Heron's Formula - NCERT Solutions

Exercise 10.1
Examples
Summary

Exercise 10.1

1. A traffic signal board, indicating 'SCHOOL AHEAD', is an equilateral triangle with side 'a'. Find the area of the signal board, using Heron's formula. If its perimeter is 180 cm, what will be the area of the signal board?

For an equilateral triangle with side a:

Semi-perimeter, s = (a + a + a)/2 = 3a/2

Using Heron's formula:

Area = √[s(s-a)(s-a)(s-a)] = √[s(s-a)³]

Area = √[(3a/2)(3a/2 - a)³] = √[(3a/2)(a/2)³] = √[(3a/2)(a³/8)] = √[3a⁴/16] = (a²√3)/4

When perimeter = 180 cm, side a = 180/3 = 60 cm

Area = (60²√3)/4 = (3600√3)/4 = 900√3 cm²

2. The triangular side walls of a flyover have been used for advertisements. The sides of the walls are 122 m, 22 m and 120 m (see Fig. 10.6). The advertisements yield an earning of ₹5000 per m² per year. A company hired one of its walls for 3 months. How much rent did it pay?

Sides: a = 122 m, b = 22 m, c = 120 m

Semi-perimeter, s = (122 + 22 + 120)/2 = 264/2 = 132 m

Area = √[s(s-a)(s-b)(s-c)]

Area = √[132(132-122)(132-22)(132-120)]

Area = √[132 × 10 × 110 × 12]

Area = √[1742400] = 1320 m²

Annual earning = 1320 × 5000 = ₹66,00,000

For 3 months = (66,00,000 × 3)/12 = ₹16,50,000

3. There is a slide in a park. One of its side walls has been painted in some colour with a message "KEEP THE PARK GREEN AND CLEAN" (see Fig. 10.7). If the sides of the wall are 15 m, 11 m and 6 m, find the area painted in colour.

Sides: a = 15 m, b = 11 m, c = 6 m

Semi-perimeter, s = (15 + 11 + 6)/2 = 32/2 = 16 m

Area = √[s(s-a)(s-b)(s-c)]

Area = √[16(16-15)(16-11)(16-6)]

Area = √[16 × 1 × 5 × 10]

Area = √[800] = 20√2 m² ≈ 28.28 m²

4. Find the area of a triangle two sides of which are 18cm and 10cm and the perimeter is 42cm.

Given: a = 18 cm, b = 10 cm, perimeter = 42 cm

Third side, c = 42 - (18 + 10) = 42 - 28 = 14 cm

Semi-perimeter, s = 42/2 = 21 cm

Area = √[s(s-a)(s-b)(s-c)]

Area = √[21(21-18)(21-10)(21-14)]

Area = √[21 × 3 × 11 × 7]

Area = √[4851] = 21√11 cm² ≈ 69.65 cm²

5. Sides of a triangle are in the ratio of 12:17:25 and its perimeter is 540cm. Find its area.

Let sides be 12x, 17x, and 25x

Perimeter = 12x + 17x + 25x = 54x = 540 cm

x = 540/54 = 10

Sides: 12×10 = 120 cm, 17×10 = 170 cm, 25×10 = 250 cm

Semi-perimeter, s = 540/2 = 270 cm

Area = √[s(s-a)(s-b)(s-c)]

Area = √[270(270-120)(270-170)(270-250)]

Area = √[270 × 150 × 100 × 20]

Area = √[81,000,000] = 9000 cm²

6. An isosceles triangle has perimeter 30 cm and each of the equal sides is 12 cm. Find the area of the triangle.

Equal sides: a = 12 cm, b = 12 cm

Perimeter = 30 cm

Third side, c = 30 - (12 + 12) = 6 cm

Semi-perimeter, s = 30/2 = 15 cm

Area = √[s(s-a)(s-b)(s-c)]

Area = √[15(15-12)(15-12)(15-6)]

Area = √[15 × 3 × 3 × 9]

Area = √[1215] = 9√15 cm² ≈ 34.86 cm²

Important Examples

Example 1: Find the area of a triangle, two sides of which are 8 cm and 11 cm and the perimeter is 32 cm.

Given: a = 8 cm, b = 11 cm, perimeter = 32 cm

Third side, c = 32 - (8 + 11) = 13 cm

Semi-perimeter, s = 32/2 = 16 cm

Area = √[s(s-a)(s-b)(s-c)]

Area = √[16(16-8)(16-11)(16-13)]

Area = √[16 × 8 × 5 × 3]

Area = √[1920] = 8√30 cm² ≈ 43.82 cm²

Example 2: A triangular park ABC has sides 120m, 80m and 50m. A gardener Dhania has to put a fence all around it and also plant grass inside. How much area does she need to plant? Find the cost of fencing it with barbed wire at the rate of ₹20 per metre leaving a space 3m wide for a gate on one side.

Sides: a = 120 m, b = 80 m, c = 50 m

Semi-perimeter, s = (120 + 80 + 50)/2 = 250/2 = 125 m

Area = √[s(s-a)(s-b)(s-c)]

Area = √[125(125-120)(125-80)(125-50)]

Area = √[125 × 5 × 45 × 75]

Area = √[2109375] = 375√15 m² ≈ 1451.58 m²

Perimeter = 120 + 80 + 50 = 250 m

Fencing length = 250 - 3 = 247 m

Cost of fencing = 247 × 20 = ₹4940

Example 3: The sides of a triangular plot are in the ratio of 3 : 5 : 7 and its perimeter is 300 m. Find its area.

Let sides be 3x, 5x, and 7x

Perimeter = 3x + 5x + 7x = 15x = 300 m

x = 300/15 = 20

Sides: 3×20 = 60 m, 5×20 = 100 m, 7×20 = 140 m

Semi-perimeter, s = 300/2 = 150 m

Area = √[s(s-a)(s-b)(s-c)]

Area = √[150(150-60)(150-100)(150-140)]

Area = √[150 × 90 × 50 × 10]

Area = √[6,750,000] = 1500√3 m² ≈ 2598.08 m²

Example from text: A triangular park with sides 40 m, 32 m and 24 m

Sides: a = 40 m, b = 24 m, c = 32 m

Semi-perimeter, s = (40 + 24 + 32)/2 = 96/2 = 48 m

Area = √[s(s-a)(s-b)(s-c)]

Area = √[48(48-40)(48-24)(48-32)]

Area = √[48 × 8 × 24 × 16]

Area = √[147456] = 384 m²

Verification: Since 24² + 32² = 576 + 1024 = 1600 = 40², it's a right triangle

Area using right triangle formula = ½ × 24 × 32 = 384 m² ✓

Chapter Summary

Key Points:

  • Heron's formula is used to find the area of a triangle when all three sides are known.
  • The formula is: Area = √[s(s-a)(s-b)(s-c)]
  • Where s = semi-perimeter = (a + b + c)/2
  • This formula is particularly useful when the height of the triangle is not known.
  • Heron (also known as Hero) was a Greek mathematician who derived this formula.

Important Formulas:

  • Semi-perimeter, s = (a + b + c)/2
  • Area = √[s(s-a)(s-b)(s-c)]
  • For equilateral triangle with side a: Area = (a²√3)/4
  • For right triangle: Area = ½ × base × height

Applications:

  • Finding area of triangular parks, fields, or plots
  • Calculating area when height is not available
  • Solving real-world problems involving triangular shapes
  • Verifying area calculations using different methods