Welcome to AssamSchool. This article contains 30 Extra Questions from Heron’s Formula from Class 9 Mathematics Chapter 10. These Heron’s Formula Class 9 Extra Questions are helpful to test your understanding and knowledge of the chapter. These additional questions will not only help you to excel in the chapter but also will give you a strong foundation in Mathematics.
We have also provided step-by-step answers to the questions for reference. So, let’s start…
30 Heron’s Formula Class 9 Extra Questions
1. Find the area of a triangle with sides \(5 cm\), \(12 cm\), and \(13 cm\) using Heron’s formula.
2. True or False: Heron’s formula can be used for any type of triangle.
3. A triangular plot has sides in the ratio \(2:3:4\) and a perimeter of \(180 m\). Calculate its area.
4. The perimeter of a triangle is \(60 cm\), and two of its sides are \(15 cm\) and \(20 cm\). Find the area using Heron’s formula.
5. A company wants to paint an advertisement on a triangular wall with sides \(25 m\), \(25 m\), and \(30 m\). If painting costs \(₹200\) per \(m²\), find the total cost.
6. A triangle has sides \(7 cm\), \(15 cm\), and \(20 cm\). Verify if it’s a right-angled triangle using Heron’s formula.
7. What is the semi-perimeter of a triangle with sides \(9 cm\), \(12 cm\), and \(15 cm\)?
8. Calculate the area of an equilateral triangle with side \(8 cm\) using Heron’s formula.
9. The sides of a triangle are \(10 cm\), \(14 cm\), and \(16 cm\). Find its area and the length of the altitude corresponding to the longest side.
10. A triangular field has sides \(50 m\), \(60 m\), and \(70 m\). Find the cost of laying grass at \(₹5\) per \(m²\).
11. Prove that Heron’s formula gives the correct area for a right-angled triangle with legs \(9 cm\) and \(12 cm\).
12. A traffic signal board is an equilateral triangle with a perimeter of \(180 cm\). Find its area using Heron’s formula.
13. If the semi-perimeter of a triangle is \(15 cm\) and its sides are \(9 cm\), \(7 cm\), and \(x cm\), find \(x\).
14. The sides of a triangle are \(11 cm\), \(12 cm\), and \(13 cm\). Calculate its area and the radius of the inscribed circle.
15. A kite is shaped like a triangle with sides \(2 m\), \(2 m\), and \(3 m\). Find the area and the cost of material required if the material costs \(₹50\) per \(m²\).
16. True or False: Heron’s formula requires knowing the height of the triangle.
17. A triangle has a perimeter of \(84 cm\), and its sides are in the ratio \(7:8:9\). Find the area.
18. A flower bed is in the shape of a triangle with sides \(5 m\), \(6 m\), and \(7 m\). Find the area and the cost of fencing it at ₹20 per meter, leaving a \(1 m\) gap.
19. Find the area of a triangle with sides \(8 cm\), \(15 cm\), and \(17 cm\).
20. A triangular park has sides \(45 m\), \(60 m\), and \(75 m\). Determine if it’s a right-angled triangle and find its area using Heron’s formula.
21. A company hired a triangular wall with sides \(122 m\), \(22 m\), and \(120 m\) for \(6\) months. If the rent is ₹5000 per m² per year, calculate the total rent paid.
22. What is the semi-perimeter of a triangle with sides \(7 cm\), \(24 cm\), and \(25 cm\)?
23. An isosceles triangle has a perimeter of \(42 cm\) with equal sides of \(12 cm\) each. Find its area.
24. Two sides of a triangle are \(18 cm\) and \(10 cm\), and the perimeter is \(42 cm\). Find the area using Heron’s formula.
25. Calculate the area of a triangle with sides \(3 cm\), \(4 cm\), and \(5 cm\) using Heron’s formula.
26. The sides of a triangle are in the ratio \(12:17:25\), and the perimeter is \(540 cm\). Find its area.
27. A slide in a park has a triangular wall with sides \(15 m\), \(11 m\), and \(6 m\). Calculate the area painted if the painting costs ₹100 per m².
28. True or False: For a triangle with sides \(5 cm\), \(5 cm\), and \(8 cm\), the semi-perimeter is \(9 cm\).
29. A triangle has sides \(21 cm\), \(28 cm\), and \(35 cm\). Verify if it’s right-angled and find its area using both methods.
30. A triangular plot has sides \(3x\), \(5x\), and \(7x\) with a perimeter of \(300 m\). Find \(x\) and the area of the plot.
Answers
- Solution:
\( s = \frac{5+12+13}{2} = 15 \, \text{cm} \).
Area = \( \sqrt{15(15-5)(15-12)(15-13)} = \sqrt{15 \times 10 \times 3 \times 2} = 30 \, \text{cm}^2 \) - Solution:
True. Heron’s formula works for all triangles. - Solution:
Sides: \( 2x, 3x, 4x \). Perimeter = \( 9x = 180 \, \text{m} \Rightarrow x = 20 \)
Sides: \(40 m\), \(60 m\), \(80 m\) \( s = 90 \, \text{m} \).
Area = \( \sqrt{90 \times 50 \times 30 \times 10} = 300\sqrt{15} \, \text{m}^2 \) - Solution:
Third side = \( 60 – 15 – 20 = 25 \, \text{cm} \) \( s = 30 \, \text{cm} \)
Area = \( \sqrt{30 \times 15 \times 10 \times 5} = 150\sqrt{2} \, \text{cm}^2 \) - Solution:
\( s = \frac{25+25+30}{2} = 40 \, \text{m} \)
Area = \( \sqrt{40 \times 15 \times 15 \times 10} = 300 \, \text{m}^2 \)
Cost = \( 300 \times 200 = ₹60,000 \) - Solution:
\( s = \frac{7+15+20}{2} = 21 \, \text{cm} \)
Area (Heron) =\( \sqrt{21 \times 14 \times 6 \times 1} = \sqrt{1764} = 42 \, \text{cm}^2 \)
For right triangle: \( \frac{1}{2} \times 7 \times 15 = 52.5 \, \text{cm}^2 \)
Not equal ⇒ Not right-angled. - Solution:
\( s = \frac{9+12+15}{2} = 18 \, \text{cm} \) - Solution:
\( s = \frac{8+8+8}{2} = 12 \, \text{cm} \)
Area = \( \sqrt{12 \times 4 \times 4 \times 4} = 16\sqrt{3} \, \text{cm}^2 \) - Solution:
\( s = \frac{10+14+16}{2} = 20 \, \text{cm} \)
Area = \( \sqrt{20 \times 10 \times 6 \times 4} = 40\sqrt{3} \, \text{cm}^2 \)
Altitude = \( \frac{2 \times \text{Area}}{16} = 5\sqrt{3} \, \text{cm} \) - Solution:
\( s = \frac{50+60+70}{2} = 90 \, \text{m} \)
Area = \( \sqrt{90 \times 40 \times 30 \times 20} = 600\sqrt{15} \, \text{m}^2 \)
Cost = \( 600\sqrt{15} \times 5 \approx ₹37,273 \) - Solution:
Heron’s formula: \( s = \frac{9+12+15}{2} = 18 \, \text{cm} \)
Area = \( \sqrt{18 \times 9 \times 6 \times 3} = 54 \, \text{cm}^2 \)
Base-height: \( \frac{1}{2} \times 9 \times 12 = 54 \, \text{cm}^2 \)
Hence verified. - Solution:
Side = \( \frac{180}{3} = 60 \, \text{cm} \)
\( s = 90 \, \text{cm} \)
Area = \( \sqrt{90 \times 30 \times 30 \times 30} = 900\sqrt{3} \, \text{cm}^2 \) - Solution:
\( 15 = \frac{9+7+x}{2} \Rightarrow x = 30 – 16 = 14 \, \text{cm} \) - Solution:
\( s = \frac{11+12+13}{2} = 18 \, \text{cm} \)
Area = \( \sqrt{18 \times 7 \times 6 \times 5} = 6\sqrt{210} \, \text{cm}^2 \)
Radius \( r = \frac{\text{Area}}{s} = \frac{6\sqrt{210}}{18} = \frac{\sqrt{210}}{3} \, \text{cm} \) - Solution:
\( s = \frac{2+2+3}{2} = 3.5 \, \text{m} \)
Area = \( \sqrt{3.5 \times 1.5 \times 1.5 \times 0.5} \approx 1.5 \, \text{m}^2 \)
Cost = \( 1.5 \times 50 = ₹75 \) - Solution:
False. Heron’s formula does not require height. - Solution:
Sides: \( 7x, 8x, 9x \)
Perimeter = \( 24x = 84 \Rightarrow x = 3.5 \)
Sides: 24.5 cm, 28 cm, 31.5 cm
\( s = 42 \, \text{cm} \)
Area = \( \sqrt{42 \times 17.5 \times 14 \times 10.5} \approx 294 \, \text{cm}^2 \) - Solution:
\( s = \frac{5+6+7}{2} = 9 \, \text{m} \)
Area = \( \sqrt{9 \times 4 \times 3 \times 2} = 6\sqrt{6} \, \text{m}^2 \)
Fencing cost = \( (5+6+7-1) \times 20 = ₹340 \) - Solution:
\( s = \frac{8+15+17}{2} = 20 \, \text{cm} \)
Area = \( \sqrt{20 \times 12 \times 5 \times 3} = 60 \, \text{cm}^2 \) - Solution:
Check: \( 45^2 + 60^2 = 2025 + 3600 = 5625 = 75^2 \). Right-angled.
Area (Heron): \( \sqrt{90 \times 45 \times 30 \times 15} = 1350 \, \text{m}^2 \) - Solution:
\( s = \frac{122+22+120}{2} = 132 \, \text{m}\)
Area = \( \sqrt{132 \times 10 \times 110 \times 12} = 1320 \, \text{m}^2 \)
Rent = \( 1320 \times 5000 \times \frac{6}{12} = ₹33,00,000 \) - Solution:
\( s = \frac{7+24+25}{2} = 28 \, \text{cm} \) - Solution:
Third side = \( 42 – 24 = 18 \, \text{cm} ). ( s = 21 \, \text{cm} \)
Area = \( \sqrt{21 \times 9 \times 9 \times 3} = 54 \, \text{cm}^2 \) - Solution:
Third side = \( 42 – 18 – 10 = 14 \, \text{cm} ). ( s = 21 \, \text{cm} \)
Area = \( \sqrt{21 \times 3 \times 11 \times 7} = 21\sqrt{11} \, \text{cm}^2 \) - Solution:
\( s = \frac{3+4+5}{2} = 6 \, \text{cm} \)
Area = \( \sqrt{6 \times 3 \times 2 \times 1} = 6 \, \text{cm}^2 \) - Solution:
Sides: \( 12x, 17x, 25x \).
Perimeter = \( 54x = 540 \Rightarrow x = 10 \)
Sides: 120 cm, 170 cm, 250 cm. \( s = 270 \, \text{cm} \)
Area = \( \sqrt{270 \times 150 \times 100 \times 20} = 9000 \, \text{cm}^2 \) - Solution:
\( s = \frac{15+11+6}{2} = 16 \, \text{m} \)
Area = \( \sqrt{16 \times 1 \times 5 \times 10} = 20\sqrt{2} \, \text{m}^2 \)
Cost = \( 20\sqrt{2} \times 100 \approx ₹2828 \) - Solution:
True. \( s = \frac{5+5+8}{2} = 9 \, \text{cm} \) - Solution:
Check: \( 21^2 + 28^2 = 441 + 784 = 1225 = 35^2 \). Right-angled.
Area = \( \frac{1}{2} \times 21 \times 28 = 294 \, \text{cm}^2 \).
Heron’s formula: \( \sqrt{42 \times 21 \times 14 \times 7} = 294 \, \text{cm}^2 \) - Solution:
Perimeter = \( 15x = 300 \Rightarrow x = 20 \)
Sides: 60 m, 100 m, 140 m. \( s = 150 \, \text{m} \)
Area = \( \sqrt{150 \times 90 \times 50 \times 10} = 1500\sqrt{3} \, \text{m}^2 \)
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Abdur Rohman is an Electrical Engineer from Charaideo, Assam, who wears multiple hats as a part-time teacher, blogger, entrepreneur, and digital marketer. Passionate about education, he founded The Assam School blog to provide free, comprehensive textbook solutions, MCQs (Multiple Choice Questions), and other academic content for students from Class V to XII.