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Heron's Formula — Class 9 CBSE Mathematics MCQs with Solutions
Free Class 9 CBSE Mathematics Heron's Formula MCQs with step-by-step solutions. Practise online on Prepizo — no login needed.
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Sample questions with solutions
Q1 — easy
Heron's Formula for area of a triangle with side lengths a, b, c is:
A. ½ × base × height
B. √[s(s−a)(s−b)(s−c)] ✓ Correct
C. s(s−a)(s−b)(s−c)
D. √(a+b+c)
Solution: Area = √[s(s−a)(s−b)(s−c)].
Q2 — easy
In Heron's formula, semi-perimeter s is given by:
A. a + b + c
B. (a + b + c) / 2 ✓ Correct
C. (a + b + c) / 3
D. abc / 2
Solution: Semi-perimeter s = (a + b + c) / 2.
Q3 — easy
Area of an equilateral triangle with side length 'a' using Heron's formula is:
A. (√3 / 4) a² ✓ Correct
B. (√3 / 2) a²
C. ½ a²
D. √3 a²
Solution: Area of equilateral triangle = (√3/4) a².
Q4 — easy
Sides of a triangle are 3 cm, 4 cm, and 5 cm. Its area is:
A. 6 sq cm ✓ Correct
B. 12 sq cm
C. 15 sq cm
D. 60 sq cm
Solution: s = (3+4+5)/2 = 6. Area = √[6(3)(2)(1)] = √36 = 6 sq cm.
Q5 — easy
Perimeter of an isosceles triangle is 32 cm. Equal sides are 12 cm each. Base is:
A. 8 cm ✓ Correct
B. 10 cm
C. 12 cm
D. 6 cm
Solution: Base = 32 − (12 + 12) = 8 cm.
Q6 — easy
In above isosceles triangle (12 cm, 12 cm, 8 cm), semi-perimeter s =
A. 32 cm
B. 16 cm ✓ Correct
C. 8 cm
D. 24 cm
Solution: s = 32 / 2 = 16 cm.
Q7 — easy
Find area of equilateral triangle with side 8 cm: (take √3 = 1.732)
A. 16√3 sq cm (27.71 sq cm) ✓ Correct
B. 32√3 sq cm
C. 64 sq cm
D. 12√3 sq cm
Solution: Area = (√3/4)(8²) = 16√3 sq cm.
Q8 — easy
Sides of a triangular park are in ratio 3 : 5 : 7 and perimeter is 300 m. Side lengths are:
A. 30 m, 50 m, 70 m
B. 60 m, 100 m, 140 m ✓ Correct
C. 15 m, 25 m, 35 m
D. 90 m, 150 m, 210 m
Solution: 3x + 5x + 7x = 300 → 15x = 300 → x = 20. Sides: 60 m, 100 m, 140 m.
Q9 — easy
If side of an equilateral triangle is doubled (2a), its area becomes:
A. 2 times
B. 4 times ✓ Correct
C. 3 times
D. 8 times
Solution: Area ∝ a², so (2a)² = 4a² (4 times).
Q10 — easy
Perimeter of a triangle is 50 cm. One side is 18 cm and difference of other two sides is 4 cm. Other two sides are:
A. 14 cm and 18 cm ✓ Correct
B. 12 cm and 16 cm
C. 10 cm and 14 cm
D. 15 cm and 19 cm
Solution: Sum of two sides = 50 − 18 = 32 cm. x − y = 4 → x = 18, y = 14 cm.
Q11 — easy
If sides of a triangle are 13 cm, 14 cm, and 15 cm, semi-perimeter s =
A. 42 cm
B. 21 cm ✓ Correct
C. 28 cm
D. 14 cm
Solution: s = (13 + 14 + 15)/2 = 42/2 = 21 cm.
Q12 — easy
For triangle with sides 13, 14, 15 cm, area is:
A. 84 sq cm ✓ Correct
B. 42 sq cm
C. 168 sq cm
D. 90 sq cm
Solution: Area = √[21(8)(7)(6)] = √7056 = 84 sq cm.
Q13 — medium
Area of a right-angled triangle with base 6 cm and hypotenuse 10 cm is:
A. 30 sq cm
B. 24 sq cm ✓ Correct
C. 60 sq cm
D. 48 sq cm
Solution: Altitude = √(10² − 6²) = 8 cm. Area = ½ × 6 × 8 = 24 sq cm.
Q14 — medium
If semi-perimeter s = 12 cm and (s − a) = 5 cm, (s − b) = 3 cm, (s − c) = 4 cm, area =
A. √720
B. √720 = 12√5 sq cm ✓ Correct
C. 60 sq cm
D. 30 sq cm
Solution: Area = √[12 × 5 × 3 × 4] = √[144 × 5] = 12√5 sq cm.
Q15 — medium
Cost of turfing a triangular field of area 250 sq m at rate of ₹12 per sq m is:
A. ₹2500
B. ₹3000 ✓ Correct
C. ₹3600
D. ₹1200
Solution: Cost = 250 × 12 = ₹3000.