Heron S Formula — Karnataka (SSLC) Class 9 Mathematics Solutions (Free)
Free step-by-step Karnataka (SSLC) Class 9 Mathematics solutions for "Heron S Formula" — important questions with detailed answers, download PDF for board exam preparation.
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Heron S Formula — Textbook Questions Answered (8)
Q1: State Heron's formula for the area of a triangle: For a triangle with sides a, b, c, the semi-perimeter s = (a+b+c)/2, and area = √[s(s-a)(s-b)(s-c)]. Find the area of a triangle with sides 5 cm, 6 cm, and 7 cm.
Step 1: Heron's Formula: For a triangle with sides a, b, and c:
Semi-perimeter s = (a + b + c)/2
Area = √[s(s - a)(s - b)(s - c)]
This formula is used when the height is not given but all three sides are known.
Step 2: Given sides:
a = 5 cm
b = 6 cm
c = 7 cm
Step 3: First, verify that these sides can form a triangle using triangle inequality:
5 + 6 = 11 > 7 ✓
6 + 7 = 13 > 5 ✓
5 + 7 = 12 > 6 ✓
Step 4: Calculate semi-perimeter:
s = (a + b + c)/2
s = (5 + 6 + 7)/2
s = 18/2
s = 9 cm
Step 5: Calculate (s - a), (s - b), (s - c):
s - a = 9 - 5 = 4
s - b = 9 - 6 = 3
s - c = 9 - 7 = 2
Step 6: Apply Heron's formula:
Area = √[s(s - a)(s - b)(s - c)]
Area = √[9 × 4 × 3 × 2]
Area = √[9 × 24]
Area = √216
Step 7: Simplify √216:
216 = 36 × 6
√216 = √(36 × 6) = 6√6 cm²
Step 8: Numerical value:
√6 ≈ 2.449
Area ≈ 6 × 2.449 ≈ 14.70 cm²
Answer: Area of the triangle = 6√6 cm² ≈ 14.70 cm².
Q2: Using Heron's formula, find the area of a triangle with sides 10 cm, 10 cm, and 12 cm (isosceles triangle).
Step 1: Given sides of isosceles triangle:
a = 10 cm
b = 10 cm
c = 12 cm
Step 2: Verify triangle inequality:
10 + 10 = 20 > 12 ✓
10 + 12 = 22 > 10 ✓
10 + 12 = 22 > 10 ✓
Step 3: Calculate semi-perimeter:
s = (a + b + c)/2
s = (10 + 10 + 12)/2
s = 32/2
s = 16 cm
Step 4: Calculate the differences:
s - a = 16 - 10 = 6
s - b = 16 - 10 = 6
s - c = 16 - 12 = 4
Step 5: Apply Heron's formula:
Area = √[s(s - a)(s - b)(s - c)]
Area = √[16 × 6 × 6 × 4]
Area = √[16 × 36 × 4]
Area = √[16 × 144]
Area = √2304
Area = 48 cm²
Step 6: Verification using base-height method:
For isosceles triangle with equal sides 10 cm and base 12 cm:
Draw altitude from apex to base, which bisects the base (6 cm each).
In right triangle: 10² = 6² + h²
100 = 36 + h²
h² = 64
h = 8 cm
Area = (1/2) × base × height = (1/2) × 12 × 8 = 48 cm² ✓
Answer: Area of the isosceles triangle = 48 cm².
Q3: Find the area of a triangle with sides 13 cm, 14 cm, and 15 cm using Heron's formula. Express answer in simplest form.
Step 1: Given sides:
a = 13 cm
b = 14 cm
c = 15 cm
Step 2: Verify triangle inequality:
13 + 14 = 27 > 15 ✓
14 + 15 = 29 > 13 ✓
13 + 15 = 28 > 14 ✓
Step 3: Calculate semi-perimeter:
s = (a + b + c)/2
s = (13 + 14 + 15)/2
s = 42/2
s = 21 cm
Step 4: Calculate the differences:
s - a = 21 - 13 = 8
s - b = 21 - 14 = 7
s - c = 21 - 15 = 6
Step 5: Apply Heron's formula:
Area = √[s(s - a)(s - b)(s - c)]
Area = √[21 × 8 × 7 × 6]
Step 6: Calculate the product:
21 × 8 = 168
7 × 6 = 42
168 × 42 = 7056
Area = √7056
Step 7: Factorize 7056:
7056 = 16 × 441 = 16 × 441
441 = 21²
7056 = 16 × 21²
√7056 = √(16 × 441) = 4 × 21 = 84 cm²
Step 8: Verification by another method:
7056 = 2⁴ × 3² × 7²
√7056 = 2² × 3 × 7 = 4 × 3 × 7 = 84 cm²
Answer: Area of the triangle = 84 cm².
Q4: A triangle has sides in the ratio 3:4:5. If its perimeter is 36 cm, find its area using Heron's formula.
Step 1: Given: Sides in ratio 3:4:5, Perimeter = 36 cm
Step 2: Let the sides be 3x, 4x, and 5x (where x is a constant).
Step 3: Use perimeter to find x:
Perimeter = 3x + 4x + 5x = 36
12x = 36
x = 3
Step 4: Find the actual sides:
Side a = 3x = 3 × 3 = 9 cm
Side b = 4x = 4 × 3 = 12 cm
Side c = 5x = 5 × 3 = 15 cm
Step 5: Verify: 9 + 12 + 15 = 36 ✓
Step 6: Note that 9, 12, 15 are multiples of 3, 4, 5, which is a Pythagorean triplet.
Verification: 9² + 12² = 81 + 144 = 225 = 15²
This is a right triangle with legs 9 cm and 12 cm, and hypotenuse 15 cm.
Step 7: For a right triangle, area can be calculated as:
Area = (1/2) × leg₁ × leg₂ = (1/2) × 9 × 12 = 54 cm²
Step 8: Verify using Heron's formula:
s = (9 + 12 + 15)/2 = 36/2 = 18 cm
Area = √[18 × (18-9) × (18-12) × (18-15)]
Area = √[18 × 9 × 6 × 3]
Area = √[18 × 9 × 18]
Area = √[18² × 9]
Area = 18√9
Area = 18 × 3
Area = 54 cm² ✓
Answer: Area of the triangle = 54 cm².
Q5: Find the area of a triangle with sides 7 cm, 8 cm, and 9 cm. Express the answer both in simplified radical form and decimal form.
Step 1: Given sides:
a = 7 cm
b = 8 cm
c = 9 cm
Step 2: Verify triangle inequality:
7 + 8 = 15 > 9 ✓
8 + 9 = 17 > 7 ✓
7 + 9 = 16 > 8 ✓
Step 3: Calculate semi-perimeter:
s = (a + b + c)/2
s = (7 + 8 + 9)/2
s = 24/2
s = 12 cm
Step 4: Calculate the differences:
s - a = 12 - 7 = 5
s - b = 12 - 8 = 4
s - c = 12 - 9 = 3
Step 5: Apply Heron's formula:
Area = √[s(s - a)(s - b)(s - c)]
Area = √[12 × 5 × 4 × 3]
Area = √[12 × 60]
Area = √720
Step 6: Simplify √720:
720 = 144 × 5 = 12² × 5
√720 = √(144 × 5) = 12√5 cm²
Step 7: Calculate decimal value:
√5 ≈ 2.236
Area ≈ 12 × 2.236 ≈ 26.83 cm²
Step 8: Verification by factoring:
720 = 16 × 45 = 16 × 9 × 5 = 144 × 5
√720 = √(144 × 5) = 12√5 ✓
Answer: Area = 12√5 cm² ≈ 26.83 cm².
Q6: An equilateral triangle has side length a cm. Derive a formula for its area using Heron's formula and compare with the standard formula (√3/4)a².
Step 1: For an equilateral triangle with side a:
All sides are equal: sides = a, a, a
Step 2: Calculate semi-perimeter:
s = (a + a + a)/2 = 3a/2
Step 3: Calculate the differences:
s - a = 3a/2 - a = a/2
s - a = 3a/2 - a = a/2
s - a = 3a/2 - a = a/2
All three differences equal a/2 (property of equilateral triangles).
Step 4: Apply Heron's formula:
Area = √[s(s - a)(s - a)(s - a)]
Area = √[(3a/2) × (a/2) × (a/2) × (a/2)]
Area = √[(3a/2) × (a/2)³]
Area = √[(3a/2) × (a³/8)]
Area = √[3a⁴/16]
Area = √(3a⁴)/√16
Area = a²√3/4 cm²
Step 5: Compare with standard formula:
Standard formula: Area = (√3/4)a²
Heron's formula result: Area = (√3/4)a²
Both are identical ✓
Step 6: Example with a = 6 cm:
Area = (√3/4) × 6²
Area = (√3/4) × 36
Area = 9√3 cm²
Area ≈ 9 × 1.732 ≈ 15.59 cm²
Answer: Area of equilateral triangle with side a = (√3/4)a² = a²√3/4 cm². Formula derived from Heron's formula matches the standard formula.
Q7: A triangular field has sides 50 m, 60 m, and 70 m. Find its area using Heron's formula. If the cost of fencing is Rs 5 per meter, find the total cost of fencing the field.
Step 1: Given sides of triangular field:
a = 50 m
b = 60 m
c = 70 m
Step 2: Verify triangle inequality:
50 + 60 = 110 > 70 ✓
60 + 70 = 130 > 50 ✓
50 + 70 = 120 > 60 ✓
Step 3: Calculate semi-perimeter:
s = (a + b + c)/2
s = (50 + 60 + 70)/2
s = 180/2
s = 90 m
Step 4: Calculate the differences:
s - a = 90 - 50 = 40
s - b = 90 - 60 = 30
s - c = 90 - 70 = 20
Step 5: Apply Heron's formula:
Area = √[s(s - a)(s - b)(s - c)]
Area = √[90 × 40 × 30 × 20]
Area = √[90 × 40 × 30 × 20]
Step 6: Calculate the product:
90 × 40 = 3600
30 × 20 = 600
3600 × 600 = 2,160,000
Area = √2,160,000
Step 7: Simplify:
2,160,000 = 21,600 × 100 = 216 × 10,000
√2,160,000 = √(216 × 10,000) = 100√216 = 100 × 6√6 = 600√6 m²
Step 8: Numerical value:
√6 ≈ 2.449
Area ≈ 600 × 2.449 ≈ 1469.4 m²
Step 9: Calculate perimeter (for fencing):
Perimeter = a + b + c = 50 + 60 + 70 = 180 m
Step 10: Cost of fencing:
Cost per meter = Rs 5
Total cost = Perimeter × Cost per meter
Total cost = 180 × 5 = Rs 900
Answer: Area of the triangular field = 600√6 m² ≈ 1469.4 m²; Total cost of fencing = Rs 900.
Q8: Verify Heron's formula for a right triangle with sides 3 cm, 4 cm, and 5 cm. Compare the area from Heron's formula with the area calculated using the standard right triangle formula (1/2 × base × height).
Step 1: Right triangle with sides:
a = 3 cm (one leg)
b = 4 cm (other leg)
c = 5 cm (hypotenuse)
Step 2: Verify it's a right triangle:
3² + 4² = 9 + 16 = 25 = 5² ✓
This is a Pythagorean triplet.
Step 3: Using standard right triangle formula:
Area = (1/2) × base × height
Area = (1/2) × 3 × 4
Area = (1/2) × 12
Area = 6 cm²
Step 4: Using Heron's formula:
Calculate semi-perimeter:
s = (3 + 4 + 5)/2
s = 12/2
s = 6 cm
Step 5: Calculate the differences:
s - a = 6 - 3 = 3
s - b = 6 - 4 = 2
s - c = 6 - 5 = 1
Step 6: Apply Heron's formula:
Area = √[s(s - a)(s - b)(s - c)]
Area = √[6 × 3 × 2 × 1]
Area = √36
Area = 6 cm²
Step 7: Comparison:
Standard formula result: 6 cm²
Heron's formula result: 6 cm²
Both methods give identical result ✓
Step 8: This demonstrates that Heron's formula is a general formula that works for all triangles, including right triangles, and produces the same result as the specific formula for right triangles.
Answer: Area using Heron's formula = 6 cm²; Area using right triangle formula = 6 cm². Both methods confirm the area is 6 cm².
Related
- NCERT Solutions: Heron S Formula — the CBSE version of this same topic.
- Constructions — more Karnataka (SSLC) Class 9 Mathematics.
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- Probability — more Karnataka (SSLC) Class 9 Mathematics.
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