Class 10 Mathematics Notes Chapter 6 (Chapter 6) – Examplar Problems (English) Book

Alright class, let's focus on Chapter 6: Triangles from your NCERT Exemplar book. This is a crucial chapter, not just for your board exams, but also forms the foundation for many geometry questions in government exams. Pay close attention as we break down the key concepts.

Chapter 6: Triangles - Detailed Notes for Competitive Exams

1. Congruence vs. Similarity:

• Congruent Figures: Figures having the same shape AND the same size. (Symbol: ≅)
• Similar Figures: Figures having the same shape but NOT necessarily the same size. (Symbol: ~)
• Key Point: All congruent figures are similar, but the converse is not necessarily true. Think of two different-sized photographs of the same person – they are similar but not congruent.

2. Similarity of Triangles:

Two triangles are similar if:
(i) Their corresponding angles are equal. (∠A = ∠P, ∠B = ∠Q, ∠C = ∠R)
(ii) Their corresponding sides are in the same ratio (or proportion). (AB/PQ = BC/QR = CA/RP)

If ∆ABC ~ ∆PQR, then all these conditions hold.

3. Criteria for Similarity of Triangles:

You don't need to check all angles and all sides. These criteria are shortcuts:

• (i) AAA (Angle-Angle-Angle) Similarity Criterion: If in two triangles, corresponding angles are equal, then the triangles are similar.

  • Practical Use (AA Similarity): If two angles of one triangle are respectively equal to two angles of another triangle, then the third angles are also equal (due to angle sum property), and hence the triangles are similar. This is frequently used.
  • Example: If in ∆ABC and ∆DEF, ∠A = ∠D and ∠B = ∠E, then ∆ABC ~ ∆DEF.

• (ii) SSS (Side-Side-Side) Similarity Criterion: If the corresponding sides of two triangles are in the same ratio, then their corresponding angles are equal, and hence the triangles are similar.

  • Example: If in ∆ABC and ∆PQR, AB/PQ = BC/QR = CA/RP, then ∆ABC ~ ∆PQR.

• (iii) SAS (Side-Angle-Side) Similarity Criterion: If one angle of a triangle is equal to one angle of another triangle and the sides including these angles are proportional, then the triangles are similar.

  • Example: If in ∆ABC and ∆DEF, ∠A = ∠D and AB/DE = AC/DF, then ∆ABC ~ ∆DEF. (Note: The equal angle must be between the proportional sides).

4. Important Results based on Similarity:

If two triangles are similar (say ∆ABC ~ ∆PQR):

• Ratio of corresponding sides: AB/PQ = BC/QR = AC/PR = k (Similarity Ratio)
• Ratio of Perimeters: Perimeter(∆ABC) / Perimeter(∆PQR) = AB/PQ = k
• Ratio of corresponding Altitudes: (Altitude from A) / (Altitude from P) = AB/PQ = k
• Ratio of corresponding Medians: (Median from A) / (Median from P) = AB/PQ = k
• Ratio of corresponding Angle Bisector Segments: (Angle bisector from A) / (Angle bisector from P) = AB/PQ = k

5. Basic Proportionality Theorem (BPT) or Thales Theorem:

• Statement: If a line is drawn parallel to one side of a triangle intersecting the other two sides in distinct points, then the other two sides are divided in the same ratio.
• In ∆ABC: If DE || BC (D on AB, E on AC), then:
  • AD/DB = AE/EC
• Corollaries (Very Useful):
  • AD/AB = AE/AC
  • DB/AB = EC/AC

6. Converse of Basic Proportionality Theorem:

• Statement: If a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side.
• In ∆ABC: If AD/DB = AE/EC, then DE || BC.

7. Areas of Similar Triangles:

• Theorem: The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.
• If ∆ABC ~ ∆PQR, then:
  • Area(∆ABC) / Area(∆PQR) = (AB/PQ)² = (BC/QR)² = (CA/RP)²
• Also: The ratio of areas is equal to the square of the ratio of corresponding altitudes, medians, or angle bisector segments.
  • Area(∆ABC) / Area(∆PQR) = (Altitude₁/Altitude₂)² = (Median₁/Median₂)²

8. Pythagoras Theorem:

• Statement: In a right-angled triangle, the square of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides (legs).
• In right ∆ABC, right-angled at B: AC² = AB² + BC²

9. Converse of Pythagoras Theorem:

• Statement: In a triangle, if the square of one side is equal to the sum of the squares of the other two sides, then the angle opposite the first side is a right angle.
• In ∆ABC: If AC² = AB² + BC², then ∠B = 90°.

10. Important Results in Right-Angled Triangles:

• If a perpendicular (BD) is drawn from the vertex of the right angle (B) of a right triangle (∆ABC) to the hypotenuse (AC):
  • ∆ADB ~ ∆BDC
  • ∆ADB ~ ∆ABC
  • ∆BDC ~ ∆ABC
  • Key Results:
    • BD² = AD × DC (Geometric Mean Theorem)
    • AB² = AD × AC
    • BC² = CD × AC

Exam Tips:

• Visualize: Always draw a diagram if one isn't provided.
• Identify the Theorem: Read the problem carefully to see which concept applies – Similarity criteria, BPT, Area theorem, or Pythagoras.
• Correct Correspondence: When writing similarity (e.g., ∆ABC ~ ∆PQR), ensure the vertices correspond correctly (A↔P, B↔Q, C↔R). This is crucial for setting up correct ratios.
• Pythagorean Triplets: Memorize common triplets (3,4,5), (5,12,13), (8,15,17), (7,24,25) and their multiples to save time in calculations.
• Area Ratios: Remember the 'square' relationship for areas of similar triangles. Don't confuse it with the linear ratio of sides/perimeters/altitudes.

Multiple Choice Questions (MCQs)

Here are 10 questions to test your understanding:

1. In ∆ABC and ∆DEF, ∠B = ∠E, ∠F = ∠C and AB = 3DE. Then, the two triangles are:
   (a) Congruent but not similar
   (b) Similar but not congruent
   (c) Neither congruent nor similar
   (d) Congruent as well as similar

2. If in ∆ABC and ∆PQR, AB/QR = BC/PR = CA/PQ, then:
   (a) ∆PQR ~ ∆CAB
   (b) ∆PQR ~ ∆ABC
   (c) ∆CBA ~ ∆PQR
   (d) ∆BCA ~ ∆PQR

3. In triangle ABC, DE || BC where D is a point on AB and E is a point on AC. If AD/DB = 5/3, then AE/AC is equal to:
   (a) 3/5
   (b) 5/3
   (c) 5/8
   (d) 8/5

4. ∆ABC ~ ∆PQR. If ar(ABC) = 25 cm² and ar(PQR) = 49 cm², and BC = 5 cm, then QR is equal to:
   (a) 7 cm
   (b) 10 cm
   (c) 4.9 cm
   (d) 3.5 cm

5. In a right triangle ABC, right-angled at B, BD is perpendicular to AC. If AD = 4 cm and DC = 9 cm, then BD is equal to:
   (a) 6 cm
   (b) 13 cm
   (c) 5 cm
   (d) 36 cm

6. The lengths of the diagonals of a rhombus are 16 cm and 12 cm. The length of the side of the rhombus is:
   (a) 9 cm
   (b) 10 cm
   (c) 8 cm
   (d) 20 cm

7. If ∆ABC ~ ∆QRP, ar(ABC)/ar(PQR) = 9/4, AB = 18 cm and BC = 15 cm, then PR is equal to:
   (a) 10 cm
   (b) 12 cm
   (c) 20/3 cm
   (d) 8 cm

8. In the given figure, if ∠A = 90°, AD ⊥ BC. If AB = 5 cm, AC = 12 cm and BC = 13 cm, then AD is equal to:
   (a) 12/13 cm
   (b) 60/13 cm
   (c) 5/13 cm
   (d) 20/13 cm
   (Assume a standard diagram for a right triangle ABC with altitude AD)

9. A vertical pole of length 6 m casts a shadow 4 m long on the ground. At the same time, a tower casts a shadow 28 m long. The height of the tower is:
   (a) 40 m
   (b) 42 m
   (c) 38 m
   (d) 44 m

10. Sides of two similar triangles are in the ratio 4 : 9. Areas of these triangles are in the ratio:
    (a) 2 : 3
    (b) 4 : 9
    (c) 81 : 16
    (d) 16 : 81

Answer Key:

1. (b) Similar but not congruent (Angles are equal - AA similarity, but sides are proportional with ratio 1:3, not equal)
2. (a) ∆PQR ~ ∆CAB (Match corresponding vertices: A↔Q, B↔R, C↔P based on the given ratios AB/QR = BC/PR = CA/PQ)
3. (c) 5/8 (AD/DB = 5/3 => AD/AB = AD/(AD+DB) = 5/(5+3) = 5/8. Since DE||BC, AE/AC = AD/AB = 5/8)
4. (a) 7 cm (Ratio of areas = square of ratio of sides => ar(ABC)/ar(PQR) = (BC/QR)². So, 25/49 = (5/QR)². Taking square root, 5/7 = 5/QR => QR = 7 cm)
5. (a) 6 cm (Using the result BD² = AD × DC => BD² = 4 × 9 = 36 => BD = 6 cm)
6. (b) 10 cm (Diagonals of a rhombus bisect each other at right angles. Half diagonals are 8 cm and 6 cm. These form the legs of a right triangle with the side as hypotenuse. Side² = 8² + 6² = 64 + 36 = 100 => Side = 10 cm)
7. (a) 10 cm (ar(ABC)/ar(QRP) = (BC/RP)². Note the correspondence: BC corresponds to RP. 9/4 = (15/PR)². Taking square root, 3/2 = 15/PR => PR = (15 × 2) / 3 = 10 cm)
8. (b) 60/13 cm (Area(∆ABC) = 1/2 × AB × AC = 1/2 × 5 × 12 = 30 cm². Also, Area(∆ABC) = 1/2 × BC × AD = 1/2 × 13 × AD. So, 1/2 × 13 × AD = 30 => AD = 60/13 cm)
9. (b) 42 m (The sun's rays are parallel, forming similar triangles. Let h be the height of the tower. Height/Shadow ratio is constant. h/28 = 6/4 => h = (6 × 28) / 4 = 6 × 7 = 42 m)
10. (d) 16 : 81 (Ratio of areas = square of ratio of corresponding sides = (4/9)² = 16/81)

Revise these concepts thoroughly. Practice problems from the Exemplar book focusing on applying these theorems. Good luck!