“Similarity in Geometry” (সদৃশতা) is a foundational topic in Class 10 West Bengal Board mathematics. This chapter covers criteria of similarity for triangles, the properties of similar triangles, and their applications in solving problems. In this overview, we explore these ideas in an easy‑to‑understand, empathetic tone — helping learners feel confident about similarity concepts and exam preparations.
- What Is Similarity? (সদৃশতা কী?)
Similarity refers to figures that have the same shape but may differ in size. In triangles, two triangles are similar if their corresponding angles are equal and their corresponding sides are proportional.
Keywords: similarity, similar triangles, proportional sides, equal angles, Class 10 WB Board.
Example:
Consider ∆ABC and ∆DEF. They are similar if
- ∠A = ∠D, ∠B = ∠E, ∠C = ∠F
- AB/DE = BC/EF = AC/DF
- Criteria for Triangle Similarity (তুলনা ও সদৃশতা নির্ধারণ)
The syllabus emphasizes three standard criteria:
2.1 AA Criterion (Angle–Angle)
If two angles of one triangle are equal to two angles of another triangle, the triangles are similar.
Example:
In ∆PQR and ∆XYZ, if ∠P = ∠X and ∠Q = ∠Y, then ∆PQR ∼ ∆XYZ.
2.2 SAS Criterion (Side–Angle–Side)
If one angle of a triangle is equal to one angle of another triangle, and the sides including these angles are in proportion, then the triangles are similar.
Example:
If in ∆GHI and ∆JKL, GH/JK = HI/KL and ∠H = ∠K, then ∆GHI ∼ ∆JKL.
2.3 SSS Criterion (Side–Side–Side)
If the corresponding sides of two triangles are proportional, then those triangles are similar.
Example:
In ∆MNO and ∆PQR, if MN/PQ = NO/QR = MO/PR, then ∆MNO ∼ ∆PQR.
- Properties of Similar Triangles (সদৃশ ত্রিভুজের গুণাবলী)
When triangles are similar:
- Corresponding angles are equal.
- Corresponding sides are proportional.
- The ratio of areas of two similar triangles = square of the ratio of their corresponding sides.
Example:
If ∆ABC ∼ ∆DEF, and AB/DE = k, then area(∆ABC) / area(∆DEF) = k².
- Applications of Similarity (সদৃশতার প্রয়োগ)
4.1 Finding Unknown Lengths
Similarity helps solve for unknown sides.
Example:
In ∆LMN and ∆PQR, if ∠M = ∠Q and LM/PQ = MN/QR, and you know LM = 6 cm, PQ = 9 cm, MN = 8 cm. Then similarity gives MN/QR = 6/9 = 2/3, so QR = (8 × 3) / 2 = 12 cm.
4.2 Solving Word Problems via Triangles
Often in real‑life geometry problems — such as shadow lengths, indirect measurement (like height of a tree) — you set up similar triangles to compute heights or distances.
Example:
A pole casts a 3 m shadow when a person of height 1.5 m casts a 0.75 m shadow. Triangles formed by height and shadow are similar.
Ratio = 1.5/0.75 = 3/1.5 = 2 → height of pole/shadow of pole = 2. If the pole’s shadow measures 6 m, pole’s height = 2 × 6 = 12 m.
4.3 Area and Perimeter Relationships
Since corresponding sides are proportional, perimeters are in the same ratio. Areas scale with the square of the ratio.
Example:
If ∆ABC ∼ ∆DEF and ratio of sides AB/DE = 4/3, then ratio of perimeters = 4/3 and ratio of areas = (4/3)² = 16/9.
- Sample Problems (উদাহরণমূলক সমস্যা)
Problem 1
Given: ∠X = ∠A, ∠Y = ∠B, and XY = 6 cm, AB = 9 cm, YZ = 8 cm.
Find: ZB if ∆XYZ ∼ ∆ABC.
Solution:
Similarity ratio XY/AB = 6/9 = 2/3. So YZ/BC = 2/3 → 8/BC = 2/3 → BC = (8 × 3) / 2 = 12 cm. So ZB = BC = 12 cm.
Problem 2
Given: In ∆PQR, PQ = 5 cm, QR = 7 cm, PR = 8 cm. In ∆LMN, LM = 10 cm, MN = 14 cm. If ∆PQR ∼ ∆LMN, find LN.
Solution:
Side scale factor = LM/PQ = 10/5 = 2 or MN/QR = 14/7 = 2. So LN = PR × 2 = 8 × 2 = 16 cm.
- Summary (সংক্ষিপ্তসার)
- Similarity (সদৃশতা) means same shape, possibly different size.
- Triangles are similar by AA, SAS, or SSS.
- Corresponding angles equal; sides in proportion.
- Applications include solving for unknown sides, indirect measurement, and area/perimeter relations.
- Real‑world problems, especially shadows and heights, rely on this.
Syllabus‑aligned, clear, and supportive for learners aiming at WB Board excellence.