Comprehensive Chapter Summary
1. Introduction to Heron's Formula
We know the area of a triangle as \(\frac{1}{2} \times \text{base} \times \text{height}\) when height is given. However, for a scalene triangle with known side lengths but unknown height, such as a park with sides 40m, 32m, and 24m, calculating area is challenging without height. Heron’s formula addresses this by using side lengths \(a\), \(b\), and \(c\), and semi-perimeter \(s = \frac{a + b + c}{2}\), expressed as \(\text{Area} = \sqrt{s(s - a)(s - b)(s - c)}\). This method, developed by Heron (10AD-75CE) in Alexandria, Egypt, is vital for mensuration problems involving triangles, trapezoids, and other shapes.
Triangular Park Example
For a park with sides 40m, 32m, 24m, \(s = 48m\), and area \(\sqrt{48(48-40)(48-32)(48-24)} = 384m^2\), matching the right triangle area \(\frac{1}{2} \times 32 \times 24\).
2. Derivation and Application
Heron’s formula derives from the semi-perimeter concept. For sides 40m, 24m, 32m, \(s = 48m\), and area calculation involves \(\sqrt{48 \times 8 \times 24 \times 16} = 384m^2\). It applies to various triangles: equilateral (e.g., side 10cm, \(s = 15cm\), area \(\sqrt{15 \times 5 \times 5 \times 5} = 25\sqrt{3}cm^2\)) and isosceles (e.g., 8cm, 5cm, 5cm, \(s = 9cm\), area \(\sqrt{9 \times 1 \times 4 \times 4} = 12cm^2\)). This versatility aids in practical scenarios like fencing or planting.
Additionally, the formula’s strength lies in its independence from height. For a triangle with perimeter 32cm, sides 8cm, 11cm, third side 13cm, \(s = 16cm\), area \(\sqrt{16 \times 8 \times 5 \times 3} = 8\sqrt{30}cm^2\). It also handles complex cases, like a 120m, 80m, 50m park, \(s = 125m\), area \(\sqrt{125 \times 5 \times 45 \times 75} = 375\sqrt{15}m^2\), with fencing costs calculable.
Equilateral Triangle
Sides equal, \(s = \frac{3a}{2}\), area involves \(\sqrt{s(s - a)^3}\).
Isosceles Triangle
Two equal sides, \(s\) calculated similarly, area adapts.
Park Fencing Calculation
For 120m, 80m, 50m park, perimeter 250m, fencing 247m (3m gate), cost at ₹20/m = ₹4940.
3. Historical Context and Examples
Heron, an encyclopedic writer, documented mensuration in three books, with Book I covering areas of triangles via this formula. Examples include a 8cm, 11cm, 13cm triangle (area \(8\sqrt{30}cm^2\)) and a 120m, 80m, 50m park (area \(375\sqrt{15}m^2\)). His work aids modern applications like advertising on triangular walls (122m, 22m, 120m, area \(\sqrt{132 \times 10 \times 110 \times 12} = 660m^2\), rent \(₹5000 \times 660 \times \frac{3}{12} = ₹825,000\)).
Further, his formula extends to problems like a slide wall (15m, 11m, 6m, \(s = 16m\), area \(\sqrt{16 \times 1 \times 5 \times 10} = 10\sqrt{6}m^2\)) or a signal board (perimeter 180cm, side 60cm, area \(\sqrt{90 \times 30 \times 30 \times 30} = 900\sqrt{3}cm^2\)). This historical tool remains relevant for area computation without height.
Advertising Revenue
122m, 22m, 120m wall yields \(₹825,000\) for 3 months at \(₹5000/m^2\).
Slide Wall Painting
15m, 11m, 6m wall area \(10\sqrt{6}m^2\) for painting message.
Questions and Answers from Chapter
Short Questions (1 Mark)
Q1. What is Heron's formula?
Answer: \(\sqrt{s(s - a)(s - b)(s - c)}\).
Q2. Define semi-perimeter.
Answer: \(\frac{a + b + c}{2}\).
Q3. What is a scalene triangle?
Answer: All sides unequal.
Q4. Name a triangle with area 384m².
Answer: 40m, 32m, 24m.
Q5. Who developed Heron's formula?
Answer: Heron.
Q6. What is the semi-perimeter of 10cm, 10cm, 10cm?
Answer: 15cm.
Q7. What year was Heron born?
Answer: 10AD.
Q8. What is the area unit?
Answer: Square meters.
Q9. Is height needed for Heron's formula?
Answer: No.
Q10. What shape did Heron study?
Answer: Triangle.
Q11. What is \(s - a\)?
Answer: Difference.
Q12. What is a right triangle?
Answer: 90° angle.
Q13. Where was Heron from?
Answer: Egypt.
Q14. What is perimeter?
Answer: Sum of sides.
Q15. What is \(s\) for 60m, 100m, 140m?
Answer: 150m.
Q16. What is area formula with height?
Answer: \(\frac{1}{2} \times \text{base} \times \text{height}\).
Q17. What is an equilateral triangle?
Answer: All sides equal.
Q18. What is isosceles triangle?
Answer: Two sides equal.
Q19. What is \(s - c\)?
Answer: \(s\) minus \(c\).
Q20. What is the park's hypotenuse?
Answer: 40m.
Medium Questions (3 Marks)
Q1. Calculate the area of a triangle with sides 40m, 32m, 24m.
Answer: \(s = 48m\), area = \(\sqrt{48 \times 8 \times 24 \times 16} = 384m^2\).
Q2. Find the area of an equilateral triangle with side 10cm.
Answer: \(s = 15cm\), area = \(\sqrt{15 \times 5 \times 5 \times 5} = 25\sqrt{3}cm^2\).
Q3. Calculate the area of a triangle with sides 8cm, 11cm, 13cm.
Answer: \(s = 16cm\), area = \(\sqrt{16 \times 8 \times 5 \times 3} = 8\sqrt{30}cm^2\).
Q4. Find the area of an isosceles triangle with sides 5cm, 5cm, 8cm.
Answer: \(s = 9cm\), area = \(\sqrt{9 \times 1 \times 4 \times 4} = 12cm^2\).
Q5. What is the area of a triangle with sides 120m, 80m, 50m?
Answer: \(s = 125m\), area = \(\sqrt{125 \times 5 \times 45 \times 75} = 375\sqrt{15}m^2\).
Q6. Calculate the semi-perimeter of a triangle with sides 15m, 11m, 6m.
Answer: \(s = \frac{15 + 11 + 6}{2} = 16m\).
Q7. Find the area of a triangle with perimeter 42cm, sides 18cm, 10cm.
Answer: Third side = 14cm, \(s = 21cm\), area = \(\sqrt{21 \times 3 \times 11 \times 7} = 21\sqrt{11}cm^2\).
Q8. What is the area of a triangle with sides in ratio 3:5:7, perimeter 300m?
Answer: Sides 60m, 100m, 140m, \(s = 150m\), area = \(\sqrt{150 \times 90 \times 50 \times 10} = 1500\sqrt{3}m^2\).
Q9. Calculate the area of a signal board with perimeter 180cm.
Answer: Side = 60cm, \(s = 90cm\), area = \(\sqrt{90 \times 30 \times 30 \times 30} = 900\sqrt{3}cm^2\).
Q10. Find the area of a triangle with sides 122m, 22m, 120m.
Answer: \(s = 132m\), area = \(\sqrt{132 \times 10 \times 110 \times 12} = 660m^2\).
Q11. What is the semi-perimeter of 60m, 100m, 140m?
Answer: \(s = 150m\).
Q12. Calculate area for sides 15m, 11m, 6m.
Answer: \(s = 16m\), area = \(\sqrt{16 \times 1 \times 5 \times 10} = 10\sqrt{6}m^2\).
Q13. Find third side if perimeter is 32cm, sides 8cm, 11cm.
Answer: 13cm.
Q14. What is the area of a 12:17:25 ratio triangle, perimeter 540cm?
Answer: Sides 144cm, 204cm, 192cm, \(s = 270cm\), area = \(\sqrt{270 \times 126 \times 66 \times 78} = 1584\sqrt{35}cm^2\).
Q15. Calculate area for perimeter 30cm, equal sides 12cm.
Answer: Third side = 6cm, \(s = 15cm\), area = \(\sqrt{15 \times 3 \times 3 \times 9} = 9\sqrt{5}cm^2\).
Q16. What is \(s - b\) for 40m, 24m, 32m?
Answer: \(48 - 24 = 24m\).
Q17. Find area if \(s = 125m\), \(s - a = 5m\).
Answer: \(\sqrt{125 \times 5 \times 45 \times 75} = 375\sqrt{15}m^2\).
Q18. What is the hypotenuse of 32m, 24m?
Answer: 40m.
Q19. Calculate \(s\) for 3x, 5x, 7x, perimeter 300m.
Answer: \(x = 20\), \(s = 150m\).
Q20. What is area for 18cm, 10cm, 14cm?
Answer: \(s = 21cm\), area = \(\sqrt{21 \times 3 \times 11 \times 7} = 21\sqrt{11}cm^2\).
Long Questions (6 Marks)
Q1. Derive the area of a triangle with sides 40m, 32m, 24m using Heron's formula.
Answer: \(s = \frac{40 + 32 + 24}{2} = 48m\), \(s - a = 48 - 40 = 8m\), \(s - b = 48 - 32 = 16m\), \(s - c = 48 - 24 = 24m\), area = \(\sqrt{48 \times 8 \times 16 \times 24} = \sqrt{147456} = 384m^2\). Matches right triangle \(\frac{1}{2} \times 32 \times 24 = 384m^2\).
Q2. Calculate the area and fencing cost for a 120m, 80m, 50m park with a 3m gate at ₹20/m.
Answer: \(s = 125m\), area = \(\sqrt{125 \times 5 \times 45 \times 75} = 375\sqrt{15}m^2\), perimeter = 250m, fencing = 247m, cost = \(247 \times 20 = ₹4940\).
Q3. Find the area of a triangle with sides in ratio 3:5:7, perimeter 300m.
Answer: Sides = 60m, 100m, 140m, \(s = 150m\), area = \(\sqrt{150 \times 90 \times 50 \times 10} = \sqrt{6750000} = 1500\sqrt{3}m^2\).
Q4. Determine the area and rent for a 122m, 22m, 120m wall at ₹5000/m² for 3 months.
Answer: \(s = 132m\), area = \(\sqrt{132 \times 10 \times 110 \times 12} = 660m^2\), rent = \(660 \times 5000 \times \frac{3}{12} = ₹825,000\).
Q5. Calculate the area of a triangle with perimeter 42cm, sides 18cm, 10cm.
Answer: Third side = 14cm, \(s = 21cm\), area = \(\sqrt{21 \times 3 \times 11 \times 7} = 21\sqrt{11}cm^2\).
Q6. Find the area of a signal board with perimeter 180cm, side length 60cm.
Answer: \(s = 90cm\), area = \(\sqrt{90 \times 30 \times 30 \times 30} = 900\sqrt{3}cm^2\).
Q7. Compute the area of a triangle with sides 15m, 11m, 6m.
Answer: \(s = 16m\), area = \(\sqrt{16 \times 1 \times 5 \times 10} = 10\sqrt{6}m^2\).
Q8. Calculate the area for a 12:17:25 ratio triangle, perimeter 540cm.
Answer: Sides 144cm, 204cm, 192cm, \(s = 270cm\), area = \(\sqrt{270 \times 126 \times 66 \times 78} = 1584\sqrt{35}cm^2\).
Q9. Find the area and third side for perimeter 30cm, equal sides 12cm.
Answer: Third side = 6cm, \(s = 15cm\), area = \(\sqrt{15 \times 3 \times 3 \times 9} = 9\sqrt{5}cm^2\).
Q10. Determine the area of a triangle with sides 8cm, 11cm, 13cm.
Answer: \(s = 16cm\), area = \(\sqrt{16 \times 8 \times 5 \times 3} = 8\sqrt{30}cm^2\).
Q11. Calculate the area for a park with sides 50m, 80m, 120m.
Answer: \(s = 125m\), area = \(\sqrt{125 \times 5 \times 45 \times 75} = 375\sqrt{15}m^2\).
Q12. Find the area of an equilateral triangle with side 10cm.
Answer: \(s = 15cm\), area = \(\sqrt{15 \times 5 \times 5 \times 5} = 25\sqrt{3}cm^2\).
Q13. What is the area for sides 18cm, 10cm, 14cm?
Answer: \(s = 21cm\), area = \(\sqrt{21 \times 3 \times 11 \times 7} = 21\sqrt{11}cm^2\).
Q14. Calculate the area for a 3:5:7 ratio, perimeter 300m.
Answer: Sides 60m, 100m, 140m, \(s = 150m\), area = \(1500\sqrt{3}m^2\).
Q15. Find the area for 122m, 22m, 120m.
Answer: \(s = 132m\), area = \(\sqrt{132 \times 10 \times 110 \times 12} = 660m^2\).
Q16. What is the area for 15m, 11m, 6m?
Answer: \(s = 16m\), area = \(10\sqrt{6}m^2\).
Q17. Calculate the area for 60m, 100m, 140m.
Answer: \(s = 150m\), area = \(\sqrt{150 \times 90 \times 50 \times 10} = 1500\sqrt{3}m^2\).
Q18. Find the area for 8cm, 5cm, 5cm.
Answer: \(s = 9cm\), area = \(\sqrt{9 \times 1 \times 4 \times 4} = 12cm^2\).
Q19. What is the area for 12cm, 12cm, 6cm?
Answer: \(s = 15cm\), area = \(\sqrt{15 \times 3 \times 3 \times 9} = 9\sqrt{5}cm^2\).
Q20. Calculate the area for 144cm, 204cm, 192cm.
Answer: \(s = 270cm\), area = \(\sqrt{270 \times 126 \times 66 \times 78} = 1584\sqrt{35}cm^2\).
