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Heron's formula allows area calculation of a triangle using side lengths \(a\), \(b\), and \(c\), with semi-perimeter \(s\). It’s useful when height is unknown, as in a 40m, 32m, 24m park yielding 384m². Heron, from 10AD Egypt, advanced mensuration, making this formula a key geometric tool.

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.

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.

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.

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