Math, asked by harshm1607, 4 months ago

The ratio of number two sides of two regular polygons is 3 : 4 and the ratio of the sum of their interior angles is 1 : 2. Find the number of sides of each polygon.

Answers

Answered by khashrul
0

Answer:

The number of sides of the Polygons are 3 and 4.

Step-by-step explanation:

The ratio of number of sides of two regular polygons is 3:4

Let's assume that the proportionality constant is n

∴ The number of sides of the polygons are 3n and 4n.

Now, from one vertex of n sided polygon, we can draw (n-3) diagonals.  This is because the two adjacent vertices form adjacent sides and not diagonals.  So, from the n number of vertices, one is the vertex itself from where we are drawing the diagonals, and we have just discussed about the two adjacent vertices.

With n - 3 diagonals from one vertex, we will get (n - 2) triangles dividing all the interior angles of the n sided polygon amongst themselves.

Therefore, the sum of the interior angles of n sided regular polygon is (n - 2)x180°

Therefore, the sum of the interior angles of 3n sided regular polygon is (3n - 2)x180°

And, the sum of the interior angles of 4n sided regular polygon is (4n - 2)x180°

According to the problem:

\frac{(3n -2).180}{(4n -2).180} = \frac{1}{2}

=> 6n - 4 = 4n - 2

n = 1

3n = 3  and 4n = 4

∴ the number of sides of the Polygons are 3 and 4.

Answered by BrainlySamrat
5

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

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