Class 9 Notes

Class 9 Mathematics Notes Chapter 12 (Heron's Formula) – Mathematics Book

Philoid

Philoid

— 6 min read

Mathematics
Alright class, let's focus on Chapter 12, Heron's Formula. This is a very useful tool, especially when dealing with triangles where we know the lengths of all three sides but don't know the height. It frequently appears in various government exams based on the Class 9 syllabus, so understanding it well is important.

Chapter 12: Heron's Formula - Detailed Notes

1. Introduction: Area of a Triangle

  • We already know the standard formula for the area of a triangle:
    Area = ½ × base × height
  • This formula is easy to use when the base and the corresponding perpendicular height are known.
  • Limitation: Finding the height can be difficult or impossible if we only know the lengths of the three sides, especially for scalene triangles (triangles with all sides of different lengths).

2. Heron's Formula: The Solution

  • Heron's formula provides a way 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. The Formula

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

  • Step 1: Calculate the Semi-perimeter (s)
    The semi-perimeter is half the perimeter of the triangle.
    s = (a + b + c) / 2

  • Step 2: Apply Heron's Formula
    The area of the triangle (often denoted by A or Δ) is given by:
    Area = √[s(s - a)(s - b)(s - c)]

    • Where:
      • s is the semi-perimeter.
      • a, b, c are the lengths of the sides of the triangle.

4. Steps to Calculate Area using Heron's Formula:

  1. Identify the side lengths: Determine the values of a, b, and c.
  2. Calculate the semi-perimeter (s): Add the lengths of the sides and divide by 2. s = (a + b + c) / 2.
  3. Calculate the differences: Find the values of (s - a), (s - b), and (s - c). Ensure these values are positive; if not, recheck your calculation of 's' or check if the given side lengths can form a triangle (Triangle Inequality Theorem: sum of any two sides must be greater than the third side).
  4. Substitute into the formula: Plug the values of s, (s - a), (s - b), and (s - c) into the formula: Area = √[s(s - a)(s - b)(s - c)].
  5. Calculate the Area: Compute the final value. Remember the unit of area will be square units (like cm², m², etc.).

Example:
Find the area of a triangle whose sides are 13 cm, 14 cm, and 15 cm.

  1. a = 13 cm, b = 14 cm, c = 15 cm.
  2. s = (13 + 14 + 15) / 2 = 42 / 2 = 21 cm.
  3. s - a = 21 - 13 = 8 cm
    s - b = 21 - 14 = 7 cm
    s - c = 21 - 15 = 6 cm
  4. Area = √[21 × (8) × (7) × (6)]
  5. Area = √[(3 × 7) × (2 × 2 × 2) × (7) × (2 × 3)]
    Area = √[2² × 2² × 3² × 7²]
    Area = 2 × 2 × 3 × 7
    Area = 84 cm²

5. Application of Heron's Formula in Finding Areas of Quadrilaterals

  • Heron's formula can be used to find the area of a quadrilateral (like a field or plot of land) if we know the lengths of all four sides and one diagonal.
  • Method:
    1. Divide the quadrilateral into two triangles using the known diagonal.
    2. You now have two triangles where all three side lengths are known (the sides of the quadrilateral and the diagonal).
    3. Calculate the area of the first triangle using Heron's formula.
    4. Calculate the area of the second triangle using Heron's formula.
    5. The total area of the quadrilateral is the sum of the areas of the two triangles.

Example: Find the area of quadrilateral ABCD where AB = 3 cm, BC = 4 cm, CD = 4 cm, DA = 5 cm and AC = 5 cm.

  1. Triangle ABC: Sides are a=3, b=4, c=5.
    • s = (3+4+5)/2 = 6 cm.
    • Area(ΔABC) = √[6(6-3)(6-4)(6-5)] = √[6 × 3 × 2 × 1] = √36 = 6 cm².
  2. Triangle ADC: Sides are a=5, b=4, c=5. (This is an isosceles triangle)
    • s = (5+4+5)/2 = 14/2 = 7 cm.
    • Area(ΔADC) = √[7(7-5)(7-4)(7-5)] = √[7 × 2 × 3 × 2] = √[84] = √(4 × 21) = 2√21 cm².
    • (Approx value: √21 ≈ 4.58, so Area ≈ 2 × 4.58 = 9.16 cm²)
  3. Area(ABCD) = Area(ΔABC) + Area(ΔADC) = 6 + 2√21 cm². (Or approx 6 + 9.16 = 15.16 cm²)

Key Points for Exam Preparation:

  • Memorize the formula for semi-perimeter and the main Heron's formula.
  • Be careful with calculations, especially under the square root. Factorize the numbers inside the square root to easily find pairs.
  • Always mention the correct units (square units) for the area.
  • Understand the application for quadrilaterals – this is a common question type.
  • Remember the Triangle Inequality Theorem to check if given sides can form a triangle.

Multiple Choice Questions (MCQs)

Here are 10 MCQs based on Heron's Formula for your practice:

  1. Heron's formula is used to calculate the area of a triangle when the lengths of its ______ are known.
    a) Base and height
    b) Two sides and included angle
    c) All three sides
    d) One side and two angles

  2. The semi-perimeter 's' of a triangle with sides a, b, c is given by:
    a) a + b + c
    b) (a + b + c) / 3
    c) (a + b + c) / 2
    d) √[a + b + c]

  3. What is the semi-perimeter of a triangle with sides 8 cm, 10 cm, and 12 cm?
    a) 30 cm
    b) 15 cm
    c) 10 cm
    d) 20 cm

  4. The sides of a triangular plot are in the ratio 3:5:7 and its perimeter is 300 m. Find its area.
    a) 1500 m²
    b) 1500√3 m²
    c) 3000 m²
    d) 3000√3 m²

  5. The area of an equilateral triangle with side length 'a' can be calculated using Heron's formula. What is its semi-perimeter?
    a) a
    b) 3a
    c) 3a/2
    d) a/2

  6. Find the area of a triangle with sides 5 cm, 12 cm, and 13 cm.
    a) 60 cm²
    b) 30 cm²
    c) 65 cm²
    d) 25 cm²

  7. If the area of an equilateral triangle is 16√3 cm², what is the length of each side? (Hint: You can use the standard formula Area = (√3/4)a² or work backwards with Heron's)
    a) 4 cm
    b) 6 cm
    c) 8 cm
    d) 10 cm

  8. To find the area of a quadrilateral ABCD using Heron's formula, we need the lengths of its four sides and:
    a) The length of one angle
    b) The length of both diagonals
    c) The length of one diagonal
    d) The perimeter of the quadrilateral

  9. In Heron's formula, Area = √[s(s - a)(s - b)(s - c)], the term (s - a) represents:
    a) The perimeter minus side a
    b) Half the perimeter minus side a
    c) The perimeter divided by side a
    d) Side a minus the semi-perimeter

  10. The perimeter of an isosceles triangle is 32 cm. The ratio of the equal side to its base is 3:2. Find the area of the triangle.
    a) 32√2 cm²
    b) 48√2 cm²
    c) 32√5 cm²
    d) 48 cm²

Answer Key for MCQs:

  1. c) All three sides
  2. c) (a + b + c) / 2
  3. b) 15 cm (s = (8+10+12)/2 = 30/2 = 15)
  4. b) 1500√3 m² (Sides are 3x, 5x, 7x. 3x+5x+7x = 300 => 15x = 300 => x=20. Sides are 60, 100, 140. s = (60+100+140)/2 = 150. Area = √[150(150-60)(150-100)(150-140)] = √[150 * 90 * 50 * 10] = √[ (1510) * (910) * (510) * 10] = √[15 * 9 * 5 * 10⁴] = √[ (35) * 3² * 5 * 10⁴] = √[3³ * 5² * 10⁴] = √(3² * 5² * 10⁴ * 3) = 3 * 5 * 10² * √3 = 1500√3)
  5. c) 3a/2 (s = (a+a+a)/2 = 3a/2)
  6. b) 30 cm² (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) (Note: This is a right-angled triangle, 5²+12²=25+144=169=13², so Area = ½ * 5 * 12 = 30)
  7. c) 8 cm (Area = (√3/4)a² => 16√3 = (√3/4)a² => 16 = a²/4 => a² = 64 => a = 8)
  8. c) The length of one diagonal
  9. b) Half the perimeter minus side a
  10. a) 32√2 cm² (Let equal sides be 3x, base be 2x. Perimeter = 3x + 3x + 2x = 32 => 8x = 32 => x = 4. Sides are 12 cm, 12 cm, 8 cm. s = (12+12+8)/2 = 32/2 = 16. Area = √[16(16-12)(16-12)(16-8)] = √[16 * 4 * 4 * 8] = √[4² * 4 * 4 * (4*2)] = √[4² * 4² * 4 * 2] = √[4² * 4² * 2² * 2] = 4 * 4 * 2 * √2 = 32√2)

Study these notes carefully and practice the MCQs. Make sure you can apply the formula accurately and understand its use for quadrilaterals. Good luck!

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