Question

The ratio of number of sides of two regular polygons is 3:4 and the ratio of measures of their each interior angle is 8:9. What is the sum of the number of diagonals of both the polygons equal to?​

Answer 1

Step-by-step explanation:

The ratio of number of sides of two regular polygons is 3:4 and the ratio of measures of their each interior angle is 8:9. What is the sum of the number of diagonals of both the polygons equal to?

Ratio of sides of two regular polygons = 3 : 4

Let sides of first polygon = 3n

and sides of second polygon = 4n

Sum of interior angles of first polygon

= (2 × 3n – 4) × 90° = (6n – 4) × 90°

And sum of interior angle of second polygon

= (2 × 4n – 4) × 90° = (8n – 4) × 90°

∴ ((6n – 4) × 90°)/((8n – 4) × 90°) = 2/3

⇒ (6n – 4)/(8n – 4) = 2/3

⇒ 18n – 12 = 16n – 8

⇒ 18n – 16n = -8 + 12

⇒ 2n = 4

⇒ n = 2

∴ No. of sides of first polygon

= 3n = 3 × 2 = 6

And no. of sides of second polygon

= 4n = 4n × 2 = 8

Answer 2

Answer:

$29$

Step-by-step explanation:

Let regular Polygon A have 3x sides while regular Polygon B has 4x sides.

Each exterior angle of A = 360/3x or 120/x while that of B = 360/4x or 90/x

Each interior angle of A = 180-(120/x) or (180x-120)/x and that of B = 180-(90/x) or (180x-90)/x.

The ratio of interior angles of A to that of B = (180x-120)/x : (180x-90)/x = 8:9. Or

9(180x-120) = 8(180x-90)

1620x-1080 = 1440x-720, or

1620x-1440x = 1080–720, or

180x = 360, or

x = 2

Polygon A has 6 sides and Polygon has 8 sides.

Polygon A has 6(6–3)/2 = 9 diagonals while B has 8(8–3)/2 = 20.

The sum of the diagonals of Polygons A and B is 9+20 = 29.