A regular polygon with exterior angle of 45° is redrawn using a scale 1 : 5. If the actual length of the sides of the regular polygon is 10cm, calculate the perimeter of the scale drawing of the regular polygon.​

16 cm

Step-by-step explanation:

"regular polygon" means all sides and angles are the same.

the sum of all exterior angles must be 360 degrees.

so, how often do 45 fit into 360 ?

2 times in 90, which in turn fit 4 times into 360, so 8 times.

the polygon has 8 corners and 8 sides.

each side is originally 10cm

so the original perimeter is 8×10 = 80 cm

now, every side is reduced to 1/5th of the original size, this also reduces the perimeter by the same factor.

=> perimeter = 80×(1/5) = 80/5 = 16 x cm

control :

applying the scaling factor of 1/5th to every side transforms every side from 10 cm to 2 cm.

we still have 8 sides.

perimeter = 8 × 2 = 16 cm

correct

there are lots of ways.   pick's theorem tells us if the vertices are lattice points the area a is

a = b/2 + i - 1

where b is the number of boundary lattice points and i is the number of interior lattice points.

in the figure i count b=16, i=50 so

a = 8 + 50 - 1 = 57

let's try the shoelace formula.   the area is half the sum of the cross products of the sides.  

a(-2,6)   -2(2)-6(-5) = 26

b(-5,2)   -5(-3) - 2(-2) = 19

c(-2,-3) -2(-3) - (-3)(4) = 18

d(4,-3)   4(3) - (-3)(5) = 27

e(5,3)   5(3) - 3(1) = 12

f(1,3)     1(6) - 3(-2) = 12

a(-2,6)

each of the cross products is twice the area of the triangle with the associated side and one vertex the origin. so the area of abo=26/2=13.

area = (1/2) (26 + 19 +   18 + 27 + 12 + 12) = 57

that checks!