Question

2. Consider a polygon with n-vertices for some n > 3. We label the vertices as (a,b;), 1<j<n, as we move along the polygon counter-clockwise, so that its n-sides are line segments which connect (Q;,b;) and (@j+1,b;+1) for 1 <j<n, under the convention that (On+1, bu+1) = (01, b). (It may be helpful to consider the triangle case n = 3 first.) Show that the area enclosed by the polygon is 46;+1 – b;a;+1. j=1

Answer 1

Let the statement P(n): The area enclosed by a polygon with n vertices is given by
Σα;bj+1 – b;α+1)
where
(an+1, bn+1) = (0,61)
.

We have to prove P(n) for all natural numbers greater than or equal to 3.

For n=3, let us see the following diagram of the triangle

ΔΑ1 Α2 Α3

A(a,b) A,(a,,b,) A,(a,b)

are perpendicular to the x-axis.

A1C, A2D, 43 B

.

A1C = 61, A2D = 62, A3B = 63, DC = a1-a2, BD = 22-23, BC = a1-a3

The area of the trapezium
A1CDA
is
(A1C + A2D)DC = -(61 + b2) (a1 – 12)
.

The area of the trapezium
A2DBA
is
(A2D + 43B)BD = -(62 +63)(az – a3)
.

The area of the trapezium
A1CBA
is
(A1C + A3B)BC = 5(61 + b3)(a1 – a3)
.

From the diagram,

Area of
ΔΑ1 Α2 Α3 =
area of
A1CDA
+area of
A2DBA
-area of
A1CBA

= =(61 + b2)(a1 – a2)+

(62 +63) (a2 – a3) -

(61 +63)(a1 – 23)

(491+Cp – 1+fqºn) 31 = (€9ID — tqx + 2qen – &qen + 1960 – žqip) 4 =

So, P(3) holds true.

Let us suppose that P(m) is true for some .

m > 3)N

Then, for any polygon with m vertices, its area is
Σ α ;bj+1 – b;aj+1)
.

Now, let us consider a polygon with (m+1) vertices. Let us see the following diagram:

II III A (2.b) Am(am,bm) A (a,b) Am+1(am+1, bm+1) A,(a, b) Ą (a,b)

Then, the area of the polygon
A1-A2...Am Am+1
is the sum of the areas of polygon
A1 A₂... Am
and the area of the triangle
1.41 -Am-4m+1
.

Area of
A1 A2...Am Am+1 =
area of
A₂ A2... Amt
area of
1.41 -Am-4m+1

- Σαβ, 1 - 1, 4,50) + Σ αιθυμι - διαχει) j=0 i=1,2, (n+1)

where
(am+1,bm+1) = (a1,61)
in the first term but not in the second term, and  
(am+2, bm+2) = (a1,61)
in the 2nd term.
(“ი!» — 144p +“+ ში&p — &qZn + Inn – შიp) =

+(ajbm – ambı + ambm+1 – Am+1bm + am+16ı – aybm+1)

τη+1 Σα;bj+1 – baj+1) j=0

Thus, if P(m) is true, then P(m+1) is true.

Therefore, by principle of mathematical induction, for all , the area enclosed by a polygon with n vertices is given by

n > 3)N

Σα;bj+1 – b;α+1)
, where
(an+1, bn+1) = (0,61)
.