){180^ \circ }$where n is the number of sides of the pentagon. Then we will put the value of n and get the answer as ${540^ \circ }$.