Class 9 Maths Chapter 10 Heron’s Formula – Exercise 10.1 NCERT Solutions

Introduction

Exercise 10.1 introduces Heron’s formula, which allows us to calculate the area of a triangle when all three sides are known. This formula is especially useful when the height of the triangle is not given.

Key Formula

• Heron’s Formula: For a triangle with sides $a,b,c$:

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

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

where $s$ is the semi‑perimeter.

Solved Questions (Step by Step)

Q1. Find the area of a triangle with sides 5 cm, 6 cm, and 7 cm.

Solution:

$$s=\frac{5+6+7}{2}=9$$

$$\text{Area}=\sqrt{9(9-5)(9-6)(9-7)}=\sqrt{9\cdot 4\cdot 3\cdot 2}=\sqrt{216}=14.7{\text{cm}}^2$$

Q2. Find the area of a triangle with sides 6 cm, 8 cm, and 10 cm.

Solution:

$$s=\frac{6+8+10}{2}=12$$

$$\text{Area}=\sqrt{12(12-6)(12-8)(12-10)}=\sqrt{12\cdot 6\cdot 4\cdot 2}=\sqrt{576}=24{\text{cm}}^2$$

Q3. Find the area of a triangle with sides 7 cm, 9 cm, and 10 cm.

Solution:

$$s=\frac{7+9+10}{2}=13$$

$$\text{Area}=\sqrt{13(13-7)(13-9)(13-10)}=\sqrt{13\cdot 6\cdot 4\cdot 3}=\sqrt{936}=30.6{\text{cm}}^2$$

Q4. Find the area of a triangle with sides 8 cm, 9 cm, and 11 cm.

Solution:

$$s=\frac{8+9+11}{2}=14$$

$$\text{Area}=\sqrt{14(14-8)(14-9)(14-11)}=\sqrt{14\cdot 6\cdot 5\cdot 3}=\sqrt{1260}=35.5{\text{cm}}^2$$

Q5. Find the area of a triangle with sides 9 cm, 12 cm, and 15 cm.

Solution:

$$s=\frac{9+12+15}{2}=18$$

$$\text{Area}=\sqrt{18(18-9)(18-12)(18-15)}=\sqrt{18\cdot 9\cdot 6\cdot 3}=\sqrt{2916}=54{\text{cm}}^2$$

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

FAQs (10 for Exercise 10.1)

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

2. Q: Who discovered Heron’s formula? A: Hero of Alexandria, a Greek mathematician.

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 all triangles? A: Yes, as long as side lengths are known.

5. Q: What is the advantage of Heron’s formula? A: It avoids the need for height measurement.

6. Q: Can Heron’s formula be applied to equilateral triangles? A: Yes, it works for all types of triangles.

7. Q: What is the area of an equilateral triangle with side $a$? A: $\frac{\sqrt{3}}{4}{a}^2$.

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

9. Q: Why is Heron’s formula important? A: It simplifies area calculation in geometry and trigonometry.

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

Conclusion

Exercise 10.1 has 15 questions that strengthen your understanding of Heron’s formula and its application in calculating triangle areas. This builds the foundation for solving real‑life problems involving triangular plots and designs.

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New Syllabus-Class 9 Maths Chapter 10 Heron’s Formula – Exercise 10.1 NCERT Solutions