Question

plss find the number of sides of each polygon..

plss do not spam and give irrespective answers....



HINT: it is a question from rectilinear figures.​

Answer 1

Sum of interior angles regular polygons = (n−2)∗180

sides of two regular polygons= x and y

So, x/y=3/4

x=3y/4

Sum of interior angles regular polygons (x)=(x−2)∗180

Sum of interior angles regular polygons (y)=(y−2)∗180

(x−2)∗180/(y−2)∗180=2/3

x−2/y−2=2/3

3y/4−2/y−2=2/3

3y−8/y−2=8/3

9y−24=8y−16

y=8

x=3y/4=3∗8/4=6

Answer 2

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Question :-

The ratio between the number of sides of two regular polygons is 3: 4 and the ratio between the sum of their interior angles is 2 : 3. Find the number of sides in each polygon.

Solution :-

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