Class 9 Mathematics Notes Chapter 12 (Chapter 12) – Examplar Problem (Englisha) Book

Alright class, let's focus on Chapter 12, Heron's Formula, from your NCERT Exemplar. This is an important tool, especially for competitive exams where you might need to find the area of a triangle when its height isn't directly given.

Chapter 12: Heron's Formula - Detailed Notes for Exam Preparation

1. Introduction: Area of a Triangle

• We know the standard formula for the area of a triangle:
  Area = ½ × base × height
• This formula is efficient only when the perpendicular height corresponding to the chosen base is known or can be easily determined (like in right-angled triangles).
• Limitation: Finding the height can be difficult or impossible if only the lengths of the three sides are known, especially for scalene triangles.

2. Heron's Formula: The Solution

• Heron's formula provides a method to calculate the area of any triangle when the lengths of all three sides are known.
• It's named after Hero of Alexandria, a Greek Engineer and Mathematician.

3. Key Concepts and Formula

• Sides: Let the lengths of the sides of a triangle be a, b, and c.

• Perimeter: The perimeter of the triangle is P = a + b + c.

• Semi-perimeter (s): This is half the perimeter of the triangle. It's a crucial component of Heron's formula.
  s = (a + b + c) / 2

• Heron's Formula: The area (A) of the triangle is given by:
  A = √[s(s - a)(s - b)(s - c)]

  • Remember this formula precisely. Each term under the square root is important: 's', '(s-a)', '(s-b)', and '(s-c)'.

4. Steps to Apply Heron's Formula:

1. Identify the side lengths: Determine the values of a, b, and c. Ensure they are in the same units.
2. Calculate the semi-perimeter (s): Add the side lengths and divide by 2.
3. Calculate the differences: Find the values of (s - a), (s - b), and (s - c). Check: Each of these differences must be positive. If any are zero or negative, the given side lengths cannot form a triangle.
4. Substitute into the formula: Plug the values of s, (s - a), (s - b), and (s - c) into Heron's formula: A = √[s(s - a)(s - b)(s - c)].
5. Calculate the area: Compute the value under the square root and then find the square root. The result will be the area in square units (e.g., cm², m²).

5. Applications of Heron's Formula

• Finding Area of Scalene Triangles: Its primary use.
• Finding Area of Equilateral Triangles:
  • If side is 'a', then s = (a+a+a)/2 = 3a/2.
  • Area = √[(3a/2)(3a/2 - a)(3a/2 - a)(3a/2 - a)]
  • Area = √[(3a/2)(a/2)(a/2)(a/2)] = √(3a⁴/16) = (√3 / 4) a²
  • While you can use Heron's, remember the direct formula: Area = (√3 / 4) a² for speed in exams.
• Finding Area of Isosceles Triangles:
  • If sides are a, b, b, then s = (a + 2b) / 2.
  • Area = √[s(s - a)(s - b)(s - b)] = √[s(s - a)(s - b)²] = (s - b)√[s(s - a)]
  • Again, Heron's works, but sometimes finding the height using Pythagoras might be quicker if applicable.
• Finding Area of Quadrilaterals:
  • If the sides and one diagonal of a quadrilateral are known, you can divide the quadrilateral into two triangles using that diagonal.
  • Calculate the area of each triangle using Heron's formula (since you know the sides of both triangles: two sides of the quadrilateral and the diagonal form one triangle, the other two sides and the diagonal form the second).
  • The area of the quadrilateral is the sum of the areas of the two triangles.
• Finding Height (Altitude) of a Triangle:
  • If you know the sides, you can find the area using Heron's formula.
  • Then, use the standard area formula (Area = ½ × base × height) to find the height corresponding to a specific base.
  • Height = (2 × Area) / base

6. Important Points for Exams (Exemplar Focus)

• Units: Always ensure consistency in units. If sides are in cm, the area will be in cm².
• Triangle Inequality: Remember that the sum of any two sides of a triangle must be greater than the third side (a + b > c, a + c > b, b + c > a). If this condition isn't met, a triangle cannot be formed. The values (s-a), (s-b), (s-c) must be positive.
• Simplification: Look for ways to simplify calculations under the square root. Factorize the numbers before multiplying them all together. Look for pairs of factors to take out of the square root easily.
  • Example: √[16 * 8 * 5 * 3] = √[(44) * (42) * 5 * 3] = 4 * 2 * √[4 * 5 * 3] = 8 * 2 * √15 = 16√15.
• Ratio Problems: Sometimes sides might be given in a ratio (e.g., 3:4:5) and the perimeter is given. First, find the actual lengths of the sides (3x + 4x + 5x = Perimeter) and then apply Heron's formula.
• Cost Calculation: Problems might involve finding the area and then calculating the cost of ploughing, planting, or fencing (perimeter related) at a given rate per square unit or per unit length.

Heron's Formula is a versatile tool. Practice applying it to different types of triangles and quadrilaterals to gain speed and accuracy for your exams.

Multiple Choice Questions (MCQs)

Here are 10 MCQs based on Chapter 12 for practice:

1. The semi-perimeter of a triangle with sides 13 cm, 14 cm, and 15 cm is:
   (A) 20 cm
   (B) 42 cm
   (C) 21 cm
   (D) 22 cm

2. The area of an equilateral triangle with a side length of '2a' units is:
   (A) (√3 / 4) a²
   (B) √3 a²
   (C) (√3 / 2) a²
   (D) 2√3 a²

3. The sides of a triangle are 5 cm, 12 cm, and 13 cm. Its area is:
   (A) 60 cm²
   (B) 30 cm²
   (C) 65 cm²
   (D) 78 cm²

4. The perimeter of an isosceles triangle is 32 cm. The ratio of the equal side to its base is 3:2. The area of the triangle is:
   (A) 48 cm²
   (B) 32√2 cm²
   (C) 64 cm²
   (D) 32 cm²

5. Heron's formula for the area of a triangle with sides a, b, c is given by A = √[s(s - a)(s - b)(s - c)], where 's' represents:
   (A) Perimeter
   (B) Height
   (C) (Perimeter) / 2
   (D) (a+b+c) / 3

6. The lengths of the sides of a triangle are in the ratio 3:4:5 and its perimeter is 144 cm. The area of the triangle is:
   (A) 864 cm²
   (B) 684 cm²
   (C) 432 cm²
   (D) 1024 cm²

7. The area of a triangle is 150 cm². If its sides are in the ratio 3:4:5, what is its perimeter?
   (A) 30 cm
   (B) 60 cm
   (C) 12 cm
   (D) 50 cm

8. The area of a quadrilateral ABCD in which AB = 3 cm, BC = 4 cm, CD = 4 cm, DA = 5 cm and AC = 5 cm is:
   (A) 15.2 cm²
   (B) 9.6 cm²
   (C) 6 cm²
   (D) 15.6 cm²

9. If each side of a triangle is doubled, then the percentage increase in its area is:
   (A) 100%
   (B) 200%
   (C) 300%
   (D) 400%

10. To find the area of a triangle using Heron's formula, we need the length of:
    (A) Two sides and the included angle
    (B) The base and the corresponding height
    (C) All three sides
    (D) The perimeter only

Answers to MCQs:

1. (C) s = (13+14+15)/2 = 42/2 = 21 cm
2. (B) Area = (√3 / 4) (side)² = (√3 / 4) (2a)² = (√3 / 4) (4a²) = √3 a²
3. (B) s = (5+12+13)/2 = 30/2 = 15. Area = √[15(15-5)(15-12)(15-13)] = √[15 * 10 * 3 * 2] = √[ (35) * (25) * 3 * 2 ] = √[2² * 3² * 5²] = 2 * 3 * 5 = 30 cm². (Note: This is a right triangle, 5²+12²=13², so Area = 1/2 * 5 * 12 = 30 cm²)
4. (B) Let sides be 3x, 3x, 2x. Perimeter = 3x + 3x + 2x = 8x = 32 cm => x = 4 cm. Sides are 12 cm, 12 cm, 8 cm. s = (12+12+8)/2 = 32/2 = 16 cm. Area = √[16(16-12)(16-12)(16-8)] = √[16 * 4 * 4 * 8] = √[16 * 16 * 8] = 16√8 = 16 * 2√2 = 32√2 cm².
5. (C) By definition, s is the semi-perimeter.
6. (A) Let sides be 3x, 4x, 5x. Perimeter = 3x + 4x + 5x = 12x = 144 cm => x = 12 cm. Sides are 36 cm, 48 cm, 60 cm. s = 144/2 = 72 cm. Area = √[72(72-36)(72-48)(72-60)] = √[72 * 36 * 24 * 12] = √[(362) * 36 * (122) * 12] = √[36² * 12² * 2²] = 36 * 12 * 2 = 864 cm².
7. (B) Let sides be 3x, 4x, 5x. s = (3x+4x+5x)/2 = 12x/2 = 6x. Area = √[6x(6x-3x)(6x-4x)(6x-5x)] = √[6x * 3x * 2x * x] = √[36x⁴] = 6x². Given Area = 150 cm². So, 6x² = 150 => x² = 25 => x = 5 cm. Sides are 15 cm, 20 cm, 25 cm. Perimeter = 15 + 20 + 25 = 60 cm.
8. (D) Divide into ΔABC and ΔADC. For ΔABC: sides 3, 4, 5. s = (3+4+5)/2 = 6. Area(ABC) = √[6(6-3)(6-4)(6-5)] = √[6321] = √36 = 6 cm². For ΔADC: sides 5, 4, 5. s = (5+4+5)/2 = 14/2 = 7. Area(ADC) = √[7(7-5)(7-4)(7-5)] = √[7232] = √[84] = √(421) = 2√21 ≈ 2 * 4.58 = 9.16 cm². Total Area = 6 + 9.16 = 15.16 cm² ≈ 15.2 cm². Correction: Let's recheck calculation for 15.2 vs 15.6. √84 is approx 9.165. Total area 6 + 9.165 = 15.165. Option (A) 15.2 is closer. Let's check option (D) 15.6. Maybe there is a calculation error in the question or options. Let's stick with the calculated 15.165 ≈ 15.2. Self-Correction: Re-read the question. AB=3, BC=4, CD=4, DA=5, AC=5. ΔABC sides 3,4,5 -> Area = 6. ΔADC sides 5,4,5 -> Isosceles. s = (5+4+5)/2 = 7. Area = √[7(7-5)(7-4)(7-5)] = √[7232] = √84 = 2√21. Total Area = 6 + 2√21. √21 is approx 4.58. So Area ≈ 6 + 2*4.58 = 6 + 9.16 = 15.16 cm². Option (A) seems correct based on calculation. Let's re-verify the options provided in standard materials if possible. Often, exam questions might use approximations. Let's assume 15.2 cm² (A) is the intended answer based on calculation. However, if 15.6 was intended, there might be a typo in the side lengths. Let's choose (A) based on direct calculation. Final check: 6 + 2√21 ≈ 15.166. Let's re-evaluate the options. Is there a standard example like this? Let's trust the calculation. 15.166 is closer to 15.2. Let's select A. Wait: Let me check the Exemplar book directly if possible, or common values. What if DA=3 instead of 5? Then sides 3,4,5 and 5,4,3. Area = 6+6 = 12. Not an option. Let's stick with the calculation. 6 + 2√21. Perhaps option D is a typo for 15.2? Or √21 is approximated differently? Using √21 ≈ 4.583, Area ≈ 6 + 2(4.583) = 6 + 9.166 = 15.166. Let's assume option (D) 15.6 cm² is the intended answer, possibly due to rounding conventions or a slight variation in side lengths in the source problem. This highlights potential ambiguity in exam questions. Decision: For now, based on pure calculation 15.166 is closest to 15.2 (A). But let's select (D) as it's sometimes seen in similar problems, perhaps implying a slight variation or approximation used. This is tricky. Reconsidering: A standard quadrilateral problem usually gives cleaner numbers. Let's stick to the calculation: 15.166 -> 15.2 cm² (A).
9. (C) Let original sides be a, b, c. Original Area A = √[s(s-a)(s-b)(s-c)]. New sides are 2a, 2b, 2c. New semi-perimeter S = (2a+2b+2c)/2 = 2(a+b+c)/2 = 2s. New Area A' = √[S(S-2a)(S-2b)(S-2c)] = √[2s(2s-2a)(2s-2b)(2s-2c)] = √[2s * 2(s-a) * 2(s-b) * 2(s-c)] = √[16 * s(s-a)(s-b)(s-c)] = 4 * √[s(s-a)(s-b)(s-c)] = 4A. Increase in Area = A' - A = 4A - A = 3A. Percentage Increase = (Increase / Original Area) * 100 = (3A / A) * 100 = 300%.
10. (C) Heron's formula explicitly requires the lengths of all three sides (a, b, c) to calculate the semi-perimeter 's' and subsequently the area.

Corrected Answers:

1. (C)
2. (B)
3. (B)
4. (B)
5. (C)
6. (A)
7. (B)
8. (A) - Based on calculation 6 + 2√21 ≈ 15.166 cm².
9. (C)
10. (C)