Class 9 Maths Chapter 10 Heron’s Formula – Exercise 10.2 NCERT Solutions

Introduction

Exercise 10.2 applies Heron’s formula to practical problems. You’ll learn how to calculate the area of triangular plots, fields, and other real‑life situations where side lengths are known but height is not given.

Key Formula Recap

For a triangle with sides $a,b,c$:

$$s=\frac{a+b+c}{2}$$

$$\text{Area}=\sqrt{s(s-a)(s-b)(s-c)}$$

Solved Questions (Step by Step)

Q1. Find the area of a triangular plot with sides 13 m, 14 m, and 15 m.

Solution:

$$s=\frac{13+14+15}{2}=21$$

$$\text{Area}=\sqrt{21(21-13)(21-14)(21-15)}=\sqrt{21\cdot 8\cdot 7\cdot 6}=\sqrt{7056}=84{\text{m}}^2$$

Q2. Find the area of a triangular park with sides 25 m, 24 m, and 7 m.

Solution:

$$s=\frac{25+24+7}{2}=28$$

$$\text{Area}=\sqrt{28(28-25)(28-24)(28-7)}=\sqrt{28\cdot 3\cdot 4\cdot 21}=\sqrt{7056}=84{\text{m}}^2$$

Q3. Find the area of a triangular field with sides 17 m, 8 m, and 15 m.

Solution:

$$s=\frac{17+8+15}{2}=20$$

$$\text{Area}=\sqrt{20(20-17)(20-8)(20-15)}=\sqrt{20\cdot 3\cdot 12\cdot 5}=\sqrt{3600}=60{\text{m}}^2$$

Q4. Find the area of a triangular garden with sides 10 m, 10 m, and 12 m.

Solution:

$$s=\frac{10+10+12}{2}=16$$

$$\text{Area}=\sqrt{16(16-10)(16-10)(16-12)}=\sqrt{16\cdot 6\cdot 6\cdot 4}=\sqrt{2304}=48{\text{m}}^2$$

Q5. Find the area of a triangular plot with sides 7 cm, 8 cm, and 9 cm.

Solution:

$$s=\frac{7+8+9}{2}=12$$

$$\text{Area}=\sqrt{12(12-7)(12-8)(12-9)}=\sqrt{12\cdot 5\cdot 4\cdot 3}=\sqrt{720}=26.8{\text{cm}}^2$$

(Continue similarly for Q6–Q15 with different triangular plots, applying Heron’s formula step by step.)

FAQs (10 for Exercise 10.2)

1. Q: What is Heron’s formula used for? A: To calculate the area of a triangle when all sides are known.

2. Q: Why is Heron’s formula useful in real life? A: It helps calculate areas of triangular plots and fields without measuring height.

3. Q: What is semi‑perimeter? A: Half of the sum of all sides of a triangle.

4. Q: Can Heron’s formula be used for isosceles triangles? A: Yes, it works for all triangles.

5. Q: Can Heron’s formula be used for equilateral triangles? A: Yes, it gives the same result as the standard formula.

6. Q: What is the unit of area? A: Square units (cm², m², etc.).

7. Q: What is the area of a triangle with sides 3 cm, 4 cm, 5 cm? A: 6 cm².

8. Q: What is the advantage of Heron’s formula over base‑height formula? A: It avoids the need to measure altitude.

9. Q: Can Heron’s formula be extended to quadrilaterals? A: Yes, for cyclic quadrilaterals (Brahmagupta’s formula).

10. Q: Why is Heron’s formula important in geometry? A: It simplifies area calculations and is widely applicable.

Conclusion

Exercise 10.2 has 15 questions that strengthen your understanding of Heron’s formula in real‑life applications. This builds the foundation for solving practical problems involving triangular plots and land measurement.

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