Test your understanding of Chapter 10: Heron’s Formula with these challenging multiple-choice questions. You need to apply Heron’s formula and related concepts to solve these problems effectively!
Class 9th Maths Important MCQs with Answers from Chapter 10 – Heron’s Formula
Question 1. Which of the following triangles can have its area calculated using Heron’s formula?
a) A triangle with base and height given
b) A triangle with sides 5 cm, 12 cm, and 13 cm
c) A right triangle with base 3 cm and height 4 cm
d) A triangle with an angle of 90° and two sides given
Answer:
b) A triangle with sides 5 cm, 12 cm, and 13 cm — Heron’s formula is used when all three sides of a triangle are known, even if the height is not explicitly given.
Question 2. The semi-perimeter (s) of a triangle with sides 10 cm, 14 cm, and 18 cm is:
a) 20 cm
b) 21 cm
c) 22 cm
d) 24 cm
Answer:
b) 21 cm — Semi-perimeter is calculated as \( s = \frac{10 + 14 + 18}{2} = 21 \, \text{cm} \).
Question 3. Heron’s formula is most useful in which of the following situations?
a) When height and base of a triangle are known
b) When two sides and the included angle are known
c) When all three sides of a triangle are known
d) When the triangle is equilateral
Answer:
c) When all three sides of a triangle are known — This is the primary purpose of Heron’s formula, as it doesn’t require the height.
Question 4. Find the area of a triangle with sides 8 cm, 15 cm, and 17 cm using Heron’s formula.
a) 60 cm²
b) 64 cm²
c) 80 cm²
d) 120 cm²
Answer:
a) 60 cm² — \( s = \frac{8 + 15 + 17}{2} = 20 \), area = \( \sqrt{20(20-8)(20-15)(20-17)} = \sqrt{20 \cdot 12 \cdot 5 \cdot 3} = 60 \, \text{cm}^2 \).
Question 5. The formula \( \sqrt{s(s-a)(s-b)(s-c)} \) is named after which mathematician?
a) Euclid
b) Heron
c) Pythagoras
d) Archimedes
Answer:
b) Heron — The formula is attributed to Heron of Alexandria, a Greek mathematician.
Question 6. What is the area of an equilateral triangle with a side of 10 cm?
a) 25√3 cm²
b) 50√3 cm²
c) 75√3 cm²
d) 100√3 cm²
Answer:
b) 50√3 cm² — \( s = \frac{10 + 10 + 10}{2} = 15 \), area = \( \sqrt{15(15-10)(15-10)(15-10)} = \sqrt{15 \cdot 5 \cdot 5 \cdot 5} = 50\sqrt{3} \, \text{cm}^2 \).
Question 7. A triangular park has sides 40 m, 24 m, and 32 m. What is its semi-perimeter?
a) 24 m
b) 40 m
c) 48 m
d) 50 m
Answer:
c) 48 m — \( s = \frac{40 + 24 + 32}{2} = 48 \, \text{m} \).
Question 8. If the sides of a triangle are 13 cm, 14 cm, and 15 cm, what is its area using Heron’s formula?
a) 81 cm²
b) 84 cm²
c) 91 cm²
d) 98 cm²
Answer:
b) 84 cm² — \( s = \frac{13 + 14 + 15}{2} = 21 \), area = \( \sqrt{21(21-13)(21-14)(21-15)} = \sqrt{21 \cdot 8 \cdot 7 \cdot 6} = 84 \, \text{cm}^2 \).
Question 9. What is the ratio of the perimeter to the semi-perimeter of any triangle?
a) 1:2
b) 2:1
c) 3:1
d) 4:1
Answer:
b) 2:1 — The semi-perimeter is always half of the perimeter, so the ratio is 2:1.
Question 10. A triangular wall has sides 122 m, 120 m, and 22 m. Calculate its area using Heron’s formula.
a) 1220 m²
b) 1320 m²
c) 1440 m²
d) 1600 m²
Answer:
b) 1320 m² — \( s = \frac{122 + 120 + 22}{2} = 132 \), area = \( \sqrt{132(132-122)(132-120)(132-22)} = 1320 \, \text{m}^2 \).
Question 11. Heron’s formula is also known as:
a) Euclid’s formula
b) Archimedes’ formula
c) Hero’s formula
d) Pythagoras’ formula
Answer:
c) Hero’s formula — The formula is attributed to Heron, who is sometimes referred to as Hero.
Question 12. If the area of a triangle is 150√3 cm² and its semi-perimeter is 30 cm, what could be the length of its sides (if it is equilateral)?
a) 15 cm
b) 20 cm
c) 30 cm
d) 10 cm
Answer:
a) 15 cm — The area of an equilateral triangle is \( \frac{\sqrt{3}}{4} \times a^2 \). Solving for \( a \), \( a = 15 \, \text{cm} \).
Question 13. A triangular park has sides 50 m, 80 m, and 120 m. What is the area of the park?
a) 600 m²
b) 1000 m²
c) 1500 m²
d) 2000 m²
Answer:
c) 1500 m² — \( s = \frac{50 + 80 + 120}{2} = 125 \), area = \( \sqrt{125(125-50)(125-80)(125-120)} = \sqrt{125 \cdot 75 \cdot 45 \cdot 5} = 1500 \, \text{m}^2 \).
Question 14. What is the semi-perimeter of a triangle with sides 7 cm, 24 cm, and 25 cm?
a) 26 cm
b) 28 cm
c) 30 cm
d) 32 cm
Answer:
b) 28 cm — \( s = \frac{7 + 24 + 25}{2} = 28 \, \text{cm} \).
Question 15. If the sides of a triangle are in the ratio 3:4:5 and its perimeter is 36 cm, what is the length of its shortest side?
a) 6 cm
b) 9 cm
c) 12 cm
d) 15 cm
Answer:
b) 9 cm — The total length is divided as \( 3x + 4x + 5x = 36 \). Solving \( x = 3 \), the shortest side is \( 3 \times 3 = 9 \, \text{cm} \).
Question 16. Calculate the area of a triangle with sides 10 cm, 10 cm, and 12 cm using Heron’s formula.
a) 48 cm²
b) 52 cm²
c) 54 cm²
d) 60 cm²
Answer:
a) 48 cm² — \( s = \frac{10 + 10 + 12}{2} = 16 \), area = \( \sqrt{16(16-10)(16-10)(16-12)} = \sqrt{16 \cdot 6 \cdot 6 \cdot 4} = 48 \, \text{cm}^2 \).
Question 17. A slide in a park has a triangular side with sides 15 m, 11 m, and 6 m. What is the area of this triangular side?
a) 30 m²
b) 33 m²
c) 36 m²
d) 39 m²
Answer:
b) 33 m² — \( s = \frac{15 + 11 + 6}{2} = 16 \), area = \( \sqrt{16(16-15)(16-11)(16-6)} = \sqrt{16 \cdot 1 \cdot 5 \cdot 10} = 33 \, \text{m}^2 \).
Question 18. An equilateral triangle has a side length of 18 cm. What is its area?
a) 81√3 cm²
b) 162√3 cm²
c) 243√3 cm²
d) 324√3 cm²
Answer:
b) 162√3 cm² — For an equilateral triangle, area = \( \frac{\sqrt{3}}{4} \times a^2 \). Substituting \( a = 18 \), area = \( \frac{\sqrt{3}}{4} \times 18^2 = 162\sqrt{3} \, \text{cm}^2 \).
Question 19. If a triangle has sides 9 cm, 12 cm, and 15 cm, what is the nature of the triangle?
a) Scalene
b) Right-angled
c) Equilateral
d) Isosceles
Answer:
b) Right-angled — The sides satisfy \( 9^2 + 12^2 = 15^2 \), hence it is a right-angled triangle.
Question 20. A company hires a triangular wall for advertisement, with sides 122 m, 22 m, and 120 m. What is the cost of advertising for 1 year if the rate is ₹5000 per m²?
a) ₹65,000
b) ₹6,600,000
c) ₹5,700,000
d) ₹7,000,000
Answer:
b) ₹6,600,000 — Area = 1320 m², Cost = \( 1320 \times 5000 = ₹6,600,000 \).
Question 21. Find the area of a triangle with sides 18 cm, 24 cm, and 30 cm.
a) 144 cm²
b) 216 cm²
c) 324 cm²
d) 432 cm²
Answer:
b) 216 cm² — \( s = \frac{18 + 24 + 30}{2} = 36 \), area = \( \sqrt{36(36-18)(36-24)(36-30)} = \sqrt{36 \cdot 18 \cdot 12 \cdot 6} = 216 \, \text{cm}^2 \).
Question 22. A triangle has sides 5 m, 12 m, and 13 m. What is its height if the base is 12 m?
a) 5 m
b) 7 m
c) 10 m
d) 12 m
Answer:
c) 10 m — The triangle is right-angled; height = 5 m (perpendicular side), base = 12 m, area = \( \frac{1}{2} \times 12 \times 5 = 30 \, \text{m}^2 \).
Question 23. For a triangle with sides 13 m, 14 m, and 15 m, which formula would you use to find its area?
a) \( \frac{1}{2} \times base \times height \)
b) \( \frac{a + b + c}{2} \)
c) \( \sqrt{s(s-a)(s-b)(s-c)} \)
d) \( \frac{\sqrt{3}}{4} \times a^2 \)
Answer:
c) \( \sqrt{s(s-a)(s-b)(s-c)} \) — Heron’s formula is used when all three sides are given.
Question 24. A triangular garden has sides 60 m, 80 m, and 100 m. What is the area of the garden?
a) 2400 m²
b) 3200 m²
c) 4800 m²
d) 6000 m²
Answer:
a) 2400 m² — \( s = \frac{60 + 80 + 100}{2} = 120 \), area = \( \sqrt{120(120-60)(120-80)(120-100)} = 2400 \, \text{m}^2 \).
Question 25. If the sides of a triangle are 9 cm, 10 cm, and 17 cm, what can be concluded about the triangle?
a) It is a right-angled triangle
b) It is an equilateral triangle
c) It is an isosceles triangle
d) It is not a valid triangle
Answer:
d) It is not a valid triangle — The sum of two sides (9 + 10) is not greater than the third side (17), hence it does not form a triangle.
Question 26. A triangular field has sides 150 m, 120 m, and 200 m. Find its area using Heron’s formula.
a) 7200 m²
b) 7800 m²
c) 9000 m²
d) 10800 m²
Answer:
d) 10800 m² — \( s = \frac{150 + 120 + 200}{2} = 235 \), area = \( \sqrt{235(235-150)(235-120)(235-200)} = 10800 \, \text{m}^2 \).
Question 27. The perimeter of a triangle is 60 cm, and its sides are in the ratio 3:4:5. What is the area of the triangle?
a) 90 cm²
b) 120 cm²
c) 144 cm²
d) 180 cm²
Answer:
b) 120 cm² — Sides are 15 cm, 20 cm, and 25 cm. \( s = 30 \), area = \( \sqrt{30(30-15)(30-20)(30-25)} = 120 \, \text{cm}^2 \).
Question 28. A triangle with sides 6 cm, 8 cm, and 10 cm is:
a) Scalene
b) Right-angled
c) Equilateral
d) Isosceles
Answer:
b) Right-angled — The sides satisfy \( 6^2 + 8^2 = 10^2 \), confirming it is a right-angled triangle.
Question 29. If the perimeter of an equilateral triangle is 36 cm, what is its area?
a) 36√3 cm²
b) 54√3 cm²
c) 72√3 cm²
d) 108√3 cm²
Answer:
b) 54√3 cm² — Each side is \( \frac{36}{3} = 12 \, \text{cm} \). Area = \( \frac{\sqrt{3}}{4} \times 12^2 = 54\sqrt{3} \, \text{cm}^2 \).
Question 30. The semi-perimeter of a triangle with sides 5 cm, 6 cm, and 7 cm is:
a) 8 cm
b) 9 cm
c) 10 cm
d) 11 cm
Answer:
b) 9 cm — \( s = \frac{5 + 6 + 7}{2} = 9 \, \text{cm} \).
Question 31. What is the height of an equilateral triangle with side 10 cm?
a) 10 cm
b) 5√3 cm
c) 10√3 cm
d) 15 cm
Answer:
b) 5√3 cm — The height of an equilateral triangle is \( \frac{\sqrt{3}}{2} \times a \). Substituting \( a = 10 \), height = \( 5\sqrt{3} \, \text{cm} \).
Question 32. For a triangle with sides 15 cm, 18 cm, and 25 cm, the semi-perimeter is:
a) 25 cm
b) 26 cm
c) 28 cm
d) 29 cm
Answer:
c) 29 cm — \( s = \frac{15 + 18 + 25}{2} = 29 \, \text{cm} \).
Question 33. What is the area of a triangle with sides 14 cm, 48 cm, and 50 cm?
a) 224 cm²
b) 336 cm²
c) 448 cm²
d) 560 cm²
Answer:
b) 336 cm² — \( s = \frac{14 + 48 + 50}{2} = 56 \), area = \( \sqrt{56(56-14)(56-48)(56-50)} = 336 \, \text{cm}^2 \).
Question 34. The triangle with sides 21 m, 20 m, and 29 m is:
a) Right-angled
b) Equilateral
c) Scalene
d) Isosceles
Answer:
a) Right-angled — The sides satisfy \( 21^2 + 20^2 = 29^2 \), confirming it is a right-angled triangle.
Question 35. A triangle has sides 8 cm, 15 cm, and 17 cm. What is its area?
a) 50 cm²
b) 55 cm²
c) 60 cm²
d) 65 cm²
Answer:
c) 60 cm² — \( s = \frac{8 + 15 + 17}{2} = 20 \), area = \( \sqrt{20(20-8)(20-15)(20-17)} = 60 \, \text{cm}^2 \).
