Question

The sides of a quadrilateral are produced in order. What is the sum of the four exterior angles? $\left( {in^\circ } \right)$

Answer 1

Hint: To solve this problem we must know a few properties of a polygon:
I. Sum of a pair of an exterior and an interior angle is $180^\circ $ .
II. Sum of all the interior angles of a polygon of n sides $ = \left( {n - 2} \right) \times 180^\circ $ .

Complete step-by-step answer:
We know that the sum of an exterior and an interior angle is $180^\circ $.
So sum of all pairs of exterior and interior angles of a quadrilateral $ = 4 \times 180^\circ = 720^\circ $
We also know that sum of all interior angles of a polygon of n sides $ = \left( {n - 2} \right) \times 180^\circ $
So the sum of all interior angles of a quadrilateral $ = \left( {4 - 2} \right) \times 180^\circ = 360^\circ $
Now,
Sum of all pairs of exterior angles = Sum of all pairs of exterior and interior angles - sum of all interior angles
Putting values, we get
Sum of all pairs of exterior angles = $720^\circ - 360^\circ = 360^\circ $
Hence, the sum of all pairs of a quadrilateral is $360^\circ $ .
So, the correct answer is “ $360^\circ $”.

Note: Since sum of all the exterior angles of a polygon of n sides $ = 180^\circ \times n - \left( {n - 2} \right) \times 180^\circ = 180^\circ \times n - 180^\circ \times n + 360^\circ $ $ = 360^\circ $ . The Sum of exterior angles of any polygon is always $360^\circ $. It does not depend on the number of sides of the polygon.