Class 9 Mathematics Notes Chapter 17 (Chapter 17) – Examplar Problem (Englisha) Book

Alright class, let's focus on the key concepts for your government exam preparation. While you mentioned Chapter 17, in the standard NCERT Class 9 Exemplar book, the topic 'Heron's Formula' is typically Chapter 12. We will proceed assuming you are referring to Heron's Formula, as it's a vital topic from Class 9 for competitive exams. If you meant a different chapter, please specify.

Chapter Notes: Heron's Formula (Based on NCERT Class 9 Exemplar)

1. Introduction: Why Heron's Formula?

• We already know the basic formula for the area of a triangle: Area = (1/2) × base × height.
• However, finding the height of a triangle isn't always straightforward, especially for scalene triangles (where all sides have different lengths).
• Heron's Formula provides a way to calculate the area of any triangle when the lengths of all three sides are known, without needing to know the height or angles.

2. The Formula

• Let the lengths of the sides of a triangle be a, b, and c.
• First, calculate the semi-perimeter (s) of the triangle:
  s = (a + b + c) / 2
• Heron's Formula for the Area (A) of the triangle is:
  A = √[s(s - a)(s - b)(s - c)]

3. Steps to Apply Heron's Formula:

1. Identify Sides: Note down the lengths of the three sides: a, b, c.
2. Calculate Semi-perimeter (s): Add the lengths of the sides and divide by 2.
3. Calculate Differences: Find the values of (s - a), (s - b), and (s - c). Tip: Ensure these differences are positive; if not, recheck your 's' calculation.
4. Substitute into Formula: Plug the values of s, (s - a), (s - b), and (s - c) into the formula A = √[s(s - a)(s - b)(s - c)].
5. Calculate Area: Compute the product inside the square root and then find the square root. Pay attention to units (e.g., if sides are in meters, the area is in square meters).

4. Applications

• Area of Scalene Triangles: This is the primary use case where finding height is cumbersome.
• Area of Isosceles and Equilateral Triangles: While specific formulas exist ((√3/4)side² for equilateral, (b/4)√(4a² - b²) for isosceles with equal sides 'a' and base 'b'), Heron's formula can still be applied. It's useful if you forget the specific formulas but remember Heron's.
  • Example (Equilateral): Sides a, a, a. s = 3a/2. Area = √[(3a/2)(a/2)(a/2)(a/2)] = √[3a⁴/16] = (√3/4)a².
• Area of Quadrilaterals:
  • A quadrilateral can be divided into two triangles by drawing one of its diagonals.
  • If the lengths of all four sides and one diagonal are known, you can find the areas of the two resulting triangles using Heron's formula.
  • The total area of the quadrilateral is the sum of the areas of these two triangles.
  • Exam Tip: Sometimes, one diagonal might create a right-angled triangle, making its area calculation easier ((1/2) * base * height).

5. Key Points & Exam Tips (From Exemplar Perspective)

• Universality: Heron's formula works for all triangle types.
• Prerequisites: You must know the lengths of all three sides.
• Calculation Accuracy: Be meticulous with arithmetic, especially under the square root. Look for ways to simplify before multiplying everything out (e.g., factorize the numbers under the square root to find pairs).
• Units: Always state the area in square units (cm², m², etc.).
• Quadrilaterals: This is a common application in exams. Remember to split the quadrilateral into two triangles using a known diagonal.
• Comparing Methods: For right-angled triangles, (1/2) × base × height is faster. For equilateral triangles, (√3/4) × side² is faster. Use Heron's when these don't apply easily or as a universal backup.
• Exemplar Problems: Often involve slightly larger numbers, require careful calculation, or combine Heron's formula with other geometric concepts (like finding the altitude once the area is known: Area = (1/2) × base × height => height = 2 × Area / base).

Multiple Choice Questions (MCQs)

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

1. Heron's formula is used to calculate the area of a triangle when which of the following are known?
   (A) Lengths of two sides and the included angle
   (B) Lengths of three sides
   (C) Length of one side and two angles
   (D) Length of the base and the height

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) √(a² + b² + c²)

3. The sides of a triangle are 3 cm, 4 cm, and 5 cm. What is its area?
   (A) 6 cm²
   (B) 7.5 cm²
   (C) 10 cm²
   (D) 12 cm²
   (Hint: Recognize the type of triangle)

4. The area of an equilateral triangle with side length 'a' is:
   (A) (√3/2)a²
   (B) (1/2)a²
   (C) (√3/4)a²
   (D) (√3)a²

5. 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

6. The lengths of the sides of a triangle are 5 cm, 12 cm and 13 cm. Find the length of the altitude corresponding to the longest side.
   (A) 60/13 cm
   (B) 120/13 cm
   (C) 30/13 cm
   (D) 5 cm

7. An isosceles triangle has a perimeter of 30 cm and each of the equal sides is 12 cm. The area of the triangle is:
   (A) 9√15 cm²
   (B) 18√3 cm²
   (C) 6√15 cm²
   (D) 12√5 cm²

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) 14.8 cm²
   (C) 15 cm²
   (D) 16.2 cm²

9. If the area of an equilateral triangle is 16√3 cm², then the perimeter of the triangle is:
   (A) 48 cm
   (B) 24 cm
   (C) 12 cm
   (D) 36 cm

10. Find the area of a triangle whose sides are 9 cm, 12 cm and 15 cm.
    (A) 54 cm²
    (B) 108 cm²
    (C) 45 cm²
    (D) 60 cm²

Answer Key for MCQs:

1. (B)
2. (C)
3. (A) [s=(3+4+5)/2=6. Area=√(6(6-3)(6-4)(6-5)) = √(632*1) = √36 = 6. Also, it's a right triangle.]
4. (C)
5. (B) [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)) = √(150905010) = √(1510 * 910 * 510 * 10) = 100√(1595) = 100√(675) = 100√(2253) = 10015√3 = 1500√3]
6. (A) [Sides 5, 12, 13 form a right triangle (5²+12²=25+144=169=13²). Area = (1/2)512 = 30 cm². Longest side (hypotenuse) = 13 cm. Also, Area = (1/2)baseheight = (1/2)13h. So, 30 = (1/2)13h => h = 60/13 cm]
7. (A) [Perimeter=30. Equal sides=12, 12. Third side = 30 - 12 - 12 = 6 cm. Sides are 12, 12, 6. s=(12+12+6)/2 = 15. Area=√(15(15-12)(15-12)(15-6)) = √(15339) = √(533333) = 33√(53) = 9√15 cm²]
8. (A) [Split into ΔABC and ΔADC using diagonal AC=5. ΔABC: sides 3, 4, 5 (right triangle). Area1 = (1/2)34 = 6 cm². ΔADC: sides 5, 4, 5 (isosceles). s=(5+4+5)/2 = 7. Area2 = √(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]
9. (B) [Area = (√3/4)a² = 16√3 => a²/4 = 16 => a² = 64 => a = 8 cm. Perimeter = 3a = 3*8 = 24 cm]
10. (A) [Sides 9, 12, 15. Check for right triangle: 9²+12² = 81+144 = 225. 15² = 225. Yes, it's a right triangle. Area = (1/2)baseheight = (1/2)912 = 54 cm². Alternatively using Heron's: s=(9+12+15)/2 = 36/2 = 18. Area=√(18(18-9)(18-12)(18-15)) = √(18963) = √(299233) = √(2² * 9² * 3²) = 293 = 54 cm²]

Study these notes carefully and practice solving problems from your Exemplar book. Pay close attention to the application in quadrilaterals. Good luck!