Class 9 Maths Chapter 10 Heron’s Formula MCQs

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 = $\text{Math input error}$.

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 = $\sqrt{20(20-8)(20-15)(20-17)}=60{\text{cm}}^2$.