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Chapter 10 Class 9 Herons Formula
Serial order wise

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### Transcript

Question 2 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.

#### Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.