निর্ণয় কর। Payel wants to draw three regular polygons on the three sides of an equilateral triangle. If there is one common side between each of the two polygons, then find the number of sides of each polygon drawn?

To determine the number of sides each regular polygon must have, let's break down the problem step by step. ### Given: - An **equilateral triangle** with three sides. - **Three regular polygons** are to be drawn, each on one side of the triangle. - Each polygon shares one side with the triangle and one side with each of its adjacent polygons. ### Steps to Solve: 1. **Understand the Angles:** - An **equilateral triangle** has internal angles of **60 degrees** each. - Let each regular polygon have **n sides**. The internal angle \(\theta\) of a regular polygon is given by: \[ \theta = \frac{(n-2) \times 180^\circ}{n} \] 2. **Angles at Each Vertex:** - At each vertex of the triangle, two polygons meet. - The sum of the angles around a point is **360 degrees**. - Therefore, the sum of the triangle's angle and the two adjacent polygon angles at each vertex must equal 360 degrees: \[ 60^\circ + 2\theta = 360^\circ \] - Solving for \(\theta\): \[ 2\theta = 300^\circ \implies \theta = 150^\circ \] 3. **Determine the Number of Sides (n):** - Substitute \(\theta = 150^\circ\) into the internal angle formula: \[ 150^\circ = \frac{(n-2) \times 180^\circ}{n} \] - Multiply both sides by \(n\): \[ 150n = 180n - 360 \] - Rearrange and solve for \(n\): \[ 30n = 360 \implies n = 12 \] ### Conclusion: Each regular polygon drawn on the sides of the equilateral triangle must be a **dodecagon (12-sided polygon)**. **Answer:** Each polygon must have twelve sides.