Last updated at May 29, 2018 by Teachoo

Transcript

Example 5 Students of a school staged a rally for cleanliness campaign. They walked through the lanes in two groups. One group walked through the lanes AB, BC and CA; while the other through AC, CD and DA. Then they cleaned the area enclosed within their lanes. If AB = 9 m, BC = 40 m, CD = 15 m, DA = 28 m and ∠ B = 90°, which group cleaned more area and by how much? Find the total area cleaned by the students (neglecting the width of the lanes). Area first group = Area of ∆ABC Area second group = Area of ∆ADC Total Area = Area ∆ABC + Area ∆ADC Area ∆ABC Since AB = 9 m and BC = 40 m, ∠ B = 90°, Then ΔABC is a right angled triangle Area of ∆ABC = 1/2 × base × height = 1/2 × 40 × 9 m2 = 20 × 9 m2 = 180 m2 Area first group = Area of ∆ABC = 180 m2 Area ΔADC 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 = 28 m , b = 15 m, c = AC Since ∠ B = 90° , Applying Pythagoras theorem in Δ ABC AC2 = AB2 + BC2 AC = √(AB2+BC2) AC = √(92+402) m = √(81+1600) m = √1681 m = √412 m = 41m Thus, AC = 41 ∴ c = 41 m s = (𝑎 + 𝑏 + 𝑐)/2 Area of Δ ADC = √(42(42 −28)(42 −15) (42 −41) ) m2 Putting a = 28 , b = 15, c = 41 & s = 42 = √(42×14×27×1) m2 = √((14×3)×(14)×(9×3) ) m2 = √((14×14)×(9)×(3×3)) m2 = √((14×14)×(9)×(9)) m2 = √((14×14)×(9)×(9)) m2 = √((142)×(92) ) m2 = √((14)2) × √((9)2) = 14× 9 = 126 m2 Thus, Area second group = Area of ∆ADC = 126 m2 Area first group = Area of ∆ABC = 180 m2 So first group cleaned (180 – 126) m2, i.e., 54 m2 more than the area cleaned by the second group. Total area cleaned by all the students = Area ∆ABC + Area ∆ADC = (180 + 126) m2 = 306 m2.

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

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.