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# Example 2

Last updated at Sept. 22, 2017 by Teachoo

Last updated at Sept. 22, 2017 by Teachoo

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Transcript

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 Rs 20 per metre leaving a space 3m wide for a gate on one side First we find area needed to plant. Area to be planted = Area of ∆ABC Area of ∆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 Here, a = 50 m , b = 80 m, c = 120 m s = (𝑎 + 𝑏 + 𝑐)/2 Area of the triangle = √(𝑠(𝑠 −𝑎)(𝑠 −𝑏)(𝑠 −𝑐)) Putting a = 50 m , b = 80 m, c = 120 m & s = 125 m = √(125×(125−50)×(125−80)×(125−120)) = √(125×(75)×(45)×(5)) m2 = √((25×5)×(25×3)×(5×9)×(5)) m2 = √((25×25)×(5×5)×(9)×(3×5)) m2 = √((252)×(52)×(32)×(3×5)) m2 = √252 × √52 × √32 × √15 = 25 × 5 × 3 × √15 = 375 √15 m2 Area needed to plant = 375 √15 m2 Finding cost of fencing Number of meters to be fenced = 250 – 3 = 247 m Cost of fencing = Rs 20 per metre. Cost of fencing park = Rs 20 × 247 = Rs 4940

Class 9

Important Questions for Exam - Class 9

- Chapter 1 Class 9 Number Systems
- Chapter 2 Class 9 Polynomials
- Chapter 3 Class 9 Coordinate Geometry
- Chapter 4 Class 9 Linear Equations in Two Variables
- Chapter 5 Class 9 Introduction to Euclid's Geometry
- Chapter 6 Class 9 Lines and Angles
- Chapter 7 Class 9 Triangles
- Chapter 8 Class 9 Quadrilaterals
- Chapter 9 Class 9 Areas of parallelograms and Triangles
- Chapter 10 Class 9 Circles
- Chapter 11 Class 9 Constructions
- Chapter 12 Class 9 Herons Formula
- Chapter 13 Class 9 Surface Areas and Volumes
- Chapter 14 Class 9 Statistics
- Chapter 15 Class 9 Probability

About the Author

CA Maninder Singh

CA Maninder Singh is a Chartered Accountant for the past 8 years. He provides courses for Practical Accounts, Taxation and Efiling at teachoo.com .