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Example 4 - Kamla has a triangular field with sides 240 m - Examples

  1. Chapter 12 Class 9 Herons Formula
  2. Serial order wise
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Example 4 Kamla has a triangular field with sides 240 m, 200 m, 360 m, where she grew wheat. In another triangular field with sides 240 m, 320 m, 400 m adjacent to the previous field, she wanted to grow potatoes and onions. She divided the field in two parts by joining the mid-point of the longest side to the opposite vertex and grew potatoes in one part and onions in the other part. How much area (in hectares) has been used for wheat, potatoes and onions? (1 hectare = 10000 m2) E is the mid-point of AD So, AE = ED = 200 m Area of wheat field = Area ∆ABC Area of potato field = Area ∆AEC Area of onion field = Area ∆EDC Area ∆ABC Area of triangle = √(s(s−a)(s−b)(s −c)) Here, s is the semi-perimeter, and a, b, c are the sides of the triangle Let a = 200 m, b = 240 m, c = 360 m s = (𝑎 + 𝑏 + 𝑐)/2 Area ∆ABC = √(400(400 −200)(400 −240)(400 −360)) m2 = √(400×200×160×40) m2 = √((4×2×4×16)×106) m2 = √((4×4)×2×(16)×106) m2 = √((16)×2×(16)×106) m2 = √((16×16)×2 ×106) m2 = √(162×2 ×106) m2 =√162×√2×√106 = 16000√2 m2 = (16000√2)/10000 hectares = 1.6 x √2 hectares = 1.6 x 1.414 hectares = 2.26 hectares (approx.) Area for growing wheat = 2.26 hectares For finding Area ∆AEC & Area Δ CED we need EC But there is no possible way to find EC Taking out ∆ ACD separately Area of triangle = 1/2 × Base × Height Triangles ∆AEC & Δ CED have the same base ( as AE = ED = 200m) And they have the same height CF. So, Area ∆AEC = Area ∆CED Area ∆AEC = Area ∆CED = 1/2 Area ∆ACD Area ∆ACD Area of triangle = √(s(s−a)(s−b)(s −c)) Here, s is the semi-perimeter, and a, b, c are the sides of the triangle Let a = 240 m, b = 320 m, c = 400 m s = (𝑎 + 𝑏 + 𝑐)/2 Area of ∆ ACD = √(480(480 −240)(480 −320)(480 −400)) m2 = √(480×240×160×80) m2 = √((48)×(24)×(16)×(8)×104) m2 = √((4×12)×(2×12)×(16)×(8)×104) m2 = √((12×12)×(4)×(16)×(2×8)×104) m2 = √((12×12)×(4)×(16)×(16)×104) m2 = √((122)×(22)×(162)×104) m2 = √122×√22×√162×√104 = 12 × 2 × 16 × 102 m2 = 38400 m2 = 38400/10000 hectares = 3.84 hectares Thus, Area Δ ACD = 3.84 hectares So, Area ∆AEC = Area ∆CED = 1/2 Area ∆ACD Area ∆AEC = Area ∆CED = 1/2 × 3.84 hectares = 1.92 hectares Hence, Area for growing potatoes = 1.92 hectares Area for growing onions= 1.92 hectares

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Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He provides courses for Mathematics from Class 9 to 12. You can ask questions here.
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