Answer:
First polygon has 16 sides
Second polygon has 24 sides
Step-by-step explanation:
Let the number of sides for first polygon be a and that for the second polygon be b
formula for sum of interior angle is;
180n-360
So for individual interior angle we have;
(180n-360)/n
So for polygon A
(180a-360)/a
For polygon B
(180b-360)/b
Let’s divide both to equate to the ratio
(180a-360/a)/(180b-360)/b) = 21/22
Factor out 180
(a-2)/a * b/b-2 = 21/22
b(a-2)/a(b-2) = 21/22
22b(a-2) = 21a(b-2) •••••• (i)
For the second part
Sum of exterior angle of any polygon is 360
Individual exterior angle is 360/n
for the first polygon;
360/a ;
For the second 360/b
Thus;
b/a = 3/2
2b = 3a ••••••••(ii)
b = 3a/2
11(2b)(a-2) = 21a(b-2)
11(3a)(a-2) = 21a( 3a/2-2)
33a(a-2) = 21a(3a-4)/2
22a(a-2) = 7a(3a-4)
22a^2 - 44a= 21a^2-28a
22a^2-21a^2+28a - 44a = 0
a^2 + 28a - 44a = 0
a^2 -16a = 0
a(a-16) = 0
a = 0 (which cannot be)
or a-16 = 0
a = 16
Recall;
b = 3a/2
b = 3(16)/2
b = 24
