Two polygons of the same number of sides are similar, if (i) their corresponding angles are equal and (ii) their corresponding sides are in the same ratio (i.e., proportion).
Step-by-Step Solution
Key Concept: Similarity of polygons (including triangles) is established when: 1) each pair of corresponding interior angles are equal, and 2) the lengths of corresponding sides are proportional, i.e., the ratio of any pair of corresponding sides is the same constant (scale factor).
1. Identify the polygons – Let the two polygons be \(P_1\) and \(P_2\) having the same number of sides (say \(n\)).
2. Correspondence of vertices – Choose a vertex ordering such that the \(i^{th}\) vertex of \(P_1\) corresponds to the \(i^{th}\) vertex of \(P_2\) for \(i=1,2,\dots,n\).
3. Check angle equality – Verify that the interior angle at each vertex of \(P_1\) equals the interior angle at the corresponding vertex of \(P_2\):
$$\angle A_i = \angle B_i \quad \text{for all } i=1,2,\dots,n.$$
4. Check side proportion – Compute the lengths of the sides of both polygons. If the ratio of any pair of corresponding sides is a constant \(k\) (called the scale factor), then all corresponding sides satisfy:
$$\frac{A_iA_{i+1}}{B_iB_{i+1}} = k \quad \text{for all } i=1,2,\dots,n,$$
where \(A_{n+1}=A_1\) and \(B_{n+1}=B_1\).
5. Conclude similarity – If both conditions (equal corresponding angles and equal ratio of corresponding sides) are satisfied, the polygons are similar. Otherwise, they are not similar.
6. Example (optional) – Consider two triangles \(\triangle ABC\) and \(\triangle PQR\) with \(\angle A = \angle P, \angle B = \angle Q, \angle C = \angle R\) and side lengths satisfying \(\frac{AB}{PQ}=\frac{BC}{QR}=\frac{CA}{RP}=2\). Hence, the triangles are similar with scale factor 2.
Correct Answer: The statement is correct. Two polygons with the same number of sides are similar if and only if (i) their corresponding interior angles are equal, and (ii) the lengths of their corresponding sides are in the same ratio (i.e., proportional).