If two sides of the quadrilateral are produced in order the sum of four exterior angles so formed

THEOREM TITLE:
(Exterior angle sum property) If the sides of a quadrilateral are produced in order, the sum of four exterior angles so formed is 360º.

Proof: Let the sides of a quadrilateral be produced in order as shown in figure, forming exterior angles ∠1, ∠2, ∠3 and ∠4. Since ∠1 and ∠ form a linear pair and the sum of the angles of a linear pair is 180º. ∴∠1 + ∠A = 180º ----(i) Similarly, we have ∠2 + ∠B = 180º ----(ii) ∠3 + ∠C = 180º ----(iii) and, ∠4 + ∠D = 180º ----(iv)

→ ∠1 + ∠2 + ∠3 + ∠4 = 720º - 360º = 360º

THEOREM TITLE:

The sum of all the exterior angles formed by producing the sides of a convex polygon in the same order is equal to four right angles.
Given: A convex polygon P1 P2 P3 P4 P5.Its sides P1 P2,P2P3,P3P4, P5P1are produced in order, forming exterior angles ∠1, ∠2, ∠3, ∠4 and ∠5.
To Prove: ∠1 + ∠2 + ∠3 + ∠4 + ∠5 = 4 right angles.
Construction: Take any point O, outside the polygon. Draw OA1,OA2,OA3,OA4, and OA5 parallel to and in the same sense as P1P2,P3P4,P4 P5 and P4 P5, and P5 P1 respectively.

Proof: Since the arms of ∠ and ∠a are parallel and drawn in the same sense. ∴ ∠1 = ∠a Similarly,∠2 = ∠b, ∠3 = ∠c,∠4 = ∠d and ∠5 = ∠e ∴ ∠1 + ∠2 + ∠3 + ∠4 + ∠5 = ∠a + ∠b + ∠c + ∠d + ∠e = 360º [Sum of the angles at a point is 360º] = 4 x 90º

= 4 right angles.

➲Each exterior angle of a regular polygon of n sides is equal to ➲If there is a regular polygon of n sides (n ≥ 3), then its each interior angle is equal to right angles i.e.,

Answer

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The sides of a quadrilateral are produced in order. What is the sum of the four exterior angles?

Solution

Let $ABCD$ be a quadrilateral such that exterior angles formed by extending the sides $AD,AB,BC$ and $CD$ are angles $x,y,z$ and $w$.

Now, since, sides $AD,AB,BC$ and $CD$ are straight lines, therefore,

$\Rightarrow \angle BAD+x={180}^{\circ }\Rightarrow \angle BAD={180}^{\circ }-x$...(i) [Linear Pair].

Similarly,

$\Rightarrow \angle ABC={180}^{\circ }-y$ ...(ii)

$\Rightarrow \angle BCD={180}^{\circ }-z$ ...(iii)

$\Rightarrow \angle ADB={180}^{\circ }-w$ ...(iv)

Adding (i), (ii), (iii) and (iv), we get,

$\Rightarrow \angle BAD+\angle ABC+\angle BCD+\angle ADB=({180}^{\circ }-x)+({180}^{\circ }-y)+({180}^{\circ }-z)+({180}^{\circ }-w)$

$\Rightarrow {360}^{\circ }={720}^{\circ }-(x+y+z+w)$ [Sum of all the angles of a quadrilateral is ${360}^{\circ }$]

$\Rightarrow x+y+z+w={360}^{\circ }$

Thus, the sum of the four exterior angles is ${360}^{\circ }$.

OR

It is well established that regardless of number of exterior angles, the sum of all the exterior angles of a polygon is always ${360}^{\circ }$.

Mathematics

RD Sharma

Standard VIII