Two geometric figures are similar if they have the same shape but not necessarily the same size. All congruent figures are similar, but not vice versa. Class 10 Triangles chapter focuses on the criteria for similarity of triangles and the Pythagoras Theorem, with rigorous proofs.
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Key Concepts
- 1.Similar Triangles: Triangle ABC ~ Triangle DEF if:
- 2.Corresponding angles are equal: angle A = angle D, angle B = angle E, angle C = angle F.
- 3.Corresponding sides are proportional: AB/DE = BC/EF = CA/FD.
Criteria for Similarity:
1. AAA (AA) Similarity: If two angles of one triangle equal two angles of another, the triangles are similar.
2. SSS Similarity: If all three pairs of corresponding sides are proportional, the triangles are similar.
3. SAS Similarity: If one pair of corresponding sides is proportional and the included angles are equal, the triangles are similar.
Basic Proportionality Theorem (BPT) / Thales Theorem:
If a line is drawn parallel to one side of a triangle intersecting the other two sides, then it divides those sides in the same ratio. If DE || BC in triangle ABC, then AD/DB = AE/EC.
Converse of BPT: If a line divides two sides of a triangle in the same ratio, it is parallel to the third side.
Areas of Similar Triangles:
If triangle ABC ~ triangle DEF, then:
Area(ABC)/Area(DEF) = (AB/DE)2 = (BC/EF)2 = (CA/FD)2.
Pythagoras Theorem: In a right-angled triangle, the square of the hypotenuse equals the sum of squares of the other two sides.
BC2 = AB2 + AC2 (if right angle at A).
Converse of Pythagoras Theorem: If in a triangle BC2 = AB2 + AC2, then the angle at A is 90 degrees.
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Worked Examples
In triangle ABC, DE || BC with D on AB and E on AC. AD = 3 cm, DB = 5 cm, AE = 6 cm. Find AC.
By BPT: AD/DB = AE/EC. So 3/5 = 6/EC. EC = 10 cm.
AC = AE + EC = 6 + 10 = 16 cm.
Triangle ABC ~ Triangle DEF. AB = 4 cm, BC = 6 cm, AC = 5 cm, DE = 8 cm. Find EF and DF.
Scale factor = DE/AB = 8/4 = 2.
EF = 2 x BC = 12 cm. DF = 2 x AC = 10 cm.
Prove that areas of similar triangles are in the ratio of the squares of corresponding sides.
This is a standard NCERT proof. If triangle ABC ~ triangle DEF, draw altitudes AM and DN.
Area(ABC)/Area(DEF) = (1/2 x BC x AM)/(1/2 x EF x DN) = (BC/EF) x (AM/DN).
From similarity, AM/DN = AB/DE = BC/EF. So ratio = (BC/EF)2.
The areas of two similar triangles are 25 cm2 and 64 cm2. If one side of the first is 5 cm, find the corresponding side of the second.
Area ratio = (5/x)2 = 25/64. So 5/x = 5/8. x = 8 cm.
In triangle PQR, right-angled at Q, PQ = 3 cm, QR = 4 cm. Find PR.
PR2 = PQ2 + QR2 = 9 + 16 = 25. PR = 5 cm.
A ladder 13 m long leans against a vertical wall. Its foot is 5 m from the wall. How high on the wall does the ladder reach?
Height2 = 132 - 52 = 169 - 25 = 144. Height = 12 m.
In triangle ABC ~ triangle DEF, AB/DE = 3/4. If Area(ABC) = 27 cm2, find Area(DEF).
Area ratio = (3/4)2 = 9/16. 27/Area(DEF) = 9/16. Area(DEF) = 27 x 16/9 = 48 cm2.
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Key Theorems
Key formulas
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Common mistakes
- Confusing similarity (same shape) with congruence (same shape AND size).
- In BPT, applying the ratio to non-corresponding parts of the divided sides.
- In Pythagoras, squaring the legs and adding (not multiplying or subtracting) to get hypotenuse squared.
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Summary
Triangles chapter centres on the power of proportionality in geometry. BPT and similarity criteria allow us to find unknown sides and angles, while the area ratio theorem and Pythagoras theorem are directly applicable in measurement problems. All theorems should be understood with their proofs for NCERT examinations.